chapter 5. acid/base reactions
TRANSCRIPT
1
Chapter 5. Acid/Base Reactions
5. Log C – pH Diagrams (Graphical Solutions)6. Titration7. Buffers and Buffer Intensity8. Carbonate System
2017 Water Chemistry
5. Log C – pH Diagrams pH as an independent variable
Develop equations for the concentration of a given species as a function of pH and known quantities (e.g., Kw, KA, CT)
a. Introduction to log C-pH diagram : A weak monoprotic acid example
i. Draw [H+] line
-log [H+] = pH
log [H+] = -pH
ii. Draw [OH-] line
[H+][OH-] = Kw
log [H+] + log [OH-] = log Kw
log [OH-] = pH – pKw
2
iii. Draw [HAc] and [Ac-] lines
[HAc] = [Ac-] =
If pH << pKA or [H+] >> KA
log [HAc] = log CT log [Ac-] = log CT – pKA + pH
If pH >> pKA or [H+] << KA
log [HAc] = log CT + pKA – pH log [Ac-] = CT
If pH = pKA, [H+] = KA
log [HAc] = log CT – log 2 = log CT – 0.3 log [Ac-] = log CT – 0.3
][
][0
HK
HCC
A
TT
][1
HK
KCC
A
ATT
0.02 M HAc Example
- KA = 1.8*10-5; pKA = 4.74- log [HAc] = - 1.70; 3.05 – pH- log [Ac-] = -6.44 + pH; -1.70
System point: where pH = pKA & log C = log CT
CBE or PCE : [H+] = [OH-] + [Ac-]
Look at intersections
- pH 7
- pH 3.2; [Ac] >> [OH-], [H+] = [Ac-]
3
b. Rapid method for construction of logC-pH diagram
i. Plot diagonal [H+] and [OH-] lines.
ii. Draw a horizontal line corresponding to
log CT.
iii. Locate system point.
- pH = pKA & log C = log CT
- make a mark 0.3 units below system point
iv. Draw 45 lines (slope 1) below CT line, and aimed at system point.
v. Approximate curved sections of species lines 1 pH unit around system point.
vi. Repeat steps as necessary for more complex graphs.
#3-5 for additional pKAs of polyprotic acids.
#2-5 for other acid/base pairs.
c. LogC-pH diagram for a monoprotic weak base
NH3 + H2O = NH4+ + OH-
+ H+ + OH- = H2O NH3 + H+ = NH4
+
- the same as drawing a diagram for conjugate acid- pH for the system point = pKA = pKW – pKB = 14 – 4.74 = 9.26
- 0.01 M NH3
4
Example 4.13 : 0.01 M NH4Cl- Can add log [Cl-]- H+, OH-, NH3, NH4
+, Cl-
CBE : [H+] + [NH4+] = [OH-] + [Cl-]
PCE : [H+] = [OH-] + [NH3]; Usually easier When [H+] = [NH3], pH = 5.8
Checking CBE: [NH4+] = [Cl-]; H+, OH- negligible
Example 4.12 : 0.02 M Potassium acetate (KAc)- Can add log [K+] - H+, OH-, HAc, Ac-, K+
CBE: [H+] + [K+] = [OH-] + [Ac-]PCE: [H+] + [HAc] = [OH-]
- Point when [H+] = [OH-], not good- Good when [HAc] = [OH-], pH of 8.5- CBE satisfied.
d. LogC-pH diagram for a weak acid and weak base
- Mixtures of 0.1 M acetic acid (HAc) and 0.1M ammonium hydroxide (NH4OH)
- They may exist in relatively high concentration in anaerobic digesters- Superimpose the curves for each material.
