Chapter 17Chapter 17

Acids, Bases and Buffers

Acids, Bases and Buffers

Overview•strong acid : strong base

•strong acid : weak base

•weak acid : strong base

•weak acid : weak base

•common ion effect

•buffers Henderson-Hasselbalch Equation

•titration curves

•strong acid : strong base at the equivalence point (OH - = H+) the pH = 7

•strong acid : weak base at the equivalence point, pH < 7

•weak acid : strong base at the equivalence point, pH > 7

•weak acid : weak base at the equivalence point, pH depends on the relative value of the Ka and the Kb of the acid and base

example: weak acid : strong base solution

H2O + HCN H3O+ + CN - Ka

H3O+ + OH - 2H2O 1/Kw HCN + OH - H2O + CN -

Ka /Kw

Ka /Kw is large so rxn. will be, essentially, complete

WA SB

example: strong acid : weak base solution

H2O + CN - HCN + OH -

Kb

H3O+ + OH - 2H2O 1/KwCN - + H3O+ H2O + HCN

Kb /Kw

Kb /Kw is large so rxn. will be, essentially, complete

SAWB

example: weak acid : weak base solution

H2O + HCN H3O+ + CN - Ka

H2O + NH3 OH - + NH4

+ Kb

pH depends upon relative magnitude of Ka and Kb

Common Ion Effect:

HCN + H2O H3O+ + CN -

Initial addition of CN - (as NaCN) shifts equilibrium, decreasing H3O+ thereby increasing the pH

How does this affect the pH quantitatively?

HCN + H2O H3O+ + CN - 0.20 M 0 0.10 M initail

-x + x + x change0.20 - x x 0.10 + x equil.

4.9 x 10 -10 = (x)(0.10 + x) 0.20 - x

x = 9.8 x 10 -10 M = [H3O+] = pH = 9.0(vs. 9.9 x 10 - 6 M = pH = 5.0 with no added NaCN)

Buffers:

•Resist change in pH on addition of small amounts of strong acid or base

•Composed of : A weak acid and the salt of its conj. base or A weak base and the salt of its conj. acid

•Most effective when pH is 1 of the pKa

HX H+ + X-

[H+] = Ka [HX] [X-]

• adding OH- causes inc. in X- dec. in HX OH- + HX H2O + X-

• adding H+ causes dec. in X- inc. in HX H+ + X- HX

• as long as amt. of OH- or H+ is small compared to HX & X-, ratio changes little

[H+] = Ka [HX] [X-]

-log [H+] = - log Ka -log [HX] [X-]

pH = pKa - log [HX] [X-]

pH = pKa + log [X- ] [HX]

Henderson-Hasselbalch Equation

Henderson-Hasselbalch Equation:

•Calculates pH of buffer solutions

• Can be used only for buffer solutions

•Can be used only when the equilibrium approximation can be used

•pH = pKa + log ([X -] / [HX]) where HX is the weak acid and X - is its conjugate base

Titration Curves:

•strong acid : strong base

Calculation of pH after addition of aliquots of base

HCl + NaOH H2O + NaCl pH 50 mL 0.10 0 mL 0.10 M 1.0 “ 10 mL 1.2 “ 20 mL 1.4

“ 49 mL 3.0“ 50 mL 7.0“ 55 mL 11.7“ 80 mL 12.4“ 100 mL 12.5

Note: large increase in pH near equivalence pt.

12

10

8

6

4

2

50

Volume of Base, mL

pH Equivalence Pt.

50 mL 0.10 M HCl titrated with 0.10 M NaOH

Titration Curves:

•Weak acid : strong base

Calculation of pH after addition of aliquots of base

HC2H3O2 + NaOH H2O + NaC2H3O2 pH 50 mL 0.10 0 mL 0.10 M 2.9 “ 10 mL 4.1 “ 20 mL 4.6

“ 25 mL half equivalence point [ pH = pKa] 4.7

“ 49 mL 6.4“ 50 mL equivalence point

8.7“ 55 mL 11.7“ 80 mL 12.4“ 100 mL 12.5

Note: large increase in pH near equivalence pt.

12

10

8

6

4

2

50

Volume of Strong Base, mL

pH

Equivalence Pt.

50 mL 0.10 M HC2H3O2 titrated with 0.10 M NaOH

half equivalence point = pKa

25

Strong Base Titrated with Strong Acid

12

10

8

6

4

2

50

Volume of Strong Acid, mL

pH

Equivalence Pt.

50 mL 0.10 M NH3 titrated with 0.10 M HCl

half equivalence point = pKb

25

12

10

8

6

4

2

200

Volume of Base, mL

pH

second equivalence pt.

