CH. 17ACID -- BASE EQUILIBRA

& BUFFERS 17.1Common ion effect

17.1 CalculationHenderson- Hasselbalch eqnBuffers how works ion effect pH range

17.3Titration Curves indicators SA/SB - SA/WB WA/SB - Poly

Calculation pH

17.4SolubilityCalculation Ksp (Qsp) precipation

17.5Solubility; pHComplex Ions

17.6 & .7Ion GroupsSeparation

COMMON ION EFFECT

The shift in the position of an equilibrium on addition of a subst that provides an ion in common w/ one of the ions already involved in the equilibrium process.Also, decr solubility of soluble aqueous portion of ionic cmpd

Dissolve HNO2 in NaNO2 soln (H2O)

LeChatilier

add NO2-; from NaNO2

)aq(-12(aq)3 )l(2)aq(2 NO OH OH HNO

H3O+ + NO2-

H3O+ lowers pH; shifts away NO2- already formed

BUFFERSresist es in pHcomposed of: WA & CB or WB & CA

Add sm amt base to buffer soln Acidic component neutralize base added

BIO SYS: blood pH 7.4 control buffer of H2O/HCO3

-

Add sm amt acid to buffer soln Basic component neutralize acid added

Fig. 16.7 pg 664

HENDERSON -- HASSELBALCH

To prep buffer soln select WA w/ pKa close to 1 pH unit of buffer soln

How pH affects % dissoc WApH of buffer soln not depend on soln vol but on pKa & molar amt WA-CB

EX. Titrant 0.100 M NaOH and neutralize 40.0 ml of 0.100 M butanoic acid (HC4H7O2) Ka = 1.54*10-4

Find pH add 20.00 ml of titrant Find pH at equivalence pt & determine best indicator to use

Acid

BaseLog pK pH a

Base

Acid * K ]OH[ a3

Base

AcidLog- LogK- ]OH[Log- a3

1st determine # mmol HC4H7O2 present

mol HC4H7O2 = 40.0 mL * (0.100 mmol/1 mL) = 4.00 mmol need 4.00 mmol NaOH to reach equiv pt means need 40.0 ml of 0.100 M NaOH

b) Add 20.00 ml NaOH, also midpoint so pH = pKa moles WA: 4.00 mmol mol NaOH added: 20.00 ml*0.100 M= 2.00 mmol

2.0 mmol/60 ml = 0.0033 Mratio of [HC4H7O2]/[C4H7O2

-] = 0.033/0.033 = 1

a) No base, NaOH, added; butanoic WA: Ka = x2/[HA] x2 = (1.54*10-4)*(0.100) = .00392 pH = -Log(0.00392) = 2.41

3.81 0.00 3.81

033.0

0.033Log pK pH a

c) Add 40.0 ml NaOH @ equiv pt

All HC4H7O2 neutralized, find [C4H7O2-]

[C4H7O2-] = mmol/tot mL

pH = -Log(5.56*10-9) = 8.25

indicator: PHENOLPHTHALEIN

mmol C4H7O2- = 40.00 ml * (0.100 mmol/ml) = 4.0 mmol

[C4H7O2-] = 4.0 mmol/80.0 ml = 0.05 M

9-11

14-

-274b

w

11-4

14-

-274

Wb

10*5.56 )05.0)(10*49.6(

10*1

]OHC[*K

K ]H[

10*6.49 10*54.1

10*1

]OH[C

K K

soln basic anion WA,

TITRATION CURVESMonoprotic Acids

pH

Vol SB

7

SA - SB

TITRATION CURVESMonoprotic Acids

pH

Vol SB

9

WA - SB

TITRATION CURVESMonoprotic Acids

pH

Vol SA

3.5

SA - WB

TITRATION CURVESPolyprotic Acids

pH

Vol Base

H2SO4

HSO4- pKa1

SO4-2 pKa2

Problem25.0-mL of 0.145 M HCl soln is titrated by 0.200 M KOH. How many mL of base must be added to reach the end point.