4.74 9.26
5
Example 4.14 NH4Cl + HAc Case 0.01 M each
- Species: H+, OH-, HAc, Ac-, NH4+, Cl-
- CBE: [H+] + [NH4+] = [OH-] + [Ac-] + [Cl-]
- PCE:H+
PRL H2O HAc NH4
OH- Ac- NH3
[H+] = [OH-] + [Ac-] + [NH3]
[H+] = [OH-] does not work[H+] = [Ac-] OK. pH = 2.8
6. Titrationa. Titration of strong acids and bases
- When strong base is titrated,+ pH changes very little at first, then slowly declines to a pH of about 10.+ essentially vertical between pH values of 10 and 4.+ equivalence point or stoichiometric end point lies between these values.
- Equivalence point is reached when the equivalents of base (or acid) added equal the equivalents of acid (or base) initially present.
Titration: procedure by which ameasured amount of chemical orreagent is added to a solution in orderto bring about a desired and measuredchange.
6
VV
CVOHH
VV
VC
0
00
0
][][
][][)1(000
00
HOHVC
CV
VV
CV
Example: Strong acid HCl is titrated with strong base NaOH
NaOH + HCl H2O + Na+ + Cl-
CBE: [Na+] + [H+] = [OH-] + [Cl-]C0 = conc. of HClV0 = initial volume of HClC = conc. of NaOHV = vol. of NaOH added
Moving terms related to titration to the left,
f = equivalence fraction. At the equivalence point, f = 1.
][][
)1(0
00
H
H
Kf
VV
CV w
Example 4.18. 500 mL of 0.01 HCl is titrated with 0.5 M NaOHCalculate pH values at different points in the titration.
- V negligible. 50 times concentrated. Very small addition.- [OH-] negligible at lower pH region
][][)1(
000
00
HOHVC
CV
VV
CV
][)1(000
00 HVC
CV
V
CV
7
Example 4.18. ContinuedCalculate pH values at different points in the titration.
- V negligible. 50times concentrated. Very small addition.- [OH-] negligible at lower pH region
5 mL addition: 50 % neutralized. Use CV/C0V0. pH 2.39 mL: 90% neutralized. pH 3.010 mL: 100% neutralized. Equivalence point! Have only NaCl and water. NaCl is
neutral, pH is 7.
After the equivalence point,
11 mL: pH = 11When pH = 4, 99% neutralized. Practically the titration is complete.
][][)1(
000
00
HOHVC
CV
VV
CV
][)1(000
00 HVC
CV
V
CV
][)1(
000
00
OH
K
VC
CV
VV
CV W
b. Titration of weak acids and bases
- Frequently encountered in natural water systems.
- When weak acids are titrated with strong bases, the character of the titration curve depends on whether the acid is monoprotic or polyprotic.
- Look at different KA (pKA) values. Shapes (buffering well at pKA)
- H3PO4 example - multiprotic
8
Titration of a weak monoprotic acid with a strong base
HA = H+ + A-
[H+][OH-] = Kw 2)
HA + B+ + OH- → B+ + A- + H2O(B+ : cation associated with the strong base)
CBE : [H+] + [B+] = [A-] + [OH-] 3)
- Simultaneous solution of 1), 2), 3) can determine [H+]- Introduce C0, V0, C, V
- MB on weak acid:[HA] + [A-] = = CT,A: it changes as you add base.
1) ][
]][[AK
HA
AH
VV
VC
0
00
VV
VC
0
00
- MB on B+
[B+] =
- Introduce ionization fraction
- CBE becomes
[B+] [A-] = 1* CT,A
- Derive equation for f by multiplying (Vo+V)/CoVo
Example 4. 19
VV
CV
0
1
00
01
1
00
00
)1()]([
)][
1()]([
A
A
K
H
VC
VVA
H
K
VC
VVHA
)(][][0
001
0 VV
VCOHH
VV
CV
)])([][
(00
01
00 VC
VVH
H
Kf
VC
CV w
9
c. Estimation of pH during titration
Case: Weak acid. [HA] = Co. Titration with a strong base
i. Beginning of titration
HA + B+ + OH- → B+ + A- + H2O
CBE: [H+] + [B+] = [A-] + [OH-]
[H+] + [B+] = [A-] + [OH-]
[H+] [A-] KA equation
[A-] << C0
Example 4.20-21.