Diprotic acid, H2C2O4, titrated with 0.10 M NaOH

first equivalence point = pKa

100

Indicators:

Indicators are generally weak acids

HInd + H2O H3O+ + Ind -

Ka Ind = [H3O+][Ind -] [HInd]

[H3O+] = [HInd] this ratio

controls color

Ka [Ind -]

Solubility Equilibria

•Solubility and Solubility Product

•Precipitation of Insoluble Salts Ksp and Q

•Common Ion Effect and Solubility

•Simultaneous Equilibria

•Solubility and pH

•Solubility and Complex Ions

• Separations and Qualitative Analysis

Solubility Product Constant:The equilibrium constant expression for the solution of a solid

AgCl(s) Ag+(aq) + Cl

-(aq)

Ksp = [Ag+][Cl -]

All rules for equilibrium constants and expressions apply

Solubility Product

Examples:

CaF2(s) Ca2+(aq) +

2F -(aq)

Ksp = [Ca2+][F -] 2

Ag3PO4(s) 3Ag+(aq) +

PO43-

(aq)

Ksp = [Ag+] 3[PO43-]

Ca2+

F -

F -

F - F - Ca2+

solid CaF2(s) dissolved Ca2+(aq) & F-

(aq) ions

CaF2(s) Ca2+(aq) + 2F -

(aq)

The quantity of CaF2(s) dissolved is reflected by the quantity of Ca2+

(aq) ions in solutionThe Molar Solubility of CaF2(s) is equal to the [Ca2+] at the eq. point = a saturated solution

Calculate the Ksp from experimental data:Prob: [Ba2+] = 7.5 x 10 -3 M in saturated BaF2. Calculate Ksp.

BaF2(s) Ba2+(aq) + 2F

-(aq)

x 2x

7.5 x 10 -3 2(7.5 x 10 -3)Ksp = (7.5 x 10 -3)(15.0 x 10 -3)2 = 1.6 x 10 -6

Estimating Solubility:

Much like doing equilibrium problems:

Determine the molar solubility of CaCO3 if the Ksp is 3.8 x 10 -9 at 25C

CaCO3(s) Ca2+(aq) +

CO32-

(aq) initial 0 0change +x +xequil. x

x x2 = 3.8 x 10 -9 x = 6.2 x 10 -5 M = [Ca2+] = [CO3

2-]

Prob: Calculate the molar solubility of Mg(OH)2 if the Ksp is 1.5 x 10 -11

Mg(OH)2(s) Mg2+(aq) +

2OH -(aq)

initial 0 0change +x +2xequil. x 2x

(x)(2x)2 = 4x3 = 1.5 x 10 -11 x = 1.6 x 10 -4 M

[Mg2+] = 1.6 x 10 -4 M [OH -] = 3.2 x 10 -4 M

Ksp = [Mg2+][OH -]2 (x) (2x)

Which is more soluble in water?

AgCl or Ag2CrO4

AgCl(s) Ag+(aq) + Cl -

(aq) Ksp = 1.8 x 10 -10

Ag2CrO4(s) 2Ag+(aq) + CrO4 2- (aq) Ksp =

9.0 x 10 -12

molar solubility = 1.3 x 10

-5 M

molar solubility = 1.3 x 10

-4 M

Direct comparisons of Ksp’s can only be used if the ion ratios are the same

Which is more soluble in water?

AgCl Ksp = 1.8 x 10 -10 or

AgCN Ksp = 1.2 x 10 -16

Mg(OH)2 Ksp = 1.5 x 10 -11 or

Ca(OH)2 Ksp = 7.9 x 10 -9

Q, Reaction Quotient

•Q = Ksp system is at equilibrium

•Q > Ksp system not at equilibrium solid forms (rxn. shifts )

•Q < Ksp system not at equilibrium solid dissolves (rxn. shifts )Prob: PbI2(s) (Ksp = 8.7 x 10 -9) placed in solution

where [Pb2+] = 1.1 x 10 -3 M. Is the solution saturated?

PbI2(s) Pb2+(aq) +

2I -(aq)

Q = 5.3 x 10 -9 < Ksp

No, more will dissolve

Concentrations required for precipitation:Prob: What is the minimum conc. of I - that can cause precipitation of PbI2 from a 0.050 M solution of Pb(NO3)2? Ksp (PbI2) = 8.7 x 10 -9.

Ksp = [Pb2+][I -]2

8.7 x 10 -9 = [I -]2

0.050

[I -] = 4.2 x 10 -4 M

How much Pb2+ remains when [I -] = 0.0015 M

[Pb2+] = 8.7 x 10 -9 = 3.9 x 10 -3 M

(0.0015)2

Ksp and Precipitation:

Prob: You have 100.0 mL of 0.0010 M AgNO3. Does AgCl precipitate if you add 5.0 mL of 0.025 HCl?