Want: mL KOH Data: 25.0-mL HCl @ 0.145 mol/L 0.200 mol/L KOH HCl + KOH --- H2O + KCl 1 mol HCl = 1 mol KOH

mol 0.200

L *

HCl mol 1

KOH mol 1 *

L

mol 0.145*HCl L 0.025 KOH mL-X

KOH mL-18.1 L

mL 1000 *L 018125.0

ProblemWhat is the pH at each point in the titration of 25.00-mL of 0.100 M HAcwith 0.100 M NaOH? a) before NaOH added b) after 10.00-mL c) after 12.50-mL d) after 25.00-mL (these are volumes added to original 25 ml)

a. Find initial [H3O+]; use Ka (1.8*10-5) HC2H3O2 + H2O - H3O+ + C2H3O2

-

Eq: 0.100-x x x

)10*8(0.100)(1. ][H x x-0.100

x K 5-

2

a = 1.3*10-3 pH = -log(1.3*10-3) = 2.89

b. 35-mL tot vol; amt neutralize= 0.025 L*(0.100 mol/L) = 0.0025 mol base added = 0.01 L*(0.100 M) = 0.001 mol

HC2H3O2 + OH - H2O + C2H3O2-

I 0.0025 0

C -0.001 +0.001

E 0.0015 0.001Convert to M using vol: .0015/.035 = .0429 M .001/.035 = .0286 M

4.56 0.18 - 4.74 0429.0

0286.0log 4.74

[HA]

][Alog pK pH

-

a

c. 37.5-mL tot vol; NaOH added=0.0125 L*(0.100 mol/L) = 0.00125 mol

HC2H3O2 + OH - H2O + C2H3O2-

I 0.0025 0

C -0.00125 +0.00125

E 0.00125 0.00125Convert to M using vol: .00125/.0375 = .0333 M .00125/.0375 = .0333 M

4.74 0.00 - 4.74 0333.0

0333.0log 4.74

[HA]

][Alog pK pH

-

a

d. Add 25-mL at eq.pt, neutralization complete; 50.00-mL tot vol amt C2H3O2

- present =0.0025 mol/0.050 L = 0.05 M

C2H3O2- + H2O - HC2H3O2 + OH-

Eq: 0.05-x x x Kb = Kw/Ka = 5.6*10-10

Kb = x2/0.05 x = 2.53*10-6 pOH = 5.28 then pH = 8.72

e. Follow up. What happens beyond the equivalence pt? pH is determined from excess base present. - add 26.0 mL, what is the pH? tot vol 51.00 mL

(0.026 L)*(0.100 M) = 0.0026 mol 0.0026 – 0.0025 = 0.0001 mol OH- excess

M = 0.0001/.051 = 1.96*10-3

11.3 2.7 - 14.0 pH 2.7 )10*log(1.96- pOH -3

Problem25.0-mL of 0.145 M HCl soln is titrated by 0.200 M KOH. How many mL of base must be added to reach the end point.

Want: mL KOH Data: 25.0-mL HCl @ 0.145 mol/L 0.200 mol/L KOH HCl + KOH --- H2O + KCl 1 mol HCl = 1 mol KOH

mol 0.200

L *

HCl mol 1

KOH mol 1 *

L

mol 0.145*HCl L 0.025 KOH mL-X

KOH mL-18.1 L

mL 1000 *L 018125.0

ProblemWhat is the pH at each point in the titration of 25.00-mL of 0.100 M HAcwith 0.100 M NaOH? a) before NaOH added b) after 10.00-mL c) after 12.50-mL d) after 25.00-mL (these are volumes added to original 25 ml)

a. Find initial [H3O+]; use Ka (1.8*10-5) HC2H3O2 + H2O - H3O+ + C2H3O2

-

Eq: 0.100-x x x

)10*8(0.100)(1. ][H x x-0.100

x K 5-

2

a = 1.3*10-3 pH = -log(1.3*10-3) = 2.89

b. 35-mL tot vol; amt neutralize=0.025L*(0.100 mol/L) = 0.0025 mol base added = 0.01 L*(0.100 M) = 0.001 mol