- Check the [H+] [A-] assumption from the pH value obtained.
- It assumption is not correct, use the approximation
AKHA
AH
][
]][[])[(][][ 0
ACKHAKH AA
0][ CKH A
)log(2
1log
2
1log
2
1]log[ 00 CpKCKHpH AA
])[(][ 0 ACKH A
ii. Midpoint of titration
HA + B+ + OH- B+ + A- + H2O
- [A-] increases and [HA] decreases- When 50% completed, [B+] ; diluted slightly
Conc. of B+ & A- become significant[H+] << [B+], [OH-] << [A-]
CBE: [H+] + [B+] = [A-] + [OH-]
- CBE becomes [B+] [A-] 1)
[HA] 2)
- 1) & 2) to mass action equation
pH pKA; indicates the value of expressing the ionization constant in the pKA form.
0
2
1C
0
2
1C
AK
C
CH
0
0
2
1
)2
1]([
AKH ][
02
1C
10
iii. Equivalence point of titration
- [B+] = Co, [H+] << [OH-] ; HA + B+ + OH- B+ + A- + H2O
CBE: [H+] + [B+] = [A-] + [OH-]
Thus [HA] [OH-] =
[A-] = Co
pH
- Depends on concentration of acid and equilibrium constant.- The weaker the acid and the higher its conc., the higher the equivalence point pH.
][OH ][
][OH ][][-
0
-0
AC
ACB
][ H
KW
AKHA
AH
][
]][[A
W
K
H
K
AH
][
]][[
AWKK
A
H
][
][
1
)(log2
10 wA pKpKC
[HA] ][ MB from but 0 AC
pH of weak base during titration with strong acid
- Can be determined in a similar fashion.
- Beginning of titration:
- Midpoint of titration:
- Equivalence point of titration:
- The inflection point at the equivalence point in the titration of weak acids or bases with ionization constants less than about 10-7 M or pK values greater than 7 is not sharp and so is difficult to detect accurately.
0log2
1
2
1CpKpKpH Bw
Bw pKpKpH
)log(2
10CpKpKpH Bw
11
Titration curves for weak bases
- Weak bases- Salts of weak acids are alkaline, NaAc, Na2CO3
- Simply acid titration curves in reverse; curve is a mirror image of that for the acid.
Example 4.22
- pH of the equivalence point for each ionization is approximately equal to the average of the pKA value for that ionization and the pKA value for the following ionization.
Example 4.23
d. Use of log C-pH diagram for titrations
- Acetic acid example: 0.01 M M HAc (HAc + B+ + OH- B+ + Ac- + H2O)
CBE:
[H+] + [B+] = [Ac-] + [OH-]
- Beginning of titration,
[H+] [Ac-]
- Midpoint,
[B+] [Ac-] = [HAc]
=> located at pKA; pH regardless of Co
- Endpoint,
[B+] = Co = [HAc] + [Ac-] (to CBE)
[H+] + [HAc] + [Ac-] = [Ac-] + [OH-]
[H+] + [HAc] = [OH-]
- Effect of changes in the conc. of weak acid: observed by moving the horizontal [HAc] and [Ac-] curves up or down.
- Study more examples in textbook.
02
1C
12
7. Buffers and Buffer Intensity Buffers: Substances in solution that offer resistance to changes in pH (as
acids or bases are added to or formed within the solution)
- Commonly a mixture of an acid and its conjugate base. Best to have a reservoir of each so there is a resistance to change in both direction (HA for base, A- for acid).
- Poorly buffered waters are susceptible to acid precipitation.
Buffer intensity: the amount of strong acid or base required to cause a small
shift in pH.
- Midpoint in the titration curve => slope is at a minimum => intensity greatest.
- At the midpoint, the solution contains equal quantities of the ionized and the unionized acid or base.
