Chapter 16 Aqueous Ionic Equilibria

Chapter 16Aqueous Ionic Equilibria

Chemistry II

Buffers: Solutions that Resist pH Change

• buffers are solutions that resist changes in pH when an acid or base is added

• they act by neutralizing the added acid or base

• but just like everything else, there is a limit to what they can do, eventually the pH changes

• many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion

Making an Acid Buffer

How Acid Buffers WorkHA(aq) + H2O(l) A⇌ −

(aq) + H3O+(aq)

• buffers work by applying Le Châtelier’s Principle to weak acid equilibrium

• buffer solutions contain significant amounts of weak acid molecules, HA – reacts with added base to neutralize it

• the buffer solutions also contain significant amounts of conjugate base anion, A− - combine with added acid to make HA and keep the H3O+ constant

H2O

How Buffers Work

HA ⇌ + H3O+A−A−

AddedH3O+

newHA

HA

H2O

HA

How Buffers Work

HA ⇌ + H3O+

A−

AddedHO−

newA−

A−

pH of a Buffer Solution

• Consider a solution of a weak acid (HA) and its conjugate base (A-):

• Acetic acid/Sodium acetate mixture,

CH3COOH(aq) + H2O(l) H⇌ 3O+(aq) + CH3COO-(aq)

• Ionization of CH3COOH is suppressed by CH3COO- since CH3COO- in this system shifts equilibrium left

8

Common Ion Effect HA(aq) + H2O(l) A⇌ −

(aq) + H3O+(aq)

• adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts equilibrium to the left

• this causes the pH to be higher than the pH of the acid solution

lowering the H3O+ ion concentration

pH of a Buffer Solution

• Use ICE table technique

Ex 16.1 - What is the pH of a buffer that is 0.100 M CH3COOH and 0.100 M CH3COONa?

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

CH3COOH + H2O CH⇌ 3COO- + H3O+

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change

equilibrium

[HA] [A-] [H3O+]

initial 0.100 0.100 0

change

equilibrium


represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+xx

0.100 x 0.100 + x x

x

xxKa

100.0

100.0

COOHCH

]OH][COO[CH

3

3-

3

CH3COOH + H2O CH⇌ 3COO- + H3O+

x

xxKa

100.0

100.0

OHHC

]OH][OH[C

232

3-232

determine the value of Ka

since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq

= [A−]init solve for x

x 5108.1

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 x0.100 x


0.100 +x

100.0

100.0

OHHC

]OH][OH[C

232

3-232 x

Ka

Ka for CH3COOH = 1.8 x 10-5



check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init

%5%018.0%1001000.1

108.11

5

the approximation is valid

x = 1.8 x 10-5

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 x


Alternatively solve (0.100 + x)x = 1.8 x 10-5

0.100 – x

Rearrange: x2 + 0.100018x -1.8 x 10-6 = 0

Use quadratic formula:

x = -b ± √b2 – 4ac = 1.8 x 10-5 and -0.10

2a


x = 1.8 x 10-5

substitute x into the equilibrium concentration definitions and solve

M 100.0108.1100.0100.0OHHC 5232 x

M 108.1]OH[ 53

x

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 1.8E-5

M 100.0108.1100.0100.0]OHC[ 5232 x

0.100 + x x 0.100 x


substitute [H3O+] into the formula for pH and solve

74.4108.1log

OH-logpH5

3

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x


Henderson-Hasselbalch Equation

• simplified buffer pH calculation – use the Henderson-Hasselbalch Equation

• calculates pH of buffer from Ka and initial concentrations of the weak acid (HA) and salt of the conjugate base (A-) as long as the “x is small” approximation is valid

initial

initiala [acid]

base][logp pH K

initial

initiala [HA]

]A[logp pH

K

Deriving the Henderson-Hasselbalch Equation

][A

[HA]]OH[

HA

]OH][[A

-3

3-

a

a

K

K

][A

[HA]log]OHlog[

-3 aK

]Olog[H- pH 3

][A

[HA]loglog]OHlog[

-3 aK

][A

[HA]loglogpH

-aK

aK log- pKa

][A

[HA]logppH

-aK

[HA]

][Alog

][A

[HA]log

[HA]

][AlogppH

-

aK

Ex 16.2 - What is the pH of a buffer that is 0.050 M HC7H5O2 and 0.150 M NaC7H5O2?