HC2H3O2 + OH - H2O + C2H3O2-

I 0.0025 0

C -0.001 +0.001

E 0.0015 0.001Convert to M using vol: .0015/.035 = .0429 M .001/.035 = .0286 M

4.56 0.18 - 4.74 0429.0

0286.0log 4.74

[HA]

][Alog pK pH

-

a

c. 37.5-mL tot vol; NaOH added=0.0125 L*(0.100 mol/L) = 0.00125 mol

HC2H3O2 + OH - H2O + C2H3O2-

I 0.0025 0

C -0.00125 +0.00125

E 0.00125 0.00125Convert to M using vol: .00125/.0375 = .0333 M .00125/.0375 = .0333 M

4.74 0.00 - 4.74 0333.0

0333.0log 4.74

[HA]

][Alog pK pH

-

a

d. Add 25-mL at eq.pt, neutralization complete; 50.00-mL tot vol amt C2H3O2

- present =0.0025 mol/0.050 L = 0.05 M

C2H3O2- + H2O - HC2H3O2 + OH-

Eq: 0.05-x x x Kb = Kw/Ka = 5.6*10-10

Kb = x2/0.05 x = 2.53*10-6 pOH = 5.28 then pH = 8.72

SOLUBILITY EQUILIBRA

Principals: examines quantitative aspects of solubility & ppt process

Ksp = solubility constant Cr2(SO4)3(s) --------> 2 Cr+3(aq) + 3 SO4-2(aq)

solid342

324

23

c)SO(Cr

SOCr Q

Qc + solid = Ksp

324

23sp SOCr K

Ksp

1. temp dependant 2. used to calc solubility of cmpd

Comparing Ksp

cmpds w/ same total # ions formed, then, higher Ksp is greater the solubility

Write dissoc eqn, # mols each ionWrite Ksp expressionFind Molarity - convert solubility to molar

CALCULATIONS Find Ksp

Solubility of silver dichromate @ 15oC is 8.3*10-3 g/100 mL

272

21sp OCrAg K Ag2Cr2O7(s) --------> 2 Ag+1(aq) + Cr2O7

-2(aq)

[Cr207-2] = 2 [Ag+]

0.0192 = 2*0.0192 = 0.0384

Find Solubility

Ksp = [0.0384]2 [0.0192] = 2.83*10-5

What is the molar solubility of iron III hydroxide in water? Ksp = 6.3*10-10

Write dissoc eqn, ion expression, ICE table

M 0.0192 g 431.8

mol 1*

L 1

mL 1000*

mL100

g 0.0083 M

Ksp = [Fe+3] [OH-]3

Ksp = [x] [3x]3 = 27x4

Ksp = 27x4

Fe(OH)3(s) --------> Fe+3(aq) + 3 OH-1(aq)

I ------ 0 0C -x +x +3xE -x x 3x

M10*2.20 27

10*6.3

27

K x 3-4

-104

sp

Ignore: solid

PPT FORMATION -- will or not form ???

Qsp = Ksp soln satur, no Qsp< Ksp soln unsat, no pptQsp > Ksp ppt form till soln satur

Will ppt form, if so, what is it?0.20 L of 0.050 M Na3PO4 & 0.10 L of 0.20 M Ca(NO3)2

Ions present: Na+1 PO4-3 Ca+2 NO3

-1

solubility rules: Ca3(PO4)2 (s) the ppt, insoluble

Ca3(PO4)2 (s) <----> 3 Ca+2 (aq) +2 PO4-3 (aq)