- pH at the midpoint in the titration => pKA or pKW-pKB for weak bases. Thus use Table 4.1 & 4.2;
- Buffers effective at a pH near the pK value; but effective within 1 pH unit of pK
][
]][[
HA
AHKA
][
][1
][
1
HA
A
KH A
Henderson-Hasselbalch equation (Equilibrium for a weak acid):
- pH of a buffer solution made of a weak acid and its salt depends upon the
ratio of salt concentration to acid concentration.
- Solution to weak acid/conjugate base problem using equil., MBE, and CBE
gives the same answer.
- pKA2 for phosphoric acid (H2PO4- = H+ + HPO4
2-) is 7.21
Useful buffer at neutral pH.
Commonly used in analytical tests.
][
][log
acid
saltpKpH A
13
dpH
dC
dpH
dC AB
Example 4.24.
1. pH of a buffer containing 0.01M acetic acid & 0.01 M sodium acetate
pH = = pKA + log (0.01/0.01) = 4.74 + 0 = 4.74
2. apH after HCl is added to this solution to give a concentration of 0.001 M.
Salt neutralizes HCl to produce acid.
pH = 4.74 +
Buffer intensity (index) calculation
; number of moles of acid or base required to produce a given
change in pH; highest at the midpoint; low at the endpoint
65.409.074.4
001.001.0
001.001.0log
][
][log
acid
saltpKpH A
][][
][
HK
KCH
H
KC
A
ATwB
dpH
Hd
Hd
dC
dpH
dC BB ][
][
303.2
]ln[]log[
HHpH
][
1
]d[H
]dln[H ;
][303.2
][
303.2
]ln[
HH
HdHddpH ][303.2
][
HdpH
Hd
][][303.2
Hd
dCH
dpH
dC BB
Buffer Index - HAc example
- Can be readily determined from a titration curve.
Also can be calculated.
- CT moles of monoprotic acid & CB moles of base:
HAc + NaAc
CBE: CB + [H+] = [OH-] + [A-]
- Use ionization fraction expression for [A-]
Also,
Since
- Differentiating CB with respect to [H+] and substituting into the above eq. yields,
2])[(
][][
][303.2
HK
HKCH
H
K
A
ATW
14
Example 4.25
- Buffer containing 0.2 M HAc and 0.1 M acetate. pH = 5.0
- Amount of NaOH (mol/L) required to increase the pH to 5.1?
CT = 0.2+0.1 = 0.3 M
[H+] = 10-5
Use
= 0.16 mol/L of base per pH unit
NaOH required = dpH = (0.16)(5.1-5.0) = 0.016 mol/L
2])[(
][][
][303.2
HK
HKCH
H
K
A
ATW
8. Carbonate System
a. Significance
- Regulating pH in natural waters (major acid-conjugate base systems)
- Sources: volcanism, combustion, respiration.
- Sinks: photosynthesis, autotropic metabolism, precipitation
- Major forms of carbon on earth
15
Mineral Carbonation of Carbon Dioxide
• Primary process by which carbon dioxide is removed from the atmosphere>99% world’s carbon reservoir is locked up as limestone & dolomite rock – CaCO3 & MgCO3
• Thermodynamically favourable, but kinetically slow
Mineral carbonation refers to the conversion of silicates to solid carbonates, mimicking the natural process by which CO2 is removed from the atmosphere
~1012 tonnes CO2in atmosphere
~1 billion tonnes/year
CO2
CO2(g) + CaSiO3 weathering→ CO32-(aq) + Ca2+(aq) + SiO2 mineralisation→ CaCO3(s)
~1018 tonnes CO2in carbonate rock
OMAN: 70,000km3 of 30% olivine; sufficient to mineralise centuries of global CO2 emissions.
Combine Mineral Carbonation with Wastes Treatment
16
b. Carbonate species and their equilibria
- Gas transfer and acid-base reactions
pKA1 = 6.37, pKA2=10.33
CO2(g) = CO2(aq) K= [CO2(aq)]/Pco2=KH = 10-1.5 (M/atm)
KH = Henry’s law constant
Henry’s law (p.25): Gas transfer (gas-liquid) equilibrium. The mass of any gas that will dissolve in a given volume of a liquid, at constant temperature, is directly proportional to the pressure that the gas exerts above the liquid.
