solutions manual to basics of analytical and chemical equilibria - brian m. tissue

Brian M. Tissue, Basics of Analytical Chemistry and Chemical Equilibria, (J. Wiley, New York, 2013).

Chapter 1: End-of-Chapter Solutions 1. In order from increasing to decreasing precision:

pipet (2 mm i.d.) - most precise buret (1 cm i.d.) graduated cylinder (2.5 cm i.d.) - least precise

The precision of the volume measurement increases as the diameter of the glassware gets smaller. 2.

a) %-RSD = 0.02 mL = 0.2 % ⇒ 0.00500 ± 0.00001 M 10.0 mL

b) %-RSD = 0.1 mL = 1 % ⇒ 0.00500 ± 0.00005 M 10.0 mL

c) %-RSD = 1 mL = 10 % ⇒ 0.0050 ± 0.0005 M 10.0 mL 3. The calculations were very precise, they have to be since space travel is difficult. Unfortunately they were inaccurate due to a gross error in calculations that failed to convert values in metric units (newtons) with values in Imperial units (pound-force). 4. The one-point calibration with pH=7 buffer is rapid and will be accurate for pH measurements near pH 7. The disadvantage of the one-point calibration is that measurements at low pH or high pH could be erroneous due to an incorrect slope in the calibration function of the pH meter. The two-point calibration is more time consuming, but will provide more accurate data over the full range of the pH meter. Figure 1.5 shows an example of the error introduced by extrapolating a one-point calibration.


5. Plots of Table 1.11 data forcing trendline through zero. diamonds: 266 nm squares: 440 nm Note that the blank for the 266 nm data is obscured by the square marker in both charts.

Plots of Table 1.11 data as given. diamonds: 266 nm squares: 440 nm

Sensitivities: 266 nm 440 nm zero intercept 0.034 μM−1 0.012 μM−1 non-zero intercept 0.032 μM−1 0.012 μM−1 There is a 6-% difference in the slopes at 266 nm depending on whether or not the trendline is forced through zero. Looking at the data, the 1.2 μM point appears low compared to the trendline. The scatter in the 440 nm data points appears random. The take-home message is that calibration curves should be constructed from at least 5 or 6 measurements.

y = 0.0344x

y = 0.0117x

0

0.01

0.02

0.03

0.04

0 0.2 0.4 0.6 0.8 1 1.2

Abso

rban

ce

Concentration (μM)

y = 0.0321x + 0.002

y = 0.0118x - 4E-05

0

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Concentration (μM)


6. The individual volume measurements give a mean and standard deviation of: 1.022 ± 0.007 mL. The density is: 2.3291 g = 2.278 g/mL 1.022 mL The RSD of the volume measurements is: 0.007 mL = 0.0071 1.022 mL The final result should have the same RSD (2.278 g/mL × 0.0071 = 0.016 g/mL) to give: density = 2.278 ± 0.016 g/mL (2.28 ± 0.02 g/mL is also correct) 7. a) For either primary standard we need (0.1000 L)(0.150 M) = 0.0150 mol of Na.

• For NaCl (f.w. = 58.443 g/mol): (58.443 g/mol)(0.0150 mol Na) = 0.8766 g NaCl

• For Na2C2O4 (f.w. = 133.998 g/mol):

133.998 g/mol 1 mol Na2C2O4 0.0150 mol Na = 1.0050 g Na2C2O4 2 mol Na (note an additional significant figure for the higher formula weight standard)

b) We need (0.05000 L)(0.0100 mol/L)(58.443 g/mol) = 0.0292 g Na. Since we can weigh to 0.0001 g, the uncertainty is

0.0001 g × 100 % = 0.3 % 0.0292 g c) We need (1.000 L)(0.500 M)(58.443 g/mol) = 29.2215 g of Na. Weighing will not be a limiting factor, so the volume measurement, 0.1 %, is the largest source of uncertainty. d) A weight measurement equivalent to 0.1 % give that the balance weighs to 0.0001 g is:

0.1 % = 0.0001 g × 100 % x g x = 0.1000 g For NaCl (f.w. = 58.443 g/mol):

0.0001 g = 1.711 mmol 58.443 g/mol The volume needed to make 0.1 mM is:

0.100 mM = 1.711 mmol x L

x = 17.11 L This result is a large volume, which shows why common lab practice is to make a concentrated stock solutions that can be diluted to the desired concentrations.


8. The actual volume delivered by the “10-mL” pipet is (10.021 g)/(0.99707 g/mL) = 10.050 mL. The error is 0.05 mL or

0.05 mL × 100 % = 0.5 %. 10.050 mL 9. a) There should be few if any interferences in drinking water, so blanks to check for contamination and a spike to check for detection limit should be sufficient. b) “Field” blanks and field spikes are necessary due to the large number of sample preparation steps that will be necessary for such a complicated sample matrix. c) In addition to the usual blanks, a “field” spike will be very important to determine if there is any loss of analyte. 10. a) There should be few if any interferences in drinking water, so a calibration curve using standards of lead in 0.1 M nitric acid should be sufficient. b) The complexity of the sample matrix warrants using the standard addition method for calibration. Any matrix effects should affect the standard addition equally to the matrix effects of the unknown amount of Pb in the test solutions. c) Given the possible loss of analyte during sample processing, using an internal standard will be the preferred method of calibration. An element similar to Pb that is not present in the sample should be chosen for the internal standard. Using an internal standard requires some knowledge of the sample composition. 11. (a) Your spreadsheet will look something like:

Notes:

Data: 78.9378.7779.0978.52

mean = 78.8275std dev = 0.243088%-RSD = 0.308379 =(C9/C8)*100std err = 0.12154495% C.I. 0.386753 =3.182*C9/SQRT(4)


• To understand the formulas copied to the right of the results, cell C10 contains the standard deviation result and cell C9 has the mean.

• The mean can gain an additional significant figure due to the size of the sum of the data: 78.828. However, the standard deviation of 0.243 shows that extended significant figures in the result have little meaning.

• The result can be reported as 78.83 ± 0.24. or 78.8 ± 0.2. (b) In the following example, the formulas are written to accommodate up to 10 data points. It is not necessary to revise formulas if data is added to the set:

12. (a) The left figure is Excel’s default scaling on the y axis. Note how it exaggerates the differences in the values compared to the right plot that is scaled from 0 to 100.

Mean and standard deviation: 98.37 ± 0.96 (an additional significant figure is gained due to the size of the sum of the data, but it can also be reported as 98.4 ± 1.0).

Results FormulasData: 78.93 N = 5 =COUNT(G$5:G$14)

78.77 sum = 394.11 =SUM(G$5:G$14)79.09 mean = 78.82 =I6/I578.52 std dev = 0.21 =STDEV(G$5:G$14)78.8 %-RSD = 0.27 =(I8/I7)*100

std err = 0.09 =I8/SQRT(I5)95% C.I. 0.26 =2.776*I8/SQRT(I5)

96.5

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b)

Mean and standard deviation: 94.95 ± 0.75 (an additional significant figure is gained due to the size of the sum of the data, but it can also be reported as 95.0 ± 0.8) c) Q (0.63 and 0.11) is less than Qc for both cases, so the values must be retained. In the first case, N is small and the criteria for rejection is high. In the second case, the outlier is not very different from other data points. d) As an example, if you must divide by a sum by N, it is more robust to use a spreadsheet’s COUNT function to get N rather than entering it numerically since it might change on adding or removing data points.

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13. a) Your spreadsheet might look something like the following screen capture:

b) Note that the automatic scaling in spreadsheets will show approximately 0.4 to 0.5 on the y-axis, exaggerating the difference in the values compared to plotting the y-scale from 0 to 0.5:

(c)

Data Set 1

Run # Voltage deviation deviation^21 0.453 -0.0034 1.176E-052 0.444 -0.0124 1.545E-043 0.457 0.0006 3.265E-074 0.448 -0.0084 7.104E-055 0.451 -0.0054 2.947E-056 0.495 0.0386 1.488E-037 0.447 -0.0094 8.890E-05

N (#): 7 sum d^2: 1.844E-03sum/N: 0.4564 std. dev. 0.0175average: 0.4564 stdev: 0.0175

0.00

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Volta

ge

Run #

Data Set 1: 0.456 ± 0.018 V

Q= (outlier-closest) / (outlier-farthest)Q= (0.495-0.457) / (0.495-0.444)Q= 0.745098

Qc=0.57Q > Qc so data point 6 may be discarded


Q is 0.745, larger than Qc of 0.57 for seven data points. The value may be rejected, and the new average is:

0.450 ± 0.005 V is also correct. 14. (a)

Note that the randomness in the scatter of the residuals indicates a linear model is appropriate for this data.

mean 0.4500 Vstd. dev. 0.0046 V

y = 0.0488x + 0.0118R² = 0.9997

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Data Set 2

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Data Set 2Deviations


(b) Below are two different ways to obtain this data:

(c)

Note that the units are not easy to show in the spreadsheets, they are: slope = 0.0488 ± 0.0004 V ppb−1 intercept = 0.0118 ± 0.0197 V

LINEST resultsConc. (ppb) Signal (V) deviation (d) zero line slope intercept

0.00 0.002 -0.010 0 0.04881 0.011765.00 0.259 0.003 0 0.00042 0.01969

10.00 0.489 -0.011 0 0.99971 0.0356425.00 1.284 0.052 0 13568.93 450.00 2.407 -0.045 0 17.23112 0.00508

100.00 4.903 0.010 0

unknown signal: 0.999 unknown conc: 20.2 ppbestimated uncertainty: ± 0.3 ppb

Conc. (ppb) Signal (V) deviation (d) x^2 x-d^2 y-d^2 (x-d)*(y-d)0.00 0.002 -0.010 0.00 1002.78 2.4191 49.255.00 0.259 0.003 25.00 711.11 1.6857 34.62

10.00 0.489 -0.011 100.00 469.44 1.1413 23.1525.00 1.284 0.052 625.00 44.44 0.0747 1.8250.00 2.407 -0.045 2500.00 336.11 0.7219 15.58

100.00 4.903 0.010 10000.00 4669.44 11.1935 228.62

sum(x)= sum(y)= sum(x^2)= Sxx= Syy= Sxy= sy=190.00 9.34 13250.00 7233.33 17.2362 353.04 0.0356

N = 6avg(x)= avg(y)= std.dev. R.S.D.(%)31.67 1.56 m = 0.0488 0.0004 0.86

b = 0.0118 0.0197 167.44

unknown: 0.999 V 20.227 ± 0.559 ppb

± 2.76 R.S.D.(%)


Chapter 2: End-of-Chapter Solutions 1. a) methane: gas, nonpolar b) octane: liquid, nonpolar c) water: liquid, polar d) beeswax: solid, nonpolar e) acetic acid: liquid, polar f) sodium acetate: solid, ionic, soluble in water g) K2CrO4: solid, ionic, soluble in water h) methyl tert-butyl ether (MTBE): liquid, moderate polararity i) ethanol: liquid, moderate polararity j) perchloric acid: liquid, polar (ionic in water) k) NaOH: solid, ionic, soluble in water l) KCl: solid, ionic, soluble in water m) CaCO3: solid, ionic, insoluble in water 2. a) dissolve aspirin tablets in water; filter through coffee filter; dry and weigh insoluble starch b) filter gravel from mixture with window screen, weigh either gravel or sand c) weigh water sample; allow water to evaporate; weigh remaining salt d) extract caffeine with organic solvent; allow solvent to evaporate and weigh remaining caffeine (repeat with SPE cleanup after extraction; drying down and weigh each fraction to determine if any other extractable components are present) 3. a) acetic acid - predominantly in water (αS(aq) = 0.6) b) hexane - predominantly in octanol c) 1-hexanol - predominantly in octanol d) methanol - predominantly in octanol e) oleic acid - predominantly in octanol 4. For KD of 1200 and equal volumes of water and octanol, αS(aq) = 8.3×10−4. For 100 mL of each solution this becomes 0.083 mL octanol in 100 mL of water. Using the density of octanol (0.8 g/mL) the solubility is approximately 0.06 g/100 mL solution. (Note that this approach is an approximate prediction since the αS(aq) calculation is dependent on relative volumes.)