Assume [HA]eq = [HA]init and [A−]eq = [A−]init (since Ka is small)

Substitute into the Henderson-Hasselbalch Equation

Check the “x is small” approximation

HC7H5O2 + H2O C⇌ 7H5O2 + H3O+

][HA

][AlogppH

-

aK

050.0

0.150log781.4pH

Ka for HC7H5O2 = 6.5 x 10-5

781.4105.6log

logp5

aa KK

4.66pH

54.66-3

-pH3

102.210]OH[

10]OH[

%5%044.0%100050.0

102.2 5

How Much Does the pH of a Buffer Change When an Acid or Base Is Added?

• though buffers do resist change in pH when acid or base are added to them, their pH does change

• calculating the new pH after adding acid or base requires breaking the problem into 2 parts

1) a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other

added acid reacts with the A− to make more HA added base reacts with the HA to make more A−

2) an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]

Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has

0.010 mol NaOH added to it?

If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for it with A−.

Construct a stoichiometry table for the reaction

CH3COOH + OH− CH⇌ 3COO + H2O

HA A- OH−

mols Before 0.100 0.100 0

mols added - - 0.010

mols After

Part 1: stoichiometry

Original pH = 4.74

Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it?

Fill in the table – tracking the changes in the number of moles for each component


HA A- OH−

mols Before 0.100 0.100 ≈ 0


mols After 0.090 0.110 ≈ 0



using the added chemical as the limiting reactant, determine how the moles of the other chemicals change

add the change to the initial number of moles to find the moles after reaction

divide by the liters of solution to find the new molarities


HA A- OH−

mols Before 0.100 0.100 0.010

mols change

mols End

new Molarity

-0.010-0.010

+0.010

00.1100.090

M 090.0L 1.00

mol 090.0]HA[

0.090

M 110.0L 1.00

mol 100.0]A[ -

0.110



Write the reaction for the ionization of the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0, and using the new molarities of the [HA] and [A−]


[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change

equilibrium

Part 2: Equilibrium

[HA] [A-] [H3O+]

initial 0.090 0.110 0

change

equilibrium


represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+xx

0.090 x 0.110 + x x

x

xxKa

090.0

110.0

OHHC

]OH][OH[C

232

3-232


Part 2: Equilibrium

x

xxKa

090.0

110.0

OHHC

]OH][OH[C

232

3-232

determine the value of Ka

since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq

= [A−]init solve for x

x 51074.1

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.090 0.110 x0.090 x


0.110 +x

090.0

110.0

OHHC

]OH][OH[C

232

3-232 x

Ka

Ka for HC2H3O2 = 1.8 x 10-5

090.0

110.0108.1 5 x

Part 2: Equilibrium


Ka for HCH3COH = 1.8 x 10-5

check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init

%5%016.0%100100.9

1074.12

5

the approximation is valid

x = 1.47 x 10-5

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x



x = 1.47 x 10-5

substitute x into the equilibrium concentration definitions and solve

M 090.01074.1090.0090.0OHHC 5232 x

M 1074.1]OH[ 53

x

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x


M 110.01074.1110.0110.0]OHC[ 5232 x

0.110 + x x 0.090 x

Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has

0.010 mol NaOH added to it?

substitute [H3O+] into the formula for pH and solve

83.41074.1log

OH-logpH5

3

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x


Compared to 4.74 without addition of base


check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

the values match

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x


5

5

232

3-232

108.1090.0

1074.1110.0

OHHC

]OH][OH[C

aK

Ka for HC2H3O2 = 1.8 x 10-5

Ex 16.3 – Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water?

• Compare to adding 0.010 mol NaOH to 1.0 L pure water

[OH-] = 0.010 mol / 1.0 L = 0.010 M

pOH = -log[OH-] = 2.00

pH + pOH = 14.00

pH = 14.00 - pOH = 12.00 vs 4.83 for buffered soln.

• Using Henderson-Hasselbalch equation:


find the pKa from the given Ka

Assume the [HA] and [A-] equilibrium concentrations are the same as the initial

CH3COOH + H2O CH⇌ 3COO + H3O+

547.4

108.1log

logp5

aa KK


[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x



Substitute into the Henderson-Hasselbalch Equation

Check the “x is small” approximation

CH3COOH + H2O CH⇌ 3COO + H3O+

][HA

][AlogppH

-

aK

83.4

090.0

0.110log547.4pH

%5%016.0%100090.0

1074.1 5

pKa for CH3COOH = 4.745

54.83-3

-pH3

1074.110]OH[

10]OH[

Buffering Effectiveness

• a good buffer should be able to neutralize moderate amounts of added acid or base

• however, there is a limit to how much can be added before the pH changes significantly