AgCl(s) Ag+(aq) +

Cl -(aq)

AgNO3(aq) Ag+(aq) + NO3

-(aq)

HCl(aq) + H2O H3O+(aq) + Cl -

(aq)

1.0 x 10 -4 mol 1.25 x 10 -4 mol 0.105 L 0.105 L

9.5 x 10 -4 M 1.2 x 10 -3 M

Q = (9.5 x 10 -4)(1.2 x 10 -3)Q = 1.1 x 10 -6 > Ksp

will precipitate

Solubility and Common Ion Effect:CaF2(s) Ca2+

(aq) + 2F -

(aq)

Adding extra Ca2+ or F - shifts equilibrium

causing a decrease in solubility of CaF2(s)

shift toward solid CaF2

Molar Solubility of CaF2 (no added F -):

CaF2(s) Ca2+(aq) + 2F -

(aq)

initial 0 0change +x +2x

equil. x 2x

x = 2.1 x 10 -4 M = [Ca2+]

Ksp = 3.9 x 10 -11 = (x)(2x)2molar solubility is 2.1 x 10 -4 moles

CaF2 / L

Molar Solubility of CaF2 (with 0.010 M NaF):

CaF2(s) Ca2+(aq) + 2F -

(aq)

initial 0 0.010change +x +2x

equil. x 0.010 + 2x

Ksp = 3.9 x 10 -11 = (x)(0.010 + 2x)2

x = 3.9 x 10 -7 M = [Ca2+]molar solubility is 3.9 x 10 -7 moles CaF2 / L with added NaF which suppresses solubility

What happens if acid is added to CaF2?

CaF2(s) Ca2+(aq) + 2F -

(aq)

H3O+ + F - HF + H2O SA WB

The second rxn. has the effect of removing F - from the first equilibrium, affecting the solubility of CaF2

H+ addition shifts rxn, increasing solubility

complete reaction

Which of the following would be more soluble in acid solution?

PbCl2 CaCO3 Mg(OH)2

the stronger the conj. base the more soluble

if the anion is a hydrolyzing conjugate base, the stronger base it is, the more soluble the salt is in acid solution

Cl - + H2O HCl + OH - K very small

CO32- + H2O HCO3

- + OH - Kb 2.1 x 10 -4

OH - + H2O H2O + OH - K very large

For example:

CaCO3(s) Ca2+(aq) + CO3

2-(aq) Ksp

3.8 x 10 -9

CO32-

(aq) + H2O HCO3- + OH - Kb

2.1 x 10 -4

OH - + H3O+ 2H2O Kw-1 1.0

x 10 14 CaCO3(s) + H3O+ Ca2+ + HCO3

- + H2O

K = 79.8

Which of the following would be more soluble in acid solution?

Ca10(PO4)6(OH)2

Ca10(PO4)6F2

hydroxy apatite - tooth enamel

fluoro apatite - fluoridated tooth enamel

Formation of Complex Ions:•Complex ions are large, polyatomic ions

•Most complex ions have large K values

•Formation of complex ions can affect solubility of some salts

•Kf is the formation constant

•Most transition metals form stable complex ions- the transition metal is a Lewis Acid

Ag+ + 2NH3(aq) Ag(NH3)2+ Kf

1.7 x 10 7

x M 0.20 M (0.010 - x) M

K = 1.7 x 10 7 = [Ag(NH3)2+] = 0.010

[Ag+][NH3]2 (x)(0.20)2

For example:Ex. 17.14 Calc. [Ag+] present at eq. when conc. NH3 added to 0.010 M AgNO3 to give eq. [NH3] = 0.20 M. Neglect vol. change.

x = 1.5 x 10 -8 M = [Ag+]

Selective Precipitation:

Prob: 0.050 M Mg2+ & 0.020 M Cu2+. Which will ppt first as OH - is added?

Mg(OH)2 Mg2+ + 2OH- Ksp = 1.8 x 10 -11

Cu(OH)2 Cu2+ + 2OH- Ksp = 2.2 x 10 -20

[OH-] = (Ksp / [M2+])1/2 Cu(OH)2

What concentration of OH- is necessary?

[OH-] = (Ksp / [Cu2+])1/2

= (2.2 x 10 -20 / 0.020)1/2

= 1.0 x 10 -9 M

[OH-] = (Ksp / [Mg2+])1/2

= (1.8 x 10 -11 / 0.050)1/2

= 1.9 x 10 -5 M

takes less OH-

AgCl(s) Ag+(aq) + Cl -

(aq) Ksp 1.8 x 10 -10

Ag+ + 2NH3(aq) Ag(NH3)2+ Kf

1.7 x 10 7

AgCl(s) + 2NH3(aq) Ag(NH3)2+ +

Cl -

K = 3.1 x 10 -3 = [Ag(NH3)2+][Cl -]

[NH3]2

formation

For example:

Prob: Does 100 mL of 4.0 M aqueous ammonia completely dissolve 0.010 mol of AgCl suspended in 1.0 L of solution?

AgCl(s) + 2NH3(aq) Ag(NH3)2+

+ Cl -

[NH3]2 = (1.0 x 10 -2)(1.0 x 10 -2) = (0.032)1/ 2 = 0.18 M

(3.1 x 10 -3)

we have 0.4 moles of NH3 available which is plenty to provide the 0.020 mol necessary to form the complex plus 0.16 mol to achieve an equilibrium concentration of 0.18 M [NH3]

K = 3.1 x 10 -3 = [Ag(NH3)2+] [Cl - ]

[NH3]2

0.010 moles 0.020 moles 0.010 moles

0.010 moles