5. The general expression is:

┌ αS(aq) = │

└

1 ┐n │ ┘ 1 + KD' Vorg

Vaq a)

┌ 0.10 = │

└

1 ┐1 │ ┘ 1 + KD' 30 mL

100 mL KD' = 30 (b)

┌ αS(aq) = │

└

1 ┐4 │ ┘ 1 + 30 7.5 mL

100 mL αS(aq) = 0.0090 (or 0.9 %) (c)

┌ αS(aq) = │

└

1 ┐n │ ┘ 1 + 30 10 mL

100 mL The easiest way to solve this question is to calculate αS(aq) as a function of n:

n αS(aq) 1 0.25 2 0.062 3 0.016 4 0.0039 5 0.00098

Doing 5 extractions leaves slightly less than 0.1 % of the solute in the aqueous phase and meets the state criterion. 6. For KD' = 66 and two sequential extractions (n = 2):

┌ 0.001 = │

└

1 ┐2 │ ┘ 1 + 66 Vorg

100 mL Vorg = 46.4 mL The total volume of organic phase will be twice this amount or 92.8 mL.


7. Adding acid to lower the pH below the pKa of phenol will protonate the phenol so that it is a neutral molecule and will be soluble in organic solvents. To avoid extracting acetic acid, the pH should not be lowered to within two pH units of 4.76. Acidifying to a pH near 8 should work well. 8.

• A field blank to check for contamination during sample collection • A lab blank to check for cross-contamination or instrument carryover. • A field spike to check for analyte loss. • Addition of an internal standard before sample preparation for calibration.

9. A strong anion-exchange stationary phase consist of a quaternary ammonium species that cannot lose it’s positive charge. Anions attracted to the stationary phase can be eluted by lowering the pH so that the anions are protonated and become neutral. A weak anion-exchange stationary phase consist of an ammonium species that can lose it’s positive charge at high pH. Anions attracted to the stationary phase can be eluted by raising the pH so that the stationary phase is neutralized. The carbonate buffer system is a convient eluant for anion-exchange SPE. Adjusting the pH of the buffer can increase the mobile phase concentration of CO3

2−, which can displace monovalent anions from the stationary phase. 10. a) A buffer solution at an acidic pH will keep the amino acids present as cations. Lowering the pH with time will elute the more strongly retained components faster. b) normal-phase chromatography uses a nonpolar solvent, e.g., hexane or a mixture, e.g., hexane-isopropanol (99:1) c) reverse-phase chromatography uses a polar solvent, e.g., water:acetonitrile mixture (mixtures are common because they allow fine tuning of the mobile phase to optimize separations) d) a buffer solution at a basic pH to prevent the anions from becoming protonated. Raising the pH with time will elute the more strongly retained components faster. 11. (a) Normal-phase partition chromatography usually uses a non-polar mobile phase such as hexane. For these alcohol solutes, methanol will have low solubility in hexane and an alternate mobile phase is needed. A possible choice is acetonitrile. The solutes will elute in the order from least polar to most polar: butanol, propanol, ethanol, methanol.


(b) Short chain alcohols are soluble in water, so a suitable mobile phase will be a water:acetonitrile mixture. The elution orde will be from most polar to least polar: methanol, ethanol, propanol, butanol. 12. Find the answer to this question by searching manufacturer literature. 13. The solution concentration of each analyte can be predicted from:

(SRM concentration)×(SRM amount)/(volume) which gives μg/mL or ppm in the test solution. The results are:

Comparing these calculations to the method measurement ranges, we find:

Cd: measurable with no dilution Cu: dilute by 5 or more Fe: measurable with no dilution Pb: detectable with no dilution (below LOQ, detectable but cannot be quantitated) Zn: measurable with no dilution

14. A polar bear of course – smile, it’s a joke. 15. (a) The online literature gives a psilocin pKa near 8.5, so it is a cation at pH = 6. (b) The measurement will be erroneously low if the sample is heated too high. The method must be validated by running through the procedure with certified reference materials or spiked samples of known psilocin content. (c) The concentration of psilocin in the test solution is: psilocin = 500 μg

4130 82110 psilocin = 25.1 μg

SRM (μg/g) SRM amount (g) volume (mL) solution (ppm)Cd 2.48 0.3 25 0.02976Cu 71.6 0.3 25 0.8592Fe 205.8 0.3 25 2.4696Pb 0.038 0.3 25 0.000456Zn 1424 0.3 25 17.088


For a sample weight of 200 mg, the concentration in ppm is: 25.1 μg psilocin = 126 ppm psilocin 0.200 g sample

(d) The internal standard will correct for any inefficiency in the SPE cleanup step. The internal standard will not correct for systematic errors due to incomplete conversion of psilocybin to psilocin or loss of psilocin due to degradation. 16. (a) hexane (non-polar organic) – may be retained to some extent (b) sodium acetate (mostly an anion at pH = 6) – not retained (c) sodium benzoate (mostly an anion at pH = 6) – not retained (d) phosphoric acid (mostly an anion at pH = 6) – not retained (e) acetone (moderate polarity organic) – should be retained 17. The limit of detection (LOD) and the lower limit on a control chart are two completely different things. What they have in common is that they are each a measurement from a given instrument or procedure. The LOD is the concentration that gives a measurement that is just detectable, usually three times the noise level. The limits on a control chart are usually three times the standard deviation of measurements of a standard. The standard will usually be at a concentration that is similar to samples being measured and usually much higher than the LOD. 18. A large number of sampling sites should be selected at random to provide the highest confidence in the results. Given obvious evidence of a soda spill, judgmental sampling is appropriate. You sample around the soda bottle to determine the extent of the spill. 19. Since the ore has been mixed to some extent, a systematic sampling plan is reasonable. Two samples should be taken from each rail car. The sampling sites within each car should be far apart to catch heterogeneity. Depending how the cars are loaded, samples might be taken from the front and rear of a car of from the top and bottom of the ore load. The systematic approach is useful to try to determine the degree of heterogeneity in a sample batch.


Chapter 3: End-of-Chapter Solutions 1. The equilibrium constant expressions are:

K1 = [A][B] K2 = [A]2[B]2 [C] [C]2

K2 = [A]2[B]2 = [A][B] [A][B] = (K1)(K1) = K12 [C]2 [C] [C]

2. Some possibilities are:

• Vinegar is 8 % acetic acid, CH3COOH. The equilibrium on reacting with water is:

CH3COOH(aq) + H2O ⇌ CH3COO−(aq) + H3O+(aq)

Ka' = [CH3COO−][ H3O+] [CH3COOH]

• Clorox and many cleaners contain sodium hypochlorite; the sodium salt of hypochlorous

acid, HClO:

ClO−(aq) + H2O ⇌ HClO(aq) + OH−(aq)

Kb' = [HClO][OH−] [ClO−]

• Disodium EDTA, the disodium salt of ethylenediamminetetraacetic acid, is very common

in soaps and shampoos. It complexes the Ca2+ and Mg2+ found in hard water, which would otherwise form a precipitate (soap scum) with surfactants.

EDTA4−(aq) + Ca2+(aq) ⇌ Ca(edta)2−(aq)

Kf' = [Ca(edta)2−] [EDTA4−][Ca2+]

• Soluble lead acetate, Pb(CH3COO)2, is contained in hair darkening products. The Pb2+

ions react with sulfur in hair protein to form lead sulfide (PbS) precipitates that adhere to hair. The subsequent equilibrium is:

PbS(s) ⇌ Pb2+(aq) + S2−(aq)

Ksp' = [Pb2+][S2−]


3. (a) 0.05 M KMnO4 + 0.05 M FeCl2 MnO4

− can Fe2+ can undergo a redox reaction, K+ and Cl− are spectator ions:

MnO4−(aq) + 5Fe2+(aq) + 8H+(aq) ⇌ Mn2+(aq) + 5Fe3+(aq) + 4H2O

K' = [Mn2+][Fe3+]5 [MnO4

−][Fe2+]5[H+]8

(b) 0.05 M HCl + 0.05 M NaOH These species are a strong acid and a strong base, respectively. Cl− and Na+ are spectator ions, the H+ and OH− undergo a neutralization reaction, and the resulting equilibrium is:

2H2O ⇌ H3O+(aq) + OH−(aq) Kw' = [H3O+][OH−]

(c) 0.05 M HCl + 0.05 M NH3 Another neutralization reaction, but in this case involving a weak base. The reaction product is NH4

+, a weak acid, and the equilibrium is:

NH4+(aq) + H2O ⇌ NH3(aq) + H3O+(aq)

Ka' = [NH3][ H3O+] [NH4

+] d) 0.05 M HCl + 0.05 M AgNO3 a precipitation reaction:

AgCl(s) ⇌ Ag+(aq) + Cl−(aq) Ksp' = [Ag+][Cl−]

e) 0.05 M CaCl2 + 0.05 M (NH4)2C2O4 a precipitation reaction:

CaC2O4(s) ⇌ Ca2+(aq) + C2O42−(aq)

Ksp' = [Ca2+][C2O42−]

4. First convert the number of grams of product to number of moles: 2.947 g MgNH4PO4·6H2O = 0.01201 mol MgNH4PO4·6H2O 245.4 g/mol Convert to mol of P:

0.01201 mol MgNH4PO4·6H2O 1 mol P 0.01201 mol P 1 mol MgNH4PO4·6H2O


Convert mol of P to grams of P: (0.01201 mol P) (30.974 g/mol) = 0.3719 g P Find weight percent:

wt-% = 0.3719 g P × 100 % = 10.75 % 3.459 g test portion 5. Sample calculation for the first measurement: 0.1225 g Fe2O3 = 0.7671 mmol Fe2O3 159.69 g/mol

0.7671 mmol Fe2O3 2 mol Fe 55.845 g Fe = 0.08568 g Fe 1 mol Fe2O3 mol Fe

wt-% = 0.08568 g Fe × 100 % = 17.88 % 0.4793 g test portion Repeating for the other samples and finding the average gives: 17.77 ± 0.10 % or 17.8 ± 0.1 % 6. (1.9482 g)/(143.32 g/mol) = 0.013593 mol and (0.013593 mol)(58.443 g/mol NaCl) = 0.79441 g NaCl a)

wt-% = 0.79441 g NaCl × 100 % = 0.8045 % 98.75 g test portion b) parts per thousand = 0.8045 % × 10 = 8.045 ‰ c)

[NaCl] = 0.013593 mol NaCl 1000 mL = 0.1390 M (98.75 g)/(1.01 g/mL) 1.0 L 7. (413.6 s)(0.04825 C/s) = 19.96 C

19.96 C 0.0002068 mol e− 96,485 C/mol

(0.0002068 mol e−)(1 mol I2/2 mol e−) 1 mol vitamin C = 0.0001034 mol vitamin 1 mol I2


(0.0001034 mol vitamin C)(176.06 g=mol) = 0.01821 g vitamin C

wt-% = 0.01821 g vitamin C 100 % = 91.0 % 0.0200 g sample 8. (a) The analyte is Fe2+ and the titrant is MnO4

−:

5Fe2+(aq) + MnO4−(aq) + 8H+(aq) ⇌ 5Fe3+(aq) + Mn2+(aq) + 4H2O

(b)

(0.03491 L)(0.002045 M MnO4−) 5 mol Fe2+ = 3.5695×10−4 mol Fe 1 mol MnO4

− (c) (3.5695×10−4 mol Fe)(55.845 g/mol) = 0.01993 g Fe d) The precipitate contained (3.5695×10−4 mol FePO4)(150.816 g/mol) = 0.05383 g FePO4. The amount of AlPO4 is then (0.0663 g − 0.05383 g) = 1.022×104 mol 121.953 g/mol which equals 6.119×10−4 g Al. e) weight percent for each element is:

wt-% = 0.01993 g Fe ×100 % = 0.03984 % Fe 50.02 g sample

wt-% = 6.119×10−4 g Al ×100 % = 0.00122 % Al 50.02 g sample 9. (a) (0.1000 M)(0.03500 L) = 3.50 mmol H+ = 3.50 mmol OH−

[OH−] = 3.50 mmol OH− = 0.0700 M 0.05000 L (b) phenol red. Neutralization of a strong base with a strong acid produces a titration end point pH near 7.