• the buffering capacity is the amount of acid or base a buffer can neutralize

• the buffering range is the pH range the buffer can be effective

• the effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base

Buffering Effectiveness

• Consider 2 buffers both with pKa = 5.00• Calculate change in pH based on addition of 0.010 mol NaOH for two

different 1.0 L solutions

Both solutions have 0.20 mol total acid and conjugate base

Solution 1: equal amounts (0.10 mols HA and 0.10 mols A-)Solution 2: more acid (0.18 mol HA and 0.020 mol A-)

HA A- OH−



mols After 0.17 0.030 ≈ 0

Effect of Relative Amounts of Acid and Conjugate Base

Buffer 10.100 mol HA & 0.100 mol A-

Initial pH = 5.00


Initial pH = 4.05pKa (HA) = 5.00

][HA

][AlogppH

-

aK

09.5

090.0

0.110log00.5pH

after adding 0.010 mol NaOHpH = 5.09

HA + OH− A⇌ + H2O

HA A- OH−



mols After 0.090 0.110 ≈ 0

25.4

17.0

0.030log00.5pH


%8.1

%1005.00

5.00-5.09

Change %

%0.5

%1004.05

4.05-4.25

Change %

a buffer is most effective with equal concentrations of acid and base

HA A- OH−



mols After 0.49 0.51 ≈ 0

%6.3

%1005.00

5.00-5.18

Change %

HA A- OH−



mols After 0.040 0.060 ≈ 0

Effect of Absolute Concentrations of Acid and Conjugate Base


Initial pH = 5.00


Initial pH = 5.00pKa (HA) = 5.00

][HA

][AlogppH

-

aK

02.5

49.0

0.51log00.5pH


HA + OH− A⇌ + H2O

18.5

040.0

0.060log00.5pH


%4.0

%1005.00

5.00-5.02

Change %

a buffer is most effective when the concentrations of acid and base are largest

Effectiveness of Buffers

• a buffer will be most effective when the [base]:[acid] = 1equal concentrations of acid and base

• effective when 0.1 < [base]:[acid] < 10

• a buffer will be most effective when the [acid] and the [base] are large

Buffering Range• we have said that a buffer will be effective when

0.1 < [base]:[acid] < 10

• substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum pH at which the buffer will be effective

][HA

][AlogppH

-

aK

Lowest pH

1ppH

10.0logppH

a

a

K

KHighest pH

1ppH

10logppH

a

a

K

K

therefore, the effective pH range of a buffer is pKa ± 1

when choosing an acid to make a buffer, choose one whose is pKa is closest to the pH of the buffer

Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25?

Chlorous Acid, HClO2 pKa = 1.95

Nitrous Acid, HNO2 pKa = 3.34

Formic Acid, HCHO2 pKa = 3.74

Hypochlorous Acid, HClO pKa = 7.54

Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium salt to make a

buffer with pH 4.25?

Chlorous Acid, HClO2 pKa = 1.95

Nitrous Acid, HNO2 pKa = 3.34

Formic Acid, HCHO2 pKa = 3.74

Hypochlorous Acid, HClO pKa = 7.54

The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range.

Ex. 16.5b – What ratio of NaCHO2 : HCHO2 would be required to make a buffer with pH 4.25?

Formic Acid, HCHO2, pKa = 3.74

][HA

][AlogppH

-

aK

][HCHO

][CHOlog51.0

][HCHO

][CHOlog74.325.4

2

2

2

224.3

][HCHO

][CHO

1010

2

2

51.0][HCHO

][CHOlog

2

2

to make the buffer with pH 4.25, you would use 3.24 times as much NaCHO2 as HCHO2

Buffering Capacity

• buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness

• the buffering capacity increases with increasing absolute concentration of the buffer components

• as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves

• buffers that need to work mainly with added acid generally have [base] > [acid]• buffers that need to work mainly with added base generally have [acid] >

[base]

Titration• in an acid-base titration, a solution of unknown concentration (titrant)

is slowly added to a solution of known concentration from a burette until the reaction is complete when the reaction is complete we have reached the endpoint of

the titration

• an indicator may be added to determine the endpoint an indicator is a chemical that changes color when the pH

changes

• when the moles of H3O+ = moles of OH−, the titration has reached its equivalence point

Titration

Titration Curve

• a plot of pH vs. amount of added titrant

• the inflection point of the curve is the equivalence point of the titration

• prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH

• the pH of the equivalence point depends on the pH of the salt solution equivalence point of neutral salt, pH = 7 equivalence point of acidic salt, pH < 7 equivalence point of basic salt, pH > 7

• beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH

Titration Curve:Unknown Strong Base Added to Strong Acid

Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)