This form more common (careful about the unit). KH = 31.6 (atm/M)equil
gasH C
PK
b. Carbonate species and their equilibria - continued
CO2(aq) + H2O = H2CO3 KCO2 = = 10-2.8 (hydration equil.)
H2CO3 = H+ + HCO3- KH2CO3 =
- It is difficult to distinguish between H2CO3 and CO2 by analytical procedure =>
Hypothetical species introduced
[H2CO3*] = [H2CO3] + [CO2(aq)] [CO2(aq)]
( [CO2(aq)] = 630 [H2CO3] : total analytical conc. is essentially dissolved
carbon dioxide)
KA1 =
KA2 =
5.3
32
3 10][[
]][[
COH
HCOH
][
][
)(2
32
aqCO
COH
33.10
3
23 10][
]][[
HCO
COH
37.6
32
3
)(2
32322*
32
3 10][[
]][[
][
][
][[
]][[
COH
HCOH
CO
COHKK
COH
HCOH
aqCOHCO
17
c. Calculation of carbonate species in open and closed systems
i. Closed system: logC-pH diagram (for polyprotic acids/bases)
0.01 M H2CO3 Example
CT = [H2CO3*] + [HCO3-] + [CO3
2-] = 10-2
37.6
*32
3 10][[
]][[
COH
HCOH 33.10
3
23 10][
]][[
HCO
COH
2211
0*32
][][1
][
H
KK
H
K
CCCOH
AAA
TT
][
][1
][2
1
13
H
K
K
H
CCHCO
A
A
TT
21
2
2
223
][][1
][
AAA
TT
KK
H
K
H
CCCO
- Diagram is constructed just as for monoprotic acids and bases, except that the slope of line for [CO3
2-] changes from +1 to +2 when the pH drops below pKA1.
+ The last term in the carbonate eq. becomes dominant when [H+] >> KA1
- How do you determine equilibrium pH?
Use proton condition:
[H+] = [OH-] + [HCO3-] + 2[CO3
2-] ; same as charge balance
2211
0*32
][][1
][
H
KK
H
K
CCCOH
AAA
TT
][
][1
][2
1
13
H
K
K
H
CCHCO
A
A
TT
][][
1
][
21
2
2
223
AAA
TT
KK
H
K
H
CCCO
pHpKpKCCO AAT 2log]log[ 2123
18
ii. Open system : Log C- pH diagram for carbonate species in equilibrium with the CO2 gas (0.000316 atm) in the atmosphere
- Other examples: NH3, H2S
[H2CO3*] [CO2(aq)] = Pco2/KH = 10-3.5/31.6 = 10-5 M; independent of pH
From the first dissociation eq.,
[HCO3-] =
log [HCO3-] = 5 – pH + 6.37 = pH - 11.37
Similarly, log [CO32-] = 2pH – 21.7
CT = [H2CO3*] + [HCO3
-] + [CO32-]
Predominance
+ When pH < pKA1, [H2CO3*] predominant
+ When pKA1 < pH < pKA2, [HCO3-] predominant
+ When pH > pKA2, [CO32-] predominant
- CT increases rapidly as pH increases above pKA1
][
10 15
H
K A
Open system – Examples
i. Deionized water
CBE or PCE (Assuming H2CO3* as an initial species)
[H+] = [OH-] + [HCO3-] + 2[CO3
2-]
pH of 5.68 (Clean rain in equilibrium with CO2)
Exact Solution in Example 4.16
ii. 10-5 M NaHCO3
Let’s do charge balance.