(c) At 30.0 mL of titrant, the endpoint has not been reached. OH−, Ca2+, Cl− (H+ is present at < 10−7 M) (d) At 40.0 mL of titrant, the endpoint has been passed. H+, Ca2+, Cl− (OH− is present at < 10−7 M) (e) Add a known amount of strong acid and back titrate the excess acid with a standard base. 10. Back titration result:

(0.01104 L)(0.1177 M SCN−) 1 mol Ag+ = 1.2994×10−3 mol Ag+ 1 mol SCN− Total Ag+ added to test portion:

(0.03555 L)(0.1015 M Ag+) = 3.6083×10−3 mol Ag+

Amount of Ag+ reacted with Cl−: 3.6083×10−3 mol Ag+ − 1.2994×10−3 mol Ag+ = 2.3089×10−3 mol Ag+ Amount of Cl− in test portion:

2.3089×10−3 mol Ag+ 1 mol Cl− = 2.3089×10−3 mol Cl− 1 mol Ag+ Concentration of Cl− in test portion: 2.3089×10−3 mol Cl− = 0.04618 M Cl− 0.05000 L 11. Back titration result:

(0.01805 L)(0.0955 M Mg2+) 1 mol EDTA = 1.7238×10−3 mol EDTA 1 mol Mg2+ Total EDTA added to test portion:

(0.02500 L)(0.1347 M EDTA) = 3.3675×10−3 mol EDTA

Amount of EDTA reacted with Ca2+: 3.3675×10−3 mol EDTA − 1.7238×10−3 mol EDTA = 1.6437×10−3 mol EDTA Amount of Ca2+ in test portion:

1.6437×10−3 mol EDTA 1 mol Ca2+ = 1.6437×10−3 mol Ca2+ 1 mol EDTA


1.6437×10−3 mol Ca2+ 1 mol CaCO3 100.087 g CaCO3 = 0.16452 g CaCO3 1 mol Ca2+ mol CaCO3

wt-% = 0.16452 g CaCO3 × 100 % = 16.45 % CaCO3 1.000 g solid 12. The average of the three titrations is 0.05620 L.

Titration result:

(0.05620 L)(0.0200 M Pb2+) 1 mol EDTA = 1.124×10−3 mol EDTA 1 mol Pb2+

1.124×10−3 mol EDTA 1 mol Cu2+ = 1.124×10−3 mol Cu2+ 1 mol EDTA Since 25.0 mL of solution was determined from the 250.0 mL that the brass was dissolved in, there is a 10-fold factor to multiply:

(1.124×10−3 mol Cu2+)(10) 63.546 g Cu = 0.71426 g Cu mol

wt-% = 0.71426 g Cu × 100 % = 67.26 % 1.062 g brass Propagating the RSD from the three titrations, (67.26 %)(0.015) = 1.0 %, gives a final result of: 67.3 ± 1.0 % Cu 13. (a)

• The initial reaction is a neutralization to convert the analyte to a new form. • Conversion of SO2 to H2SO4 is a redox reaction. • Titration of the H2SO4 with OH− is another neutralization reaction.

(b)

1 mol SO32− 1 mol H2SO3 1 mol SO2 1 mol H2SO4

1 mol H2SO3 1 mol SO2 1 mol H2SO4 2 mol OH−

trial titrant vol (L)1 0.05552 0.05713 0.056

average 0.05620stdev 0.00082RSD 0.015


Overall: 1 mol SO3

2− 2 mol OH−

(c) Since the titration is a strong acid and a strong base, phenol red is the best choice. In practice, phenolphthalein, which goes from colorless to pink at pH 8 is also suitable. (d) Ba2+ and SO4

2− form a BaSO4 precipitate. Adding the soluble BaCl2 should show a precipitate to confirm the presence of SO4

2−. The acid-base titration is not specific for sulfuric acid, so the presence of sulfate should be confirmed. 14.

advantages disadvantages Coulometry is an absolute measure and does not depend on other standards for calibration.

Like other classical methods, it is not sensitive enough for trace analysis.

Since it is an electrochemical method, it can be automated for rapid and unattended analysis.

Not all analytes can be determined by electrochemical methods.

15.

titration gravimetry Requires standard solutions and volumetric glassware.

Requires only a balance for measurement.

Requires a reaction that goes to completion and a means of detecting the end point.

Requires a reaction that converts the analyte to a pure, weighable form.

Can be automated.

16. Even with the best glassware, balance, and technique, the purity of a primary standard will always be a limit in a chemical analysis. 17.

titration gravimetry The endpoint is often difficult to observe.

Some analyte is left in solution.

If the analyst overshoots the endpoint, the result will be erroneously high.

The result will be erroneously low due to the lost analyte.


Chapter 4. End-of-Chapter Solutions 1. (a) Rearranging c = λν 650 nm:

ν = 2.998×108 m/s = 4.61×1014 Hz or 461 GHz 650×10−9 m 405 nm:

ν = 2.998×108 m/s = 7.40×1014 Hz or 740 GHz 405×10−9 m (b) Using E = hν, where h is Planck’s constant, 6.626×10−34 J s. 650 nm: E = (6.626×10−34 J s)(4.61×1014 s−1) = 3.06×10−19 J

E = 1 1 m = 15,400 cm−1 650×10−9 m 100 cm 405 nm: E = (6.626×10−34 J s)(7.40×1014 s−1) = 4.90×10−19 J

E = 1 1 m = 24,700 cm−1 405×10−9 m 100 cm Both wavelengths are in the visible spectral region and are of the same order of magnitude in energy. You will want to develop a sense of scale of the different spectral regions. The wavenumbers or inverse centimeter unit, cm−1, is a common unit in spectroscopy because of the convenient scale in the infrared through visible regions. 2. (lowest energy) radio, microwave, infrared, visible, ultraviolet, X-ray, γ-ray (highest energy) 3. The H atom has one proton and one electron. The highest charge that it can have is +1. Maybe the writer meant "highly excited." 4.

λ = 1 1 m = 3.2×10−6 m = 3.2 μm 3100 cm−1 100 cm


The energy of a 280 nm photon is

E = 1 1 m = 35,700 cm−1 280×10−9 m 100 cm Dividing this energy by the IR photon energy gives: 35,700 cm−1 = 11.5 IR photons 3100 cm−1 This example shows that the scale of energies between the infrared and the visible differs by roughly ten-fold. 5. (a) a filled antibonding orbital – No, once an orbital is full it is full and no further electrons may be added. (b) an empty antibonding orbital – Yes, an empty orbital can accept an electron. 6. Schematic of emission (left) and Raman scattering (right). The electron symbols show the occupied levels after the transition occurs. In the emission example, one electron was placed in the excited state by some means, such as absorbing collisional, thermal, or light energy. In the scattering example, the photon is shown at an angle to indicate a change in direction after interacting with the electronic system. (Raman scattering is a weak process, and most scattered photons return to their original energy level, the ground state in this example.)

7. Bond length is inversely related to bond energy. The shortest bond in the series, C−F, will be the strongest and require the highest IR energy to be excited.


8. The equation for the best fit line to the calibration data is y = (−0.8432 mL/mg)x + 0.6148 where x is glucose concentration in mg/mL and y is the unitless absorbance measurement. For a measurement of 0.521 the expression becomes 0.521 = (−0.8432 mL/mg)[glucose] + 0.6148 and the result is [glucose] = 0.111 mg/mL. 9. n-hexane. C6H14 has no non-bonding or π electrons and therefore can only have the short wavelength σ → σ* transitions. 10. (a) A = abc, so in a 5-cm cuvette A will be 5(0.333) = 1.67 AU. (b) A = −log(T) so T = 10−1.67 = 0.0216, which is equivalent to 2.16 %-transmittance. (c) In a 2-cm cuvette A will be 2(0.333) = 0.666 AU and T = 10−0.666 = 0.216 or o 21.6 %-transmittance. Note that the linear relationship for A versus b or c is true only for A, not for transmittance. 11. Absorbance is proportional to path length, so first determine the absorbance at 1.0 cm. P = 398 photons and Po = 1000 so:

A = −log 348 = 0.458 1000 The absorbance at 2.0 cm will be twice the 1.0-cm absorbance or 0.916. Now reversing our calculation:

0.916 = −log photons 1000

leads to 121 photons at 2.0 cm, and therefore a transmittance of 0.121 or percent-transmittance of 12.1 %. 12. The stain on the cuvette will appear as an absorbance in addition to the analyte absorbance, so the measurements will be erroneously higher than the true value.


13. A = 0.1 = −log(T), T = 0.79 and A = 1.0 = −log(T), T = 0.10 These transmittance values provide a range where lamp noise in subtracting two large values (for the high T region) and stray light (for the low T region) are unlikely to affect the measurement. 14. Array spectrometers acquire the whole spectrum simultaneously and are therefore very rapid. Their disadvantage is that the spectral resolution is fixed for the pixel width of the detector. Scanning instruments are more flexible and can provide higher resolution and greater sensitivity by varying the slit width and changing the type of detector, respectively. 15. Using the data point for a distance of 1.0 cm:

A = −log 348 = 0.458 1000 Using the Beer-Lambert law: 0.458 = ε(1.00 cm)(4.0×10−5 M) ε = 11,500 M−1 cm−1 The result should be the same for any data point, but doing a number of calculations and taking the mean will average random scatter. 16. (a) Inserting the unknown measurement into the equation for the calibration curve. 0.271 = −0.8432[glucose] + 0.6148 [riboflavin] = 0.408 mg/mL (b) The background of 0.009 must be subtracted from the measurement, then calculating as before: (0.271 − 0.009) = −0.8432[glucose] + 0.6148 [riboflavin] = 0.418 mg/mL 17. A very weak absorbance will attenuate the light power passing through the test portion a very small amount. The light measurements of P and Po will be nearly identical. The noise fluctuations of the lamp source will limit the ratio of P and Po that can be distinguished.


18. Note that the trendline equation does not show the units of the slope and intercept. Since the fluorescence signal is given in arbitrary units, the slope is 50.466 ppm−1. Inserting the unknown measurement into the equation for the calibration curve: 14.5 = (50.466 ppm−1)[riboflavin] + 0.7505 [riboflavin] = 0.272 ppm 19. (a) Using the Beer-Lambert law: 0.549 = ε(1.00 cm)(4.00×10−5 M) ε = 13,720 M−1 cm−1 (I carry an extra significant digit in intermediate calculations that I will drop in the final result.) (b) Again using the Beer-Lambert law and the molar absorptivity from the calibration above. 0.455 = (13720 M−1 cm−1)(1.00 cm)[ Fe(phen)3

2+]

[Fe(phen)32+] = 0.455

(13720 M−1 cm−1)(1.00 cm) [Fe(phen)3

2+] = 3.32×10−5 M (c) The water sample was diluted by 10 to add the other reagents for the absorbance measurement, so [Fe2+] = (3.32×10−5 M)(10) = 3.32×10−4 M (d) This absorbance value is above the linear range for most spectrophotometers. The test portion should be diluted until the absorbance is less than or near 1.0. (e) This absorbance measurement will be lower than a test portion that was all Fe2+. The calculated concentration will be less than the true iron concentration in the water sample.