• initial pH = -log(0.100) = 1.00

• initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10-3

• before equivalence point

NaOH added molesL 1

NaOH mol 0.100NaOH added of L

added 5.0 mL NaOH

NaOH mol 100.5L 1

NaOH mol 0.100NaOH L 0.0050

4

used HCl moles NaOH mol 1

HCl mol 1NaOH mole

5.0 x 10-4 mol NaOH

used HCl mol 100.5NaOH mol 1

HCl mol 1NaOH mol 100.5

4

4

excess HCl mol used HCl mol -HCl mol initial

excess HCl mol102.00

used HCl mol105.0 -HCl mol 102.503-

-4-3

]O[HHCl M NaOH LHCl L

excess HCl mol

3

]O[HHCl M 0.0667 NaOH L 0050.0HCl L 0.0250

HCl mol102.00

3

-3

2.00 x 10-3 mol HCl

]O-log[HpH 3 18.10667.0-logpH

excess NaOH mol105.0

used NaOH mol102.50 -NaOH mol 103.004-

-3-3

NaOH mol 1000.3

L 1


3

][OHNaOH M 0.00909 NaOH L 0300.0HCl L 0.0250

NaOH mol100.5 -4

123

14

3

1001.11009.9

101

]OH[]OH[

wK

][OHNaOH M NaOH LHCl L

excess NaOH mol


• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)

• at equivalence, 0.00 mol HCl and 0.00 mol NaOH

• pH at equivalence = 7.00

• after equivalence point

NaOH added molesL 1


added 30.0 mL NaOH5.0 x 10-4 mol NaOH xs

excess NaOH mol HCl mol initial -added NaOH mol

wK ]][OHOH[ 3

96.111001.1log- pH 9- ]Olog[H- pH 3

added 30.0 mL NaOH0.00050 mol NaOHpH = 11.96


Adding NaOH to HCl

25.0 mL 0.100 M HCl0.00250 mol HClpH = 1.00

added 5.0 mL NaOH0.00200 mol HClpH = 1.18




added 25.0 mL NaOHequivalence pointpH = 7.00



Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH

• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)

• Initial pH:[HCHO2] [CHO2

-] [H3O+]

initial 0.100 0.000 ≈ 0

change -x +x +x

equilibrium 0.100 - x x x

Ka = 1.8 x 10-4

M 1042.4]O[H

100.0100.0108.1

]HCHO[

]O][H[CHO

33

24

2

32

x

x

x

xx

Ka

37.210424.-log

]Olog[H- pH3-

3

%5%2.4%100100.0

102.4 3


• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)

• initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3

• before equivalence added 5.0 mL NaOH

NaOH mol 100.5L 1


4

HA A- OH−

mols Before 2.50E-3 0 0

mols added - - 5.0E-4

mols After 2.00E-3 5.0E-4 ≈ 0

2

2

HCHO mol

CHO mollogppH aK

14.3pH10.002

10.05log74.3pH

5-

4-

74.3108.1-log

log- p4-

aa KK

62

M 107.1][OH

0500.00500.0106.5

]CHO[

]][OH[HCHO

6

211

2

2

x

x

x

xx

Kb

96

14

3

109.5107.1

101

]OH[]OH[

wK

114

14

CHO ,

106.5108.1

101

2

a

wb K

KK

22-

2-2

2-2

-3

CHO M105.00

NaOH L102.50HCHO L102.50

CHO mol102.50

2

2

2

CHO M

NaOH LHCHO L

CHO mol



• initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3

• at equivalence added 25.0 mL NaOH

NaOH mol 1050.2L 1


3

HA A- OH−

mols Before 2.50E-3 0 0

mols added - - 2.50E-3

mols After 0 2.50E-3 ≈ 0

[HCHO2] [CHO2-] [OH−]

initial 0 0.0500 ≈ 0

change +x -x +x

equilibrium x 5.00E-2-x x

CHO2−

(aq) + H2O(l) HCHO2(aq) + OH−(aq)