Ksp = [Ca+2]3[PO4-3]2

Ksp = 1.2*10-29

Qsp = [Ca+2]3 [PO4-3]2

= (0.067)3(0.033)2 = 3.28*10-7

mols Ca+2 = 0.10 L * 0.20 M = 0.020 mols[Ca+2] = 0.020 mols/0.30 L = 0.067 M

Qsp > Ksp, so ppt will form Ca3(PO4)2 ppt till Qsp = 1.2*10-29

mols PO4-3 = 0.20 L * 0.05 M = 0.10 mols

[PO4-3] = 0.10 mols/0.30 L = 0.033 M

If Qsp = 6.7*10-35 What then??? Qsp < Ksp, no ppt

COMPLEX IONS

Central METAL ion covalent

bonded to 2+ LIGANDS

Metal ion acts as Lewis Acid,Ligand acts as Base

Can separate metal from its ore, isolate & remove toxic metal, convert metal ion to diff form

ION GROUPS

Separation of 2 competing ppting metal ions thru diff properties of solubility of cmpds

Add ppting ion till Qsp of more soluble ion close to its Ksp

This allows max amt of less soluble cmpd ppt out, while none of the more soluble

Mixture of many diff metal ions, use various techniques to sep ions into characteristic groups pg 737

Group 1 Insol Chloridesppt w/ HCl Ag+ Hg2

+2 Pb+2

Group 2 Acid-insol Sulfidesppt w/ H2SCu+2 Cd+2 As+3 Sb+3 Bi+3

Sn+2 Sn+4 Hg+2 Pb+2

Group 3 Base-insol S-2 & OH-1

ppt w/ NH4+-NH3 buffer

Zn+2 Mn+2 Ni+2 Fe+2 Co+2 as S-2

Al+3 Cr+3 as OH-

Group 4 Insol PO4-3

ppt w/ (NH4)2HPO4, NaCO3

Mg+2 Ca+2 Ba+2

Group 5 Alkali & NH4+ Ions

generally speaking, what’s left Use various techniques to ppt out specific ions from each group

Calculate the pH of a soln that is 0.350 M pyridinium chloride (C5H5NHCl) and 0.210 M pyridine (C5H5N).

WB C5H5N & C5H5NHCl contain common ion effect: C5H5NH+

I 0.210 0.350 0.0 C -x +x +x E 0.210-x 0.350+x x

C5H5N(aq) ) + H2O(l) C5H5NH+(aq) + OH-(aq)

assume 5% rule for WA

0.210

0.350x 10*1.7

210.0

x)(x)(0.350

N]H[C

]NHH][C[OH 9-

55

55

x = [OH-] = 1.02*10-9 pOH = -Log(1.02*10-9) = 8.99 pH = 14 – 8.99 = 5.01

Kb = 1.7*10-9

The addition of bromide ion will decrease the water solubility of which of the following salts?

a.BaSO4 b. Li2CO3 c. PbS d. AgBrAgBr

Which pair of compounds will form a buffer solution when dissolved in water in equimolar amounts?

a. HCl and KCl b. HNO3 and NaNO3

c. HCl and NH4Cl d. NH3 and NH4Cl

NH3 and NH4Cl

The Ka of HCN is 4.9 x 10-10. What is the pH of a buffer solution that is 0.100 M in both HCN and KCN?

a. 4.7 b. 7.0 c. 9.3 d. 14.09.3 same concentrations,-log(4.9*10-10)

The Ka of HCN is 4.9 x 10-10. What is the pH of a buffer solution that is 0.100 M in HCN and 0.200 M in KCN?

a. 7.0 b. 9.0 c. 9.3 d. 9.69.6

pH = pKa + log[KCN]/[HCN]

The Ka of HCN is 4.9 x 10-10. What is the pH of a buffer solution that is 1.00 M in HCN and 0.100 M in KCN?

a. 7.0 b. 8.3 c. 9.0 d. 9.38.3 pH = pKa + log[KCN]/[HCN]

In titrating a weak base with a strong acid, the best indicator to use would be:

a. methyl red (changes color at pH = 5).