Chapter 5. End-of-Chapter Solutions 1. (a) bucket of deionized water: H2O, H3O+, OH− (b) small amount of HNO3 added: H2O, H3O+, NO3

−, OH− Note that NO3

− is a strong electrolyte so we assume that HNO3 dissociates completely. [OH−] will be very small relative to the concentrations of the other ions in solution, but OH− is present. (c) small amount of acetic acid, CH3COOH, added: H2O, CH3COOH, H3O+, CH3COO−, OH−

CH3COOH dissociates to a small extent, so we call it a weak acid. [CH3COOH] is significant and much larger than [H3O+] or [CH3COO−]. One way to define a weak acid is simply as an acid that does not dissociate completely in aqueous solution. Again, [OH−] will be very small relative to the other concentrations in solution, but OH− is present. 2. (a) Ba(NO3)2: neutral, barium nitrate is a strong electrolyte (b) Ca(ClO4)2: neutral, calcium perchlorate is a strong electrolyte (c) KI: neutral, potassium iodide is a strong electrolyte (d) NaF: basic, fluoride ion, F−, is a weak base (e) NH4Br: acidic, ammonium ion, NH4

+, is a weak acid (f) NH4F: amphiprotic, compare Ka of NH4

+ to Kb of F−, the larger dominates, in this case the solution will be acidic (g) ammonium acetate: amphiprotic, so compare compare Ka of NH4

+ to Kb of CH3COO−, in this case Ka ≈ Kb and the solution will be close to neutral 3. (a) below pH of 2.3 the charge is +1, pH between 2.3 and 9.9 the overall charge is 0 (zwitterionic form), for pH greater than 9.9 the charge is −1 (b) extracting into an organic solvent will not occur if the amino acid is charged, so adjust the pH to between 3-9. 4. (a) 0.01 M HCl (hydrochloric acid is a strong acid and produces a lower pH than an equal amount of a weak acid) (b) 0.01 M HClO4 (perchloric acid is a strong acid and produces a lower pH than an equal amount of a weak acid) (c) 1×10−4 M HClO4 (pH = 4, 0.01 M HClO has pH 4.8)


5. (a) 0.001 M KOH (a lower concentration of strong base will have lower pH, i.e., more acidic, than higher concentration of strong base) (b) 0.01 M CH3COONa (weak base, lower pH than strong base) (c) 0.01 M CH3COONa (pH ≈ 8.4, 1×10−4 M NaOH has pOH = 4 and pH = 10) 6. (a) 0.010 M KI Ic = 0.5{(+1)2(0.010 M) +(−1)2(0.010 M)} = 0.010 M (b) 0.250 M Ca(NO3)2 Ic = 0.5{(+2)2(0.250 M) +(−1)2(0.500 M)} = 0.750 M (c) 0.250 M AlCl3 Ic = 0.5{(+3)2(0.250 M) +(−1)2(0.750 M)} = 1.50 M (we neglect the reaction of Al3+ with water) (d) 0.250 M (NH4)2SO4 Ic = 0.5{(+1)2(0.500 M) +(−2)2(0.250 M)} = 0.750 M (e) 0.250 M CH3COONa Ic = 0.5{(+1)2(0.250 M) +(−1)2(0.250 M)} = 0.250 M For d) and e) there is some reaction of NH4

+ and CH3COO− with water, but if you write the equilibria you'll see that there is no change in the number of ions in solution. 7. (a) Ic = 0.010 M, use the Debye-Hűckel equation, inserting the di for each ion. Sample calculation for OH−:

log γOH = −0.509(−1)2(0.010)0.5 1 + (3.29(0.35)(0.010)0.5

i. K+: γK = 0.899 ii. I−: γI = 0.899 iii. H3O+: γH3O = 0.914 iv. OH−: γOH = 0.900 (b) Ic = 0.750 M, use the Debye-Hűckel equation, inserting the di for each ion: i. Ca2+: γCa = 0.223 ii. NO3

−: γNO3 = 0.578 iii. H3O+: γH3O = 0.752 iv. OH−: γOH = 0.602


8. (a) Using the activity coefficients from the previous question, set up the expression for Kw and substitute activity coefficients and concentrations for the activities. Kw = (aH3O+)(aOH−) = (γH3O+)[ H3O+](γOH−)[OH−] = (γH3O+)(γOH−) Kw'

Kw' = Kw (γH3O+)(γOH−)

Kw' = 1.01×10−14 = 1.23×10−14 (0.914)(0.900) b)

Kw' = 1.01×10−14 = 2.23×10−14 (0.752)(0.602) 9. 0.2 M CH3COOH. acid dissociation is approximately 1%, Ic = 0.002 M 0.2 M CH3COONa. Ic = 0.2 M (a small amount of acetate reacts with water to form CH3COOH and OH−, since the OH− has the same charge as CH3COO−, there is no effect on ionic strength) 0.2 M CH3COOH in 0.2 M NaCl. Ic slightly higher than 0.2 M due to acid dissociation in addition to the 0.2 M NaCl 0.2 M CH3COONa in 0.2 M NaCl. Ic = 0.4 M 1.0 M CH3COOH. acid dissociation is 0.5%, Ic 0.005M (a) Ranking the solutions from lowest to highest ionic strength. 0.2 M CH3COOH, 1.0 M CH3COOH, 0.2 M CH3COONa, 0.2 M CH3COOH in 0.2 M NaCl, 0.2 M CH3COONa in 0.2 M NaCl (b) Activity coefficients decrease with increasing Ic, so the solution with the highest Ic, 0.2 M CH3COONa in 0.2 M NaCl, will produce activity coefficients farthest from the ideal case of 1.0.


10. Percent-dissociation is equal to:

%-dissoc = [H3O+] × 100 % cHA Sample calculation:

Ka' = [H3O+]2 cHA − [H3O+]

Enter Ka' and cHA, then rearrange and solve with the quadratic equation. For Ka' = 1×10−2.

Ka' = 0.01 = [H3O+]2 = 0.01 M − [H3O+]

[H3O+] = 0.0062 M

%-dissoc = 0.0062 M × 100 % = 62 % 0.01 M (a rather strong “weak acid”) Other results: (a) 3.1 % (b) 27 % (c) 62 % 11. The ionic strength is approximately 0.75 M (slightly higher due to acid dissociation). Activity coefficients are 0.75 for H3O+ and 0.62 for A−. Correct Ka to obtain Ka', then do the calculation in the same way as above. (a) 4.6 % (b) 37 % (c) 75 % You can see there can be a significant difference even for monoprotic acids at high ionic strength. 12.

CH3COOH(aq) + H2O(aq) ⇌ CH3COO−(aq) + H3O+(aq)

Ka' = [CH3COO−][H3O+] [CH3COOH]

1.75×10−5 = [H3O+]2 cHA − [H3O+]


Enter 0.0100 M for the acetic acid formal concentration, rearrange, and solve with the quadratic equation. [H3O+] = 4.10×10−4 M p[H3O+] = 3.39. 13. Ic = 0.5{(+1)2(4.1×10−4 M) + (−1)2(4.1×10−4 M)} = 4.1×10−4 M Activity coefficients are 0.978 for H3O+ and 0.977 for CH3COO−, so Ka' = 1.83×10−5. Using this value in the calculation results in an insignificant change in the result. [H3O+] = 4.19×10−4 M p[H3O+] = 3.38. 14. Activity coefficients are 0.765 for H3O+ and 0.667 for CH3COO−, so Ka' = 3.43×10−5. Using this value in the calculation results in a change of 0.15 pH units. [H3O+] = 5.69×10−4 M p[H3O+] = 3.25. 15. The equilibrium is:

CH3COO−(aq) + H2O ⇌ CH3COOH(aq) + OH−(aq)

Kb' = Kw' = [CH3COOH][OH−] Ka' [CH3COO−]

Kb' = 1.01×10−14 = 5.77×10−10 1.75×10−5

5.77×10−10 = [OH−]2 0.0100 M − [OH−]

Solve for [OH−] using the quadratic equation, then convert to [H3O+]. [OH−] = 2.40×10−6 M p[OH−] = 5.62 p[H3O+] = 14.00 − 5.62 = 8.38.


16. Activity coefficients are 0.90 for both CH3COO− and OH−. Given the form of the Kb' expression

Kb' = Kw' = [CH3COOH][OH−] Ka' [CH3COO−]

The effect of the activity coefficients cancel and Kb' = Kb. The result is the same as in question 15. 17. Using the spreadsheet allows you to do multiple calculations quickly. Find the result by tabulating p[H3O+] for each weak acid for cHA = 0.1 M, 0.01 M, 0.001 M, etc. Here I set up the calculation using acetic acid and dichloroacetic acid at cHA = 0.01 M. If cHA is high and Ka' is not large, we expect the approximate solution to give the same result as using the quadratic equation. acetic acid approximate solution:

1.75×10−5 = [H3O+]2 0.001 M

[H3O+] = 4.18×10−4 M p[H3O+] = 3.38 quadratic equation:

1.75×10−5 = [H3O+]2 0.001 M − [H3O+]

[H3O+] = 4.10×10−4 M p[H3O+] = 3.39 dichloroacetic acid

5.5×10−2 = [H3O+]2 0.001 M

[H3O+] = 2.35×10−2 M p[H3O+] = 1.63

5.5×10−2 = [H3O+]2 0.001 M − [H3O+]

[H3O+] = 0.864×10−2 M p[H3O+] = 2.06 For acetic acid the approximate calculation is very close to the quadratic result. For the strong dichloroacetic acid, the approximation introduces a significant error.


(a) ≈ 1×10−4 M (b) ≈ 0.1 M, so the quick approximation mostly fails for weak acids with relatively large Ka values except at fairly high formal concentrations. 18. 0.001 M acetic acid has a pH of 3.91 using Ka = 1.75×10−5. Setting pH = 3.81 and working backwards requires a Ka' of 2.85×10−5. Using the usual means of correcting Ka:

Ka = γ[A−]γ[H3O+] = γ2Ka' [HA]

1.75×10−5 = γ2(2.85×10−5) results in γ2 = 0.625 and γ = 0.79. Inserting values for [Na+] and [Cl−] in ionic-strengthactivity-coefficients.xls leads to an ionic strength of 0.11 M. 19. This problem is a Ka' calculation worked backwards. p[H3O+] = 3.85 [H3O+] = 10−3.85 = 1.41×10−4 M

Ka = [H3O+]2 cHA − [H3O+]

Ka = (1.41×10−4)2 0.01 − 1.41×10−4

Ka = 2.02×10−6 Repeating for the measured pH: p[H3O+] = 3.79 [H3O+] = 10−3.79 = 1.62×10−4 M

Ka' = [H3O+]2 cHA − [H3O+]

Ka' = (1.62×10−4)2 0.01 − 1.62×10−4

Ka' = 2.67×10−6


Usually we look up Ka from a reference table, calculate Ic to find activity coefficients, and use the activity coefficients to determine Ka', which provides a more realistic prediction of a weak acid equilibrium. Here we know Ka and Ka' and we can calculate the ionic strength of the solution. [H3O+] = 10−3.85 = 1.62×10−4 M

Ka = γ[A−]γ[H3O+] = γ2Ka' [HA]

γ = (Ka / Ka' )0.5 γ = (2.02×10−6/2.67×10−6)0.5 γ = 0.870 Now use the Debye-Huckel expression to find Ic

log(0.870) = −0.509(1)2(Ic)0.5 1 + (3.29(0.4)(Ic)0.5

−0.0605 = −0.509(Ic)0.5 1 + 1.32(Ic)0.5

0.0605 = 0.509(Ic)0.5 1 + 1.32(Ic)0.5

0.0605 + 0.0796(Ic)0.5 = 0.509(Ic)0.5 0.0605 = 0.429(Ic)0.5 (Ic)0.5 = 0.141 (Ic) = 0.020 M


Chapter 6. End-of-Chapter Solutions 1. Note that the question asked for monoprotic acids only. (a) pH = 5.0: acetic acid, pKa = 4.76. (b) pH = 7.0: there is not a good monoprotic organic acid in Table 5.8 to buffer at this pH. Alternate choices are the inorganic hypochlorous acid, pKa = 7.54, or the diprotic carbonic acid, pKa1 = 6.35. (c) pH = 9.0: phenol, pKa = 9.99. Phenol is not used as a pH buffer in practice due to toxicity, ammonium ion, pKa = 9.25, is an alternate inorganic choice. 2. (a) pH = 5.0: citric acid, pKa2 = 4.76. (b) pH = 7.0: carbonic acid, pKa1 = 6.35. (c) pH = 9.0: carbonic acid, pKa2 = 10.33. (A bit far from 9.0, but the best choice in Table 5.8.) 3. Σ αi = 1.0. Since alpha is a fraction, the total of all alpha’s must be one. This condition is always true and it does not depend on ionic strength. Individual fractions might change slightly as Ic changes, but the total must still be 1.0. 4. H3PO4 + OH− → H2PO4

− 0.60 mol OH− H2PO4

− + OH− → HPO42− 0.60 mol OH−

HPO42− + OH− → PO4

3− 0.30 mol OH− ----------------- total = 1.5 mol of NaOH

5. PO4

3− + H+ → HPO42− 0.050 mol H+

HPO42− + H+ → H2PO4

− 0.050 mol H+ ----------------- total = 0.10 mol of HCl


6. Contents of initial solution: (0.15 M Na+, 0.05 M PO4

3−): Ic = 0.5{(+1)2(0.15 M) + (−3)2(0.05 M)} = 0.30 M Contents of solution after adding HCl: (0.15 M Na+, 0.10 M Cl−, 0.05 M H2PO4

−): Ic = 0.5{(+1)2(0.15 M) + (−1) 2 (0.10 M) + (−1)2(0.05 M)} = 0.15 M 7. (a) Adding 0.070 moles of NaOH to a solution containing 0.040 moles of a monoprotic weak acid: excess OH−, so basic but not a buffer solution. (b) Adding 0.040 moles of NaOH to a solution containing 0.070 moles of a monoprotic weak acid: significant amounts of a weak acid and conjugate base, so a buffer solution, pH depends on pKa of HA. (c) Adding 0.040 moles of NaOH to a solution containing 0.070 moles of a potassium hydrogen phthalate: Buffer solution at pH = 5.4. (d) Adding 0.040 moles of HCl to a solution containing 0.070 moles of a potassium hydrogen phthalate: Buffer solution at pH = 2.9. 8. Using p[H3O+] = 0.5(pKa1 + pKa2): a) p[H3O+] = 0.5(7.20 + 12.35) = 9.78 b) p[H3O+] = 0.5(6.35 + 10.33) = 8.34 c) 1.0×10−8 M is less than the intrinsic [H3O+] of pure water, so p[H3O+] is expected to be 7.0 or just slightly higher. 9. From the phosphate alpha plots (Figure 6.6) at pH = 3, we see that we need H3PO4 = 0.1 and H2PO4

− = 0.9. Since we are starting with 0.500 moles of phosphate, we will need 0.0500 moles of H3PO4 and 0.450 moles of H2PO4

− to achieve this ratio. Protonating all of the PO43− to obtain

HPO42− requires 0.500 mol of HCl. Protonating all of the HPO4

2− to obtain H2PO4− requires

another 0.500 mol of HCl. Finally we add 0.050 mol of HCl to convert 0.050 mol of H2PO4− to

0.050 mol of H3PO4. The total amount of strong acid added is 0.500 mol + 0.500 mol + 0.050 mol = 1.05 mol HCl.


10. (a) halfway point (for the first acidic proton): the solution contains equal amounts of C6H4(COOH)2 and C8H5O4

−, so p[H3O+] = pKa1 = 2.95. (b) the first equivalence point: the solution contains predominantly C8H5O4

−, so p[H3O+] = 0.5(pKa1 + pKa2) = 4.18 (c) the second equivalence point: the solution contains C8H4O4

2−, so solve as a weak base problem correcting for dilution during titration:

Kb = Kw = [C8H5O4−][OH−]

3.91×10−6 [C8H4O42−]

2.58×10−9 = [OH−]2 0.0333 M − [OH−]

[OH−] = 9.28×10−6 M p[OH−] = 5.03 p[H3O+] = 14.0 − 5.02 = 8.97 11. Determining the error in the previous answer requires predicting p[H3O+] after correcting pKa for ionic strength. The solution at the halfway point contains 0.05 M Na+ and 0.05 M C8H5O4

−, and the ionic strength is 0.05 M. Activity coefficients are 0.834 and 0.854 for C8H5O4

− and H3O+, respectively. Correcting Ka:

Ka' = [H3O+]2 cHA − [H3O+]

p[H3O+] = 2.95 + log 1.12×10−3 = 1.57×10−3 (0.834)(0.854) Now using this Ka' , [H3O+] = 1.57×10−3 M and p[H3O+] = 2.80. The difference in results is 0.15 pH units. 12. The predominant form of phosphate can be viewed from the alpha plots directly. Find the desired pH on the x-axis and then determine which curve has the largest alpha value at that pH. a) pH = 4: H2PO4

− b) pH = 6: H2PO4

− c) pH = 8: HPO4

2− d) pH = 10: HPO4

2−


13. Since the pH of a saturated solution is approximately halfway between pKa1 and pKa2, the form of NTA in the solution must be H2NTA−. 14. After the neutralization reaction, the solution contains [C6H4(COOH)2] = 0.010 M and [C8H5O4

−] = 0.015 M. Using the Henderson-Hasselbalch expression gives:

p[H3O+] = 2.95 + log 0.015 M = 3.13 M 0.010 M 15. Interpolating in the raw data values in alpha-plot-3protic.xls gives αHPO42− = 4.5×10−6. The value can also be found using Equation 6.17. 16. The concentration of any given species is the fraction of that species times the total concentration: [HPO4

2−] = (4.5×10−6)(0.0500 M) = 2.2×10−7 M HPO42−

17. The values can be obtained by Henderson-Hasselbalch calculations or from alpha plot data. Some pH values are as follows: HCO3

−:CO32− p[H3O+]

0.05 9.0 0.10 9.3 0.15 9.5 0.50 10.3 0.85 11.1 0.90 11.3 0.95 11.6

18. Using alpha-plot-3protic or a direct calculation shows that αCO3 is 1.5×10−5. The [CO3

2−] concentration is 1.5×10−6 M, which is much smaller than [HCO3

−].


19. Note that the recipe specifies adjusting the pH to 7.4, which is done by adding a small amount of strong acid or strong base to this buffer solution. Due to the high ionic strength and day-to-day temperature fluctuations, it is easier to measure and adjust the pH than to try to calculate the pH. Using formula weights and combining common ions, the solution contains: 1.616 mol +1 ions 1.413 mol −1 ions 0.101 mol −2 ions in a solution volume of 1 L. Ic = 0.5{(+1)2(1.616 M) + (−1)2(1.413 M) + (−2)2(0.101 M)} = 1.72 M


Chapter 7: End−of−Chapter Solutions 1. (a) KMnO4(s): Mn(VII) in a covalent oxyanion, MnO4

−. (b) MnCl6

3−(aq): Mn(III), Cl has a −1 oxidation state, and the overall complex charge is −3. (c) Mn(H2O)6

2+(aq): Mn(II), water is neutral, and the overall charge of this ion is +2. (d) Mn(H2O)6

7+(aq): Mn(VII), this complex as written does not exist, the charge of a Mn(VII) ion is so high that it will react with water to form an oxyanion. (e) MnO2(s): Mn(IV), O has the usual −2 oxidation state, and there is no charge to this neutral solid material. 2. Ligands have unbonded electrons (usually pairs) and metal ions have empty orbitals. Coordinate covalent bonds form between the electron pairs on the ligands and the empty orbitals of the metal ions. The bonding between a base and a proton is analogous. The base has unbonded pairs of electrons and the proton has the 1s orbital empty. 3. When using the indicator, the indicator remains uncomplexed until reaching the endpoint. At the endpoint where all EDTA is complexed, metal ion will bind to the indicator. If we assume that the indicator binds either the Ca2+ analyte or the Pb2+ titrant equally, then the indicator will still change color at the endpoint. When using a Pb2+-sensitive electrode, Pb2+ titrant that displaces Ca2+ from Ca(edta)2− will cause an overshoot of the endpoint. The measurement of the excess EDTA will be erroneously high. Calculation of the Ca2+ concentration will then be erroneously low. 4. The Cl− is a strong electrolyte and has no direct effect on p[H3O+]. The pKa and the conversion to Ka for each cation is listed in the table. For 1.0 mM ionic strength, I assume pKa' = pKa. The calculation is the same as for any weak acid as done in Chapter 5:

Ka' = [H3O+]2 0.001 M − [H3O+]

and the results are tabulated in the last column of the table. The calculation for Li+ and Ca2+ gives a result higher than 7.0, so these two ions have no effect on p[H3O+]. pKa Ka p[H3O+] Li+ 13.8 1.6×10−14 7.0 (no effect) Ca2+ 12.6 2.5×10−13 7.0 (no effect) Cu2+ 7.5 3.2×10−8 5.2


5. In this question we find the concentration of the metal ion that produces p[H3O+] = 6.5 or: [H3O+] = 10−6.5 = 3.2×10−7 M Setting up the Ka' expression and solving for c:

Ka' = [H3O+]2 [Ca2+] − [H3O+]

2.5×10−13 = (3.2×10−7)2 [Ca2+] − 3.2×10−7

[Ca2+] = 0.4 M

3.2×10−8 = (3.2×10−7)2 [Cu2+] − 3.2×10−7

[Cu2+] = 3.2×10−6 M 6. (a) Mn(edta)2−: at high pH the Mn2+ can form hydroxide complexes and at low pH the EDTA can be protonated, either extreme will reduce the Mn(edta)2− concentration. (b) Mn(H2O)6

2+(aq): at high pH the Mn2+ can form hydroxide complexes to reduce the Mn(H2O)6

2+ concentration. At low pH there is no effect. (c) Cu(NH3)4

2+(aq): at high pH the Cu2+ can form hydroxide complexes and at low pH the NH3 can be protonated, either extreme will reduce the Cu(NH3)4

2+ concentration. 7. Common interferences in EDTA titrations are copper or iron. They can interfere by appearing as analyte, giving erroneously high results, or by interfering with the complexometric indicator. These metals have very large formation constants with cyanide. Adding cyanide to a test portion will mask these metals so they do not complex with the EDTA titrant. 8. (a) These pH values correspond to pOH values of 9 and 13. Using Figure 7.5 we see that at pOH = 9 we have significant fractions of Fe(OH)3, Fe(OH)2

+, and FeOH2+. At pOH = 13 there is FeOH2+ and Fe3+. (b) The concentration is found by multiplying the total concentration by the alpha fraction. At pOH = 9 we can get the Fe3+ alpha from Figure 7.6 and it is 0.001. The concentration is then (1.0×10−3 M)(0.001) = 1.0×10−6 M. The alpha at pOH = 13 is 0.92, and the concentration is (1.0×10−3 M)(0.92) = 9.2×10−4 M


(c) You are starting with 0.10 M NH4

+ and 0.10 M Cl− from the buffer and 0.005 M Fe3+ and 0.015 M Cl− from the FeCl3. At pH = 9.2 (pOH = 4.8), all of the iron is in the form of Fe(OH)3, which has no charge. To reach that form it has reacted with water to form H3O+. The NH3 of the buffer system neutralizes this H3O+ to produce 0.015 mol of NH4

+. After the solution reaches equilibrium, it is now 0.115 M NH4

+ and 0.115 M Cl−. The ionic strength is thus: Ic = 0.5{(−1)2(0.115 M) + (+1)2(0.115 M)} Ic = 0.115 M

9. The log cumulative formation constants for Ag+ and Br− are 4.38, 7.33, 8.00, and 8.73. (a) The stepwise formation constants are

104.38 = 2.4×104 107.33−4.38 = 102.95 = 8.9×102 108.00−7.33 = 100.67 = 4.7 108.73−8.00 = 100.73 = 5.4

(b) The 1,10−phenanthroline has larger formation constant values for Ag+ than does Br−, so for comparable concentrations, it will displace Br− to complex with the Ag+. 10. At the end point of a titration, there will either be complete depletion of a metal being titrated or the presence of a metal from the titrant. Complexometric indicators have different colors when a metal is bound or unbound. Taking the case of EDTA titrant and a metal ion as analyte, at the end point the EDTA complexes all of the metal ion, removing metal from a metal−indicator complex. (Suitable indicators have Kf' values lower than for the metal ion and EDTA.) Electronically, binding of the metal changes the distribution of electrons in the complex compared to the electron distribution in the absence of the metal. This change in electron distribution can shift the electronic energy levels. We see the shift in wavelength of the absorbed light as a color change. 11. Less than the maximum number of ligands might be stable in solution if the stepwise formation values are well separated. The alpha plots show that the different complexes appear to be separable for Cu2+ and Fe3+ oxalate but not for Al3+ oxalate.


12. βeff' = αM αL β1 pH = 5 pH = 6 β1' αM αL βeff' αM αL βeff' Ca2+ 1×1011 3.54×10−7 2.25×10−5 Pb2+ 2×1018 3.54×10−7 2.25×10−5 Zn2+ 3×1016 3.54×10−7 2.25×10−5 13. equilibria

Ksp = (aLa)(aIO3)3 Ksp = (γLa)[La3+](γIO3)3[IO3

−]3 Ksp = (γLa)[La3+](γIO3)3Ksp'

Ksp' = Ksp (γLa)(γIO3)3

14. (a) The Cu3(PO4)2 is insoluble and the precipitation equilibrium is:

Cu3(PO4)2(s) ⇌ 3Cu2+(aq) + 2PO43−(aq)

The NH3 can complex with copper so you also have: Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)4

2+(aq) There will also be the base hydrolysis equilibria for the ammonia and PO4

3− to form the protonated form of each. (b) Mixing these two soluble salts will form a Cu(OH)2 precipitate and the copper ammine complex:

Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH−(aq) Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)4

2+(aq) There will also be the acid−base equilibrium for the NH4

+. 15. Predict the precipitation order for the following solutions (a) The solubility products for these precipitates are Cd(OH)2: Ksp = 7.2×10−15 Cu(OH)2: Ksp = 2.2×10−20 Zn(OH)2: Ksp = 3×10−17 Since the stoichiometry is the same for all of these precipitates, we can predict the precipitation order to be Cu(OH)2 first, Zn(OH)2 second, and Cd(OH)2 last.


(b) adding PO4

3− to a solution that is 0.1 mM each of Cu2+ and Al3+. The Ksp values are Cu3(PO4)2: 1.40×10−37 and AlPO4: 9.84×10−21. The PO4

3− concentration at which each precipitate forms is: Ksp = [Cu2+]3[PO4

3−]2 1.40×10−37 = (1×10−4)3[PO4

3−]2 [PO4

3−] = 3.7×10−13 M Ksp = [Al3+][PO4

3−] 9.84×10−21 = (1×10−4)[PO4

3−] [PO4

3−] = 9.8×10−17 M So, AlPO4 will precipitate first. 16. Neglecting competing equilibria: (a) BaCO3: s = [Ba2+] = [CO3

2−] (b) Ba(OH)2: s = [Ba2+] = 0.5[OH−] (c) Pb3(PO4)2: s = 0.333[Pb2+] = 0.5[PO4

3−] (d) ZnF2: s = [Zn2+] = 0.5[F−] (e) Ca(C2O4)·H2O: s = [Ca2+] = [C2O4

2−] (the waters of hydration in the solid have no effect) 17. pH effects: (a) BaCO3: Ba2+ is a strong electrolyte so there is no effect at high pH, at low pH the CO3

2− is protonated and solubility increases (b) Ba(OH)2: Ba2+ is a strong electrolyte so there is no effect at high pH, at low pH the OH− is neutralized so more Ba(OH)2 dissolves (c) Pb3(PO4)2: high pH will increase solubility due to formation of lead hydroxide complexes and low pH will increase solubility due to protonation of PO4

3− (d) ZnF2: high pH will increase solubility due to formation of zinc hydroxide complexes and low pH will increase solubility due to protonation of F− to form HF 18.


Calculate Q, the reaction quotient, and compare it to Ksp' (since we are neglecting activity effects we will use Ksp' = Ksp = 5.3×10−9). If Q exceeds Ksp', the concentrations have exceeded the solubility limit and a precipitate will form.

Q = [Ca2+][F−]2 Q = (0.00010 M)(0.00010 M)2 = 1.0×10−12 M Q < Ksp'

So the concentrations are below the level at which a precipitate will form. 19. There are different ways to think qualitatively about the effect of ionic strength on ionic equilibria. One way is to think of the high ion concentration as “screening” the electrostatic attraction of oppositely charged ions that are involved in an equilibrium. Thus, increasing the ionic strength of an aqueous solution will reduce the tendency of ions to recombine. The result is higher ion concentrations relative to neutral species, such as precipitates, in higher ionic strength environments. If some amount of an ion involved in a precipitation equilibrium is present or added from an additional source other than the precipitate, the concentration of this \common” ion will have a direct effect on solubility. This direct effect will decrease solubility and will usually be much larger than the indirect effect caused by ionic strength affecting equilibrium constants. 20. Neglecting ionic strength effects, Ksp' = Ksp = 7.50×10−12. Neglecting competing equilibria, we need only consider:

La(IO3)3(s) ⇌ La3+(aq) + 3IO3−(aq)

Ksp' = [La3+][IO3−]3

7.50×10−12 = (s)(3s)3 = 27s4 s = 6.9×10−4 M

21. We will neglect competing equilibria, such as lanthanum hydroxide complexes, but we will correct Ksp for activity effects.

Ksp = (aLa)(aIO3)3 Ksp = (γLa)[La3+](γIO3)3[IO3

−]3 Ksp = (γLa)[La3+](γIO3)3Ksp'

Ksp' = Ksp (γLa)(γIO3)3

The ionic strength is 0.010 M (neglecting the concentrations of La3+ and IO3

− ions from the lanthanum iodate), and activity coefficients from the Debye-Hückel equation are:

γLa = 0.44 and γIO3 = 0.90

Ksp' = 7.50×10−12 = 1.9×10−11 (0.44)(0.90)3


Setting up the equilibrium problem Ksp' = [La3+][IO3

−]3 = 1.9×10−11 The equilibrium concentration of IO3

− is the total that comes from both the sodium iodate and the lanthanum iodate.

[IO3−] = 0.010 M+ 3s

Thus Ksp' = (s)(0.010 M + 3s)3 = 1.9×10−11

Try neglecting 3s compared to 0.010 M

Ksp' = (s)(0.010 M)3 = 1.9×10−11 s = 1.9×10−5 M

Checking our assumption: 3(1.9×10−5) << 0.010 M

The approximation in the calculation is reasonable and our answer is 1.9×10−5 M. 22. (a) Cd(OH)2 Ksp = 4.5×10−15 (b) Mg(OH)2 Ksp = 7.1×10−12 Since this metal hydroxide has the highest Ksp, it is predicted to be the most soluble and will dissolve first as a basic solution is made more acidic. (c) Pb(OH)2 Ksp = 1.2×10−15 (d) Zn(OH)2 Ksp = 1×10−17 Since this metal hydroxide has the smallest Ksp, it is predicted to be the least soluble and will precipitate first as an acidic solution is made more basic. 23. Since CO3

2− has the largest Kb' of these anions, it will react with water to the greatest extent. Thus for the listed insoluble Ba salts, the solubility of BaCO3 will be affected the most by a competing equilibrium. BaCO3(s) ⇌ Ba2+(aq) + CO3

2−(aq) + H2O ↕ HCO3

−(aq) + OH−(aq)


Chapter 8: End-of-Chapter Solutions 1. (a) Cl2 will be reduced to Cl−, Sn2+ will be oxidized to Sn4+ (b) Cl2 will be reduced to Cl−, I− will be oxidized to I2 2. Sn(s), Pb(s), etc. All of the metals that have an E value more negative than 0.0 V can be oxidized; thereby reducing H+(aq) to H2(g). 3. (a) no reaction (b) no reaction (c) Ni(s) + Cu2+(aq) → Ni2+(aq) + Cu(s) (d) no reaction 4. (a) 3ClO3

−(aq) + 5NO(g) + H2O(l) ⇌ 1.5Cl2(g) + 5NO3−(aq) + 2H+( aq), 15 e− transferred.

(b) CH4(g) + 2O2(g) ⇌ CO2(g) + 2H2O(g), 8 e− transferred. (c) Zn(s) + 2MnO2(s) ⇌ ZnO(s) + Mn2O3(s), 2 e− transferred. (d) 4Zn(s) + NO3

−(aq) + 6H2O + 7OH−(aq) ⇌ 4Zn(OH)42−(aq) + NH3(aq), 8 e− transferred.

5. (a) Technologically important redox processes include smelting, corrosion, batteries, fuel cells, etc. As one example, aluminum is produced by reducing aluminum oxide:

Al2O3 + 6e− → 2Al in molten cryolite (sodium aluminum fluoride) in a giant steel furnace at 960 C. There is no chemical species directly oxidized in the furnace. An electric current passed through the cell causes the reduction. If the electric current is supplied by a fossil fuel burning generating plant, then a carbon species is oxidized to carbon dioxide. (b) A significant environmental redox process is the oxidation of minerals, e.g.: FeS2(s) + 14Fe3+(aq) + 8H2O ⇌ 15Fe2+(aq) + 2SO4

2−(aq) + 16H+(aq) where the S is oxidized from −1 in disulfide, S2

2−, to the +7 oxidation state in sulfate and Fe3+ is reduced to Fe2+.


(c) Biological redox processes include cellular respiration (citric acid cycle), photosynthesis, and a whole bunch more. 6. (a) All species are at standard concentrations, so E = −(−0.257) + 0.0 = 0.257V (b) All species are at standard concentrations, so E = −(−0.763) + 0.0 = 0.763V (c) The contents of the two half-cells are identical, so there is no driving force for a reaction and E = 0.0V (d) The same species are in each half cell (so E = 0.0 V), but there is a driving force to equalize Ag+ concentrations. The reaction is: Ag(s) + Ag+(aq) (0.5 M) → Ag+(aq) (0.001 M) + Ag(s) where the concentrations in parentheses are the starting concentrations. Using the Nernst equation to calculate E for the overall equation:

E = Eo − 0.0592 V log 0.001 M 1 0.5 M

E = 0.0V − (−0.16 V) = 0.16 V 7. (a) MnO4

−(aq) + 8H+(aq) + 5Fe2+(aq) ⇌ Mn2+(aq) + 5Fe3+(aq) + 4H2O (b) From the balanced reaction, the stoichiometric factor is 1 mol MnO4

− to 5 mol Fe2+. (c) (0.02240 L)(0.150 M) = 3.36×10−3 mol MnO4

−

(3.36×10−3 mol MnO4−) 5 mol Fe2+ = 1.68×10−2 mol Fe2+ 1 mol MnO4

− 8. (a) ClO−(aq) + 2I−(aq) + 2H+(aq) ⇌ Cl−(aq) + I2(aq) + H2O (b) I2(aq) + 2S2O3

2−(aq) ⇌ 2I−(aq) + S4O62−(aq)

(c) From the combination of the two balanced reactions, the stoichiometric factor is 2 mol 2S2O3

2− to 1 mol ClO−. 9.


Since concentrations are 1.0 M, we can work with E values: Ecell = 0.340 + 0.7626 = 1.103 V.

1.103 V = RT ln(K) nF K = 1.8×1037 As expected for the large cell potential, the equilibrium constant is large. 10. The spontaneous reaction is: Pb(s) + PbO2(s) + 2SO4

2−(aq) + 4H+(aq) ⇌ 2PbSO4(s) + 2H2O Recharging drives this reaction in reverse: 2PbSO4(s) + 2H2O ⇌ Pb(s) + PbO2(s) + 2SO4

2−(aq) + 4H+(aq) During charging solid lead sulfate is converted to solid lead metal and solid lead oxide on the plates of the battery and the concentration of sulfuric acid in solution increases. 11. Vitamin C is easily oxidized. It therefore serves as a sacrificial reagent to prevent strong oxidizing agents from damaging other molecules in biological cells. Because vitamin C can undergo a one-electron oxidation to a stable intermediate species, it reacts faster than other biochemical species that require a two-electron reaction. 12. Both are limited to measurement of analytes in solution and both can be specific for a given oxidation state. Voltammetry provides more parameters to vary and can be used to study physical processes or for analyte identification. potentiometry voltammetry measurement of: voltage current experimental: passive measurement variable voltage applied set up: indicator and reference electrodes usually three electrodes analytes: ions only ions and neutral species general: electrode is selective for a given ion can be used for analyte identification 13. The advantages and disadvantages of potentiometric measurements versus wet chemical methods are similar to those of any instrumental method compared to a classical method: potentiometry wet chemical sensitivity: can be very sensitive (μM or ppm) less sensitive


selectivity: usually very selective depends on specific analytes/interferences

interferences: levels of interferences are documented

interferences can require prior removal

calibration: requires calibration standards traceable to primary standards general use: rapid, can analyze many samples rather slow due to solution

manipulation field use: easily portable mostly lab based, field kits can

measure limited sample amounts 14. (a) Mg2+ in seawater: very high ionic strength, standards must match samples (b) NO3

− in a freshwater stream: should be free of interferences and measurable directly with an ISE (c) NH4

+ in a sample of solid fertilizer: sample must be dissolved and pH controlled (d) Cl− in stainless steel: sample must be dissolved (not trivial for SS), interfering ions might require masking or removal. 15. Per our convention in the chapter, I use concentration symbols rather than activity for clarity in these expressions: (a) Ecell = Econst + (0.0296 V)log[Mg2+] (b) Ecell = Econst − (0.0592 V)log[NO3

−] (c) Ecell = Econst + (0.0592 V)log[NH4

+] (d) Ecell = Econst − (0.0592 V)log[Cl−] 16. (a) For this data the lowest concentration point is not linear with the other points, so the linear range of the calibration data is from 10−4 to 10−1 M F−. (b) The measured slope, −0.055 V, equals (RT/nF)×2.303. Inserting the constants and n = −1 (for F−):

−0.055 V = 8.3145 J mol−1 K−1(T) 2.303 96485 C/mol gives an absolute temperature of 277 K or 4 C. Either the lab was quite chilly, or there is some deviation from the theoretical slope (quite common). Since standards and samples were treated the same and measured on the same day, the deviation from the theoretical slope should not affect the accuracy of the unknown measurement.


(c) The calibration curve is y = −55.0x−149.0. Inserting 48 mV for y gives an x of −3.58. Taking the inverse log gives [F−] = 2.6 10−4 M. 17. In any potentiometric measurement there is a linear relationship between measured potential and log(a), where a is the activity of the analyte ion. The slope and intercept in this relationship are variable. The slope will change if the temperature of the experiment changes. The intercept will change if reference or junction potentials in the circuit change, which are also sensitive temperature. By doing a two-point calibration, both the slope and intercept are corrected to establish an accurate linear relationship for the pH readout. 18. An ion-selective electrode will be selective for only one specific analyte ion. Other ions will not interfere unless they are present at higher concentrations than the analyte. Cyclic voltammetry will detect all analytes present as the voltage is scanned. If voltages of different components in a mixture overlap, other components will interfere with measurement of the analyte. Voltammetry will detect neutral species in addition to ions. 19. Based on Figure 8.13: (a) The width of the transition for each analyte is approximately 60 mV. Two analytes with reduction potentials closer than 60 mV would overlap and be difficult to distinguish, so potentials > 60 mV are necessary to measure the waves separately. (b) Taking the full range of the x scale, the scan range of 700 mV (−100 mV to −800 mV) takes 350 s. The scan rate is (700 mV)/(350 s) = 2.00 mV/s. (c) The second analyte has a current signal of ≈ 7.0 A. The first analyte has a current signal of ≈ 9.0 A. Setting up a ratio for the second analyte being 1.0×10−4 M gives a concentration of 1.3×10−4 M for the first analyte. (d) The direct ratio used in the previous answer assumes that the analytes are similar, i.e., they have equal diffusion constants.


Chapter 9: End-of-Chapter Solutions 1. The following comparison provides general trends, but both atomic absorption spectroscopy (AAS) and atomic absorption spectroscopy (AES) will have analyte-specific exceptions. Due to the atomization requirement, both methods destroy the sample and measure the total concentration of a given element, i.e., neither provides information on analyte speciation. Obtaining speciation-specific results is possible by separating analytes before measurement. AAS AES scope usually single analyte simultaneous multi-element analysis sensitivity ppm range for flame AAS

can be more sensitive than AES for easily ionized elements

ppb range for ICP-OES in general AES is more sensitive than AAS

atomization sources

sufficient signal at lower temperature (simpler and cheaper)

excitation efficiency increases with temperature

2. The key difference between potentiometry and atomic spectroscopy is that an ion-selective electrode can be used in situ and it can be specific for a given speciation of an element. Atomic spectroscopy is generally more sensitive and provides total analyte concentration. The two methods are often used together to provide complementary information. potentiometry atomic absorption spectroscopy general can be portable and used in-situ

preserves sample simpler and less expensive

lab based destroys sample more expensive

sensitivity sensitive (10−5 M, 1 ppm) very sensitive ( 0.1 ppm typical) selectivity usually very selective narrow line, so intrinsically selective interferences interferences for specific ion-selective

electrodes are usually known and correctable with selectivity factors and additional measurements

high concentrations of refractory substances can reduce atomization efficiency

calibration requires calibration standards requires calibration standards speciation oxidation-state specific total analyte concentration 3. (a) Ca2+ and Mg2+ in leafy vegetables: The matrix is expected to contain multiple substances that can complex the analytes. In this case ashing the sample will reduce interferences. AES and AAS are both very sensitive for alkali and alkaline earth metals and will be the faster and preferred method. Ca2+ and Mg2+ ISEs can be used after digesting the sample and using the standard addition method for calibration.


(b) NO3− and NH4

+ in surface water: nitrate and ammonium ISEs will be the preferred analysis. These two measurements will distinguish the analytes from each other and from other nitrogen species. Surface water will be a reasonably clean matrix. (c) multiple metals: ICP-AES will provide simultaneous multi-element quantitation at the ppb level. (d) NH4

+ ISE, same reason as for the NO3− ion

(e) ClO4

− in sea water: A perchlorate ISE is suitable with correction for Cl− in the seawater. ICP-MS is suitable with prior separation of perchlorate from other anions. (f) Co in steel: Spark source AES is sensitive and rapid (does not require dissolving steel). 4. The difference in these two masses is 111.91460 u − 111.90276 u = 0.01184 u. Using this value and the approximate m of 112: m = 112 u = 9460 ∆m 0.01184 u

The required resolution is approximately 10,000. 5. A simple interpolation between the calibration data is sufficient, but the equation is also easy to determine. The slope is

(1.284 – 0.033)/(10.0 ppm) = 0.1251 ppm−1 and the calibration equation is

y = (0.1251 ppm−1)[Pb2+] +0.033. For the unknown measurement:

0.471 = (0.1251 ppm−1)[Pb2+] + 0.033 [Pb2+] = 3.50 ppm

6. The inductively coupled plasma (ICP) generates a much higher temperature than a flame, 6000-8000 K versus 1700-2700 K. The higher temperature results in greater atomization and excitation efficiencies, so detection limits are lower (1 ppb for ICP-AES vs. 0.1 ppm for flame AES).


7. (a) A plot shows that the data is linear from 0 to 50 ppm.

(b) Using the trendline shown in the plot: 0.455 = (0.0140 ppm−1)[Ca2+] [Ca2+] = 32.5 ppm Given that the blank measurement was zero, I chose to force the calibration function through zero. There is only a small difference in the result if the trendline is allowed to vary the intercept. (c) The unknown measurement is outside of the linear range of the experiment. You have two options. If you discover the situation as you make measurements, you can dilute the unknown by a factor of 2 and measure the AES signal again. If you discover the situation after the instrument is turned off, you can fit the calibration data to a 2nd-order polynomial (see plot below). The response in atomic spectroscopy is known to be non-linear at higher concentrations. For a measurement of 0.952, a matrix effect appears to create a systematic error compared to the calibration data. The analyst should be conservative and quote a result of > 50 ppm Ca.

y = 0.0140xR² = 0.9984

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 20 40 60 80 100

Abso

rban

ce

Ca concentration (ppm)

y = -1E-04x2 + 0.0181xR² = 0.9969

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 20 40 60 80 100

Abso

rban

ce

Ca concentration (ppm)


8. The details of setting a discriminator level for a detector is more advanced than we need to consider. The purpose of this question is to think about data and the resulting signal-to-noise ratio. Looking at the tabulated data, the signal of the blank, due to background, decreases rapidly as a function of DISC. The signal of the test solution is the summation due to the analyte and the blank. It also falls off rapidly versus DISC, but levels off at a higher level than the blank. For example, the analyte signal at DISC = 40 mV is 179000 cps – 5920 cps = 173000 cps. Subtracting the high background at lower DISC introduces greater uncertainty in the result. The following plot shows signal counts on the left vertical axis for the data in the table. On the right vertical axis is S/N for the open circles. The open circles are found from

S/N = (test solution – blank) √blank

where the square root of the blank serves as an estimate of the noise.

0

500

1000

1500

2000

2500

3000

0

1000000

2000000

3000000

4000000

5000000

6000000

7000000

8000000

9000000

10000000

0 5 10 15 20 25 30 35 40

S/N

Sign

al (c

ps)

DISC (mV)


9. The atomization source, a flame, furnace, or plasma, creates background light emission that will reach the detector. This background signal is not related to analyte concentration and is subtracted from the detector signal. In atomic absorption spectrometry, the absorption signal might be affected by absorption due to molecular species. A background correction technique will also remove this source of error. These background sources can vary depending on the matrix of the sample. 10. In graphite-furnace atomic spectroscopy (GFAAS), the test portion is introduced as a small plug of liquid or solid directly into the furnace tube. Since only microliters of the test portion are needed, measurements can be made when the amount of sample is very limited. The main disadvantage of GFAAS is that it measures a transient signal and it is not possible to signal average during any one measurement. The result is a less precise measurement than when using a constant nebulizer. Multiple GFAAS measurements can be made if sufficient sample is available, requiring a longer time to complete the measurement. The matrix effects in GFAAS can also require more extensive method validation for a given analyte and sample matrix. 11. Refractory sample constituents can be digested with aggressive acid and redox reagents. The downside is the potential loss of analyte in this process. Other methods include direct introduction of the sample in graphite-furnace AAS or spark source AES. These methods are rapid. They are best suited for repetitive analysis of similar samples so that matrix effects can be corrected after thorough method validation. The laser ablation sampling method is a third alternative, which has the advantage of providing spatially resolved elemental analysis on a heterogeneous sample. 12. Atomic spectra have very narrow linewidths of 0.1 nm and less. Molecular spectra consist of broad bands, often 50 to 100 nm wide. An atomic spectrometer must have a resolution on the order of 0.1 nm or less to resolve closely spaced lines. Spectrometers for molecular absorption and fluorescence will have resolution on the order of 1 nm. 13. The high vapor pressure of Hg leads to loss of analyte when handled in conventional ways. The cold-vapor technique, which keeps the analyte in solution at room temperature before measurement, minimizes this loss for more accurate Hg determinations.