Kb = 5.6 x 10-11

23.8109.5-log

]Olog[H- pH9-

3

[OH-] = 1.7 x 10-6 M

excess NaOH mol105.0

used NaOH mol102.50 -NaOH mol 103.004-

-3-3

NaOH mol 1000.3L 1


3

][OHNaOH M NaOH LHCl L

excess NaOH mol

][OHNaOH M 0.0091 NaOH L 0300.0HCl L 0.0250

NaOH mol100.5 -4



• after equivalence point

NaOH added molesL 1


added 30.0 mL NaOH

5.0 x 10-4 mol NaOH xs

excess NaOH mol HCHO mol initial -added NaOH mol 2

12

3

14

3

1001.1101.9

101

]OH[]OH[

wK

96.111001.1log- pH 9- ]Olog[H- pH 3

wK ]][OHOH[ 3

Tro, Chemistry: A Molecular Approach

65

added 30.0 mL NaOH0.00050 mol NaOH xspH = 11.96


Adding NaOH to HCHO2

added 12.5 mL NaOH0.00125 mol HCHO2

pH = 3.74 = pKa

half-neutralization

initial HCHO2 solution0.00250 mol HCHO2

pH = 2.37


pH = 3.14


pH = 3.56


pH = 3.92


pH = 4.34


added 25.0 mL NaOHequivalence point0.00250 mol CHO2

−

[CHO2−]init = 0.0500 M

[OH−]eq = 1.7 x 10-6

pH = 8.23


Titrating Weak Acid with a Strong Base

• the initial pH is that of the weak acid solution calculate like a weak acid equilibrium problem

e.g., 15.5 and 15.6• before the equivalence point, the solution becomes a buffer

calculate mol HAinit and mol A−init using reaction stoichiometry

calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−

init

• half-neutralization pH = pKa


70

Titrating Weak Acid with a Strong Base

• at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is establishedmol A− = original mole HA

calculate the volume of added base like Ex 4.8

[A−]init = mol A−/total literscalculate like a weak base equilibrium problem

e.g., 15.14

• beyond equivalence point, the OH is in excess[OH−] = mol MOH xs/total liters[H3O+][OH−]=1 x 10-14

Ex 16.7a – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume

of KOH at the equivalence point

Write an equation for the reaction for B with HA.

Use Stoichiometry to determine the volume of added B

HNO2 + KOH NO2 + H2O

KOH L .02000

KOH mol 0.200

KOH L 1

NO mol 1

KOH mol 1

NO L 1

NO mol 0.100NO L .04000

22

22

L 0400.0mL 1

L 0.001mL 0.40

mL 0.20L 0.001

mL 1L 0200.0

Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after

adding 5.00 mL KOHWrite an equation for the reaction for B with HA.

Determine the moles of HAbefore & moles of added B

Make a stoichiometry table and determine the moles of HA in excess and moles A made


KOH mol 00100.0L 1

KOH mol 200.0

mL 1

L 0.001mL 00.5

HNO2 NO2- OH−

mols Before 0.00400 0 ≈ 0


mols After ≈ 0

22 HNO mol 00400.0

L 1

HNO mol 100.0

mL 1

L 0.001mL 0.40

0.00300 0.00100

Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after

adding 5.00 mL KOH.

Write an equation for the reaction of HA with H2O

Determine Ka and pKa for HA

Use the Henderson-Hasselbalch Equation to determine the pH

HNO2 + H2O NO2 + H3O+

HNO2 NO2- OH−



mols After 0.00300 0.00100 ≈ 0

Table 15.5 Ka = 4.6 x 10-4

15.3106.4loglogp 4 aa KK

2

2

HNO

NOlogppH aK

67.20.00300

00100.0log15.3pH


74

Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence pointWrite an equation for the reaction for B with HA.

Determine the moles of HAbefore & moles of added B

Make a stoichiometry table and determine the moles of HA in excess and moles A made


HNO2 NO2- OH−



mols After ≈ 0

22 HNO mol 00400.0

L 1

HNO mol 100.0

mL 1

L 0.001mL 0.40

0.00200 0.00200

at half-equivalence, moles KOH = ½ mole HNO2

Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence point.

Write an equation for the reaction of HA with H2O

Determine Ka and pKa for HA

Use the Henderson-Hasselbalch Equation to determine the pH

HNO2 + H2O NO2 + H3O+

HNO2 NO2- OH−



mols After 0.00200 0.00200 ≈ 0

Table 15.5 Ka = 4.6 x 10-4

15.3106.4loglogp 4 aa KK

2

2

HNO

NOlogppH aK

15.30.00200

00200.0log15.3pH

Titration Curve of a Weak Base with a Strong Acid

Titration of a Polyprotic Acid

• if Ka1 >> Ka2, there will be two equivalence points in the titrationthe closer the Ka’s are to each other, the less

distinguishable the equivalence points are

titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH

Monitoring pH During a Titration

• the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+]using a probe that specifically measures just H3O+

• the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve

• if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator

Monitoring pH During a Titration

Indicators

• many dyes change color depending on the pH of the solution

• these dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution

HInd(aq) + H2O(l) Ind⇌ (aq) + H3O+

(aq)

• the color of the solution depends on the relative concentrations of Ind:HInd when Ind:HInd ≈ 1, the color will be mix of the colors of Ind and HInd when Ind:HInd > 10, the