b. bromothymol blue (changes at pH = 7).

c. phenolphthalein (changes at pH = 9).

d. none of the above. methyl red (changes color at pH = 5)

Titrating a weak acid with a strong base, best indicator to use would be:a. methyl red (changes color at pH = 5).

b. bromothymol blue (changes at pH = 7).

c. phenolphthalein (changes at pH = 9).

d. none of the above.

phenolphthalein (changes at pH = 9)

The Ksp of BaCO3 is 5.0 x 10-9. What is the concentration of barium ion in a saturated aqueous solution of BaCO3?

a. 7.1 x 10-5 M b. 2.5 x 10-9 M c. 5.0 x 10-9 M d. 1.0 x 10-8 M

7.1 x 10-5 M 5.0*10-9 = x*x5.0*10-9 = x2

The Ksp of BaF2 is 1.7 x 10-6. What is the concentration of barium ion in a saturated aqueous solution of BaF2?

a. 1.7 x 10-6 M b. 3.4 x 10-6 M c. 7.5 x 10-3 M d. 1.5 x 10-2 M

7.5 x 10-3 M

Ksp = [Ba+2][F-]2

= (x)*(2x)2

1.7*10-6 = 4x3

The Ksp of BaF2 is 1.7 x 10-6. What is the concentration of fluoride ion in a saturated aqueous solution of BaF2?

a. 1.7 x 10-6 M b. 5.7 x 10-5 M c. 7.6 x 10-3 M d. 1.5 x 10-2 M

1.5 x 10-2 M

BaF2 -- Ba+2 + 2 F-

[F-] =2x = 2(7.5*10-3)

Which of the following substances will be more soluble in acidic solution than in basic solution:(a)Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d) AgCl(s)? So, which is more soluble at low pH than high pH

Ni(OH)2 more soluble in acidic, basicity of OH-

CaCO3 dissolves in acidic, as CO3- basic anion

BaF2 dissolves, as F- basic anion

(d) The solubility of AgCl is unaffected by changes in pH because Cl– is the anion of a strong acid and therefore has negligible basicity.

Addition of _____ will increase the solubility of MgCO3.

MgCO3(s) Mg2+(aq) + CO32–(aq)

• MgCl2

• Na2CO3

• NaOH

• HCl

• KHCO3

HCl

Predict the order of precipitation of Ba2+, Pb2+, Ca2+ with the addition of NaF.

• Ca2+ then Ba2+ then Pb2+

• Pb2+ then Ba2+ then Ca2+

• Pb2+ then Ca2+ then Ba2+

• Ca2+ then Pb2+ then Ba2+

• Ba2+ then Pb2+ then Ca2+

Ksp

BaF2 1.7 x 10–6

PbF2 3.6 x 10–8

CaF2 3.9 x 10–11

Ca2+ then Pb2+ then Ba2+

Predict the order of precipitation of Cl–, CO32–, Br– upon the

addition of AgNO3.

• Br– then CO32– then Cl–

• Br– then Cl– then CO32–

• Cl– then CO32– then Br–

• Cl– then Br– then CO32–

• CO32– then Cl– then Br–

Ksp

AgCl 1.8 x 10–10

Ag2CO3 8.1 x 10–12

AgBr 5.0 x 10–13

Br– then Cl– then CO32–

A solution contains 1.0 × 10-2 M Ag+ and 2.0 ×10-2 M Pb2+. When Cl– is added to the solution, both AgCl (Ksp = 1.8× 10-10) and PbCl2 (Ksp = 1.7×10-5) precipitate from the solution. What concentration of Cl– is necessary to begin the precipitation of each salt? Which salt precipitates first?

We are given Ksp values for the two possible precipitates. Using these and the metal ion concentrations, we can calculate what concentration of Cl– ion would be necessary to begin precipitation of each. The salt requiring the lower Cl– ion concentration will precipitate first.