14. The nonmetals have atomic transitions at wavelengths shorter than 190 nm. This region is called the vacuum ultraviolet (VUV) region because working in this region requires evacuation of atmospheric gases, mainly oxygen, that absorb this light strongly. Although ICP-OES is capable of measuring nonmetals with a suitable evacuated spectrometer, ICP-MS is usually the method of choice. 15. Measuring isotopic ratios is usually done by mass spectrometry. Isotopes will differ by one or more mass units, and for stable isotopes of elements requires a mass spectrometer with moderate resolution, on the order of 500. The atomic emission lines of different isotopes will have slightly different wavelengths, but the differences are small and requires a very high resolution spectrometer to resolve.


Chapter 10: End-of-Chapter Solutions

1.

(a)

Chromatographic detector Use

ECD (electron capture detector) GC only

FID (flame ionization detector) GC only

fluorescence LC only

mass spectrometer GC and LC

RI (refractive index) detector LC only

TCD (thermal conductivity detector) GC only

UV/Vis absorption detector LC only

(b) Of the choices, only the mass spectrometer provides molecular structure information. UV-Vis

absorption can provide some identification, but many organic analytes have similar spectra.

(c) The refractive index and UV-vis absorption detectors (for LC) and the thermal conductivity

detector (for GC) are non-destructive.

2.

GC detector advantage disadvantage

TCD

(thermal conductivity detector)

universal

non-destructive

less sensitive than other

detectors

FID

(flame ionization detector)

sensitive

high dynamic range

destroys analyte

ECD

(electron capture detector)

selective for electronegative

functional groups

very sensitive for select

analytes

non-destructive

limited dynamic range

mass spectrometer provides structural

information for solute

identification

more complex maintenance

and operation

destroys analyte

There are a number of other more specialized GC detectors that are not discussed in the text.

Some examples include:

photoionization (PID): selective for aromatic compounds, e.g., benzene, touluene,

ethylbenzene, and xylenes (BTEX)

flame photometric detector (FPD): flame chemiluminescence for compounds containing

sulfur and phosphorous

nitrogen phosphorous detector (NPD): plasma ionizer for compounds containing nitrogen

and phosphorous

dry electrolytic conductivity detector (DELCD): an ECD replacement for chlorine and

bromine containing compounds


3.

(a) reverse-phase partition liquid chromatography:

n-hexanol, benzene, and n-hexane (more polar to less polar)

the nonpolar analytes will interact more strongly and be retained longer

(b) normal-phase partition liquid chromatography:

n-hexane, benzene, and n-hexanol (less polar to more polar)

the more polar analytes will interact more strongly and be retained longer

(c) capillary GC with nonpolar stationary phase

In this case we expect the analytes to elute in order of boiling point (b.p.) from lowest to

highest b.p. The expected order is:

n-hexane (b.p. = 68 C)

benzene (b.p. = 80 C)

n-hexanol (b.p. = 157 C)

The deciding factor between reversed-phase and normal-phase partition chromatography is

usually based on solubility of the solutes. Give the reasonable solubility of the listed solutes in a

water-based mobile phase, reversed-phase will be the preferred method for these analyte.

The main advantage of GC is the superior resolution compared to HPLC. If these solutes must be

separated from other components, then GC will be advantageous. Given that the alcohol is more

polar than the other solutes, the disadvantage of GC is that there may not be an optimum

stationary phase to achieve narrow symmetric peaks for all analytes.

4.

Ions in solution have an electrostatic attraction to immobilized ions on the stationary phase. The

strength of the interaction depends on the charge and size of the analyte ions. Ions of higher

charge and smaller size, which can approach the stationary phase closer, are attracted more

strongly and are retained longer on the column.

5.

(a) The analytes are anions so use the anion-exchange HPLC column. Attempting to inject

inorganic ions into a GC results in decomposition or depositing non-volatile salts in the injector.

(b) The analytes are cations so use the cation-exchange HPLC column. The rationale is the same

as for anions.

(c) The analytes are volatile organic compounds so use capillary GC column. Since the analytes

are nonpolar, choose the nonpolar polydimethyl siloxane stationary phase.

(d) Answering this question requires some research into the structure of barbiturates. Based

simply on the structures, either GC or HPLC should be suitable techniques. Earlier analytical

methods used GC analysis. A blood matrix will require an extraction step and the barbiturates are

often derivatized. The stationary phase will depend on the product of the derivatization reaction.


Since these molecules are water soluble, the C18 HPLC column is being developed as a standard

analytical method. Again SPE cleanup will be needed due to the complex blood matrix.

6.

(a) To maintain these acids as anions, the mobile phase pH must be greater than the analyte with

the highest pKa (niacin, pKa = 4.85). To be safe, a pH > 6.85 will keep > 99 % of the niacin

deprotonated.

(b) Since these analytes are weak acids, a strong anion exchange stationary phase is preferred.

The stationary phase will not change as a function of mobile phase pH. All analytes are anions at

the initial pH. As the weakly retained analytes elute from the column, the pH can be lowered to

begin protonating the more strongly retained analytes. Since the weak acids become neutral

when protonated, these analytes will then elute from the column.

7.

The large size of proteins makes them very non-volatile. If they were injected into a gas

chromatograph they would decompose rather than vaporize intact.

8.

(a) The peak height of each analyte is taken from Table 10.7. The identity of the peaks is made

by matching the retention times to the standard data in Table 10.6. The data is summarized here:

analyte retention time, tR (min) peak height (arb. units)

theobromine 1.0 12.2

theophylline 1.7 7.0

caffeine 4.3 2.9

The concentration of each analyte is determined by converting the test portion peak height to

concentration using the calibration data in Table 10.6. The calculation for each analyte is a

simple proportionality:

ctheobromine =

1.0 µg/mL

12.2 7.2

ctheobromine = 1.7 µg/mL

ctheophylline =

1.0 µg/mL

7.0 12.4

ctheophylline = 0.56 µg/mL

ccaffeine =

1.0 µg/mL

2.9 3.9

ccaffeine = 0.74 µg/mL


(b) Given the results in part (a), the weight percent of the analytes in the cocoa sample is

determined by correcting for sample preparation. The analyte was contained in 30 mL of

extracting solvent, so the amount of each analyte was:

wttheobromine = (1.7 µg/mL)(30 mL) = 50.8 µg

wttheophylline = (0.56 µg/mL)(30 mL) = 16.9 µg

wtcaffeine = (0.74 µg/mL)(30 mL) = 22.3 µg

Dividing each weight by the 50 mg sample portion and multiplying by 100 % gives:

50.8 µg × 100 % = 0.10 % theobromine

50,000 µg

16.9 µg × 100 % = 0.034 % theophylline

50,000 µg

22.3 µg × 100 % = 0.045 % caffeine

50,000 µg

9.

The total ion chromatogram (TIC) is the summation of all peaks and provides the largest

analytical signal. The largest signal provides the best signal-to-noise ratio and thus the greatest

sensitivity. Selective ion monitoring (SIM) displays the ion current signal at only one m/z. This

mode provides greater selectivity by picking out the parent ion or an intense fragment ion of the

analyte of interest. Solutes that do not have a mass fragment at the selected m/z will not appear in

the chromatogram, making it simpler to interpret and reducing the chance of interferences

affecting a quantitative measurement.

10.

Table of MS designs

mass analyzer design advantage disadvantage

magnetic-sector simultaneous detection of

several m/z – best precision

for isotope ratio

measurements

slow scan rate

quadrupole

compact and inexpensive

rapid scan rate

moderate resolution

limited scan range (≈ 5000

Da maximum)


time-of-flight

high resolution (15,000)

rapid scan rate

pulsed operation requires

interface to some ion sources

ion trap can accumulate ions

useful as an interface between

a continuous ion source and a

time-of-flight mass analyzer

moderate resolution

limited scan range (≈ 5000

Da maximum)

orbitrap highest resolution expensive

11.

The electrodes provide the electrostatic driving force to move the charged proteins (or other type

of macromolecules). Using SDS surfactant gives all of the proteins a negative charge, so they

migrate from the negative electrode towards the positive electrode. The buffer maintains the

system at a pH where the proteins are stable. In discontinuous gel electrophoresis, the change in

buffer pH “stacks” the proteins before entering the running gel. Large proteins are hindered to a

greater extent by the pore size of the gel. Smaller proteins migrate faster and travel farther in gel

electrophoresis. Since preparation of the gel can vary day-to-day, measurements are made

relative to molecular weight standards in one or more separate lanes of the gel.

12.

The biomolecules in a gel are usually visualized by adding a stain. The most common stains are

coomassie brilliant blue for proteins and ethidium bromide for nucleic acids. Silver stain works

for either type of biomolecule and provides greater sensitivity. In some applications it is possible

to irradiate the gel with ultraviolet light to detect the biomolecules using fluorescence.

13.

The acrylamide concentration and the relative amount of cross linker (usually bisacrylamide)

added during preparation of the gel determines the pore size. Since proteins have a large range of

sizes, the pore size must be appropriate to match the proteins being separated. Other additives

besides the buffer and SDS surfactant include urea to denature proteins, a reducing agent to

break disulfide bonds, and reagents to fine tune the polymerization. Most labs that use SDS-

PAGE will have detailed recipes and protocols that are optimized for their applications.

14.

When multiple analytical methods are suitable for a given class of analytes, in this case amino

acids, it can often be difficult to know what method to choose. I provide a general discussion

here to help you understand the methods that you might have discovered. In many cases the

method selection is determined by practical reasons, what instrument do you have, or by

constraints such as time or cost. A chief scientific reason is the complexity of the sample matrix.


The high resolution of GC allows rapid chromatography, but requires derivatization of the amino

acids. HPLC is readily amenable to amino acids in aqueous solution, but will usually require

longer run times than GC. Capillary electrophoresis provides the highest resolution. Despite the

greater complexity to achieve repeatable results, CE might be the best choice is the amino acids

must be separated from similar interferences.

15.

Common forms of arsenic are arsenite, As(III), arsenate, As(V), methylarsonic acid, As(V), and

dimethylarsinic acid, As(V). Since all of these compounds can form anions at high pH (≈11 or

higher), they can be separated by anion-exchange chromatography. The effluent of the

chromatography column can enter an ICP-MS to measure each arsenic species separately.

16.

I did a Google search on “capilliary electrophoresis application note” to find manufacturer

literature. There is certainly many, many more papers in the primary literature. A few examples

that I found are:

solutes # detection limit speed notes

carbonic

anhydrase and

related proteins

5 Agilent

isoelectric point

determination

antibodies < 10 Horiba detection of

glycosylation

therapeutic

proteins

< 10 Beckman-Coulter 30 min

A common aspect that I found in my limited search is the focus on biomedical analytes.

solutions manual to basics of analytical and chemical equilibria - brian m. tissue

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