[H+] + [Na+] = [OH-] + [HCO3-] + 2[CO3
2-]
[Na+] = 10-5 = [HCO3-]
Homework #4
4.38, 4.41, 4.43, 4.50
4-A. Determine pH and conc. of carbonate species in water containing 10-5 M Na2CO3 in an open system. PCO2 = 10-3.5 atm
19
d. Alkalinity and acidity (p. 549-558)
i. Alkalinity: Capacity of water to neutralize acids
- A form of acid neutralizing capacity (ANC: Example 4.17)
- Titration curve for a hydroxide-carbonate mixture
- Interpretation in most natural waters:
[Alk]tot = [HCO3-] + 2[CO3
2-] + [OH-] – [H+]
+ Net deficiency of protons with respect to CO2 (H2CO3*): proton condition concept
- In sea water we use:
[Alk]tot = [HCO3-] + 2[CO3
2-] + [B(OH)4-] + [HPO4
2-] + [H3SiO4-] + [MgOH-] + [OH-] - [H+]
Measurements of Alkalinity- Measurement by titration with a strong acid (back) to pH about 4.5.
1) Phenolphthalein endpoint: Carbonate alkalinity
H+ + OH- H2O
H+ + CO32- HCO3
-
light pink to colorless at pH about 8.3
2) Methyl orange endpoint: Total alkalinity
H+ + HCO3- H2CO3
Yellow to red at pH about 4.5, where all carbonate are as H2CO3
- Unit: meq/L or mg/L as CaCO3
ii. Acidity - Capacity of water to neutralize bases
- A form of base neutralizing capacity (BNC)
- Interpretation in most natural waters:
[Acy]tot = 2[H2CO3] + [HCO3-] + [H+] - [OH-] : Net excess of protons with respect to CO3
2-
- Measurement by titration with a strong base back to the pH [of a pure CO32- solution] (about 10.7)
20
Example
- 200 mL of a water required 1.1 mL of 0.02 N H2SO4 to titrate to the phenolphthalein endpoint
- 22.9 mL of 0.02 N H2SO4 to titrate it further to the methyl orange endpoint.
- Total and carbonate alkalinity?
Carbonate alkalinity =
Total alkalinity = as CaCO3?
Example 4.17
- A groundwater sample that was carbonated with minerals such as CaCO3(s), MgCO3(s), etc.
- Sampled without exposing the water to atmosphere. No other weak acids or bases.
- Analysis shows that CT,CO3 is 10-3 M and pH = 7.6
a) Acid Neutralizing Capacity? Alkalinity
ANC = Alkalinity = [HCO3-] + 2[CO3
2-] + [OH-] – [H+]
( [HCO3-] & [CO3
2-] can be determined if you know CT and pH)
= CT(1 + 22) + 10-6.4 – 10-7.6
= 10-3(0.943+2(1.76E-3)) + 10-6.4 – 10-7.6
= 9.47*10-4 eq/L = 47 mg/L as CaCO3
L
meq
mLq
meq
L
eqmL 11.0)
200
1)(
1000)(
02.0)(1.1(
L
meq
mLq
meq
L
eqmL 4.2)
200
1)(
1000)(
02.0)(24(
b) pH when GW is brought into equilibrium with the atmosphere (Pco2= 10-3.5 atm)
- Use CBE to develop the master equation.
CB + [H+] = [OH-] + [HCO3-] + 2[CO3
2-] + CA
CB = sum of the strong-base cations (Na+, Ca2+, etc.)
CA = sum of the strong-acid anions (Cl-, NO3-)
CB - CA: Initially present in the water and not volatile; Does not change with exposure to CO2.
CB - CA = [OH-] + [HCO3-] + 2[CO3
2-] – [ H+] = 9.47*10-4 = alkalinity
Alkalinity does not change upon CO2 exposure.
9.47*10-4 = [OH-] + [HCO3-] + 2[CO3
2-] – [H+]
To solve for [H+], obtain expressions for carbonate species that can take into account of the open system. CT changes (No fixed value for CT now) according to pH.
[H2CO3*] = PCO2/KH = 10-5 M; One fixed value. [H2CO3*] = 0 CT,CO3 => CT,CO3 = 10-5/0
9.47*10-4 = 10-14/[H+] -[H+] + 10-5/0(1 +22)
- By trial and error, pH = 8.33.
- pH increased from 7.6 to 8.33. This means CO2 leaves the system to maintain a constant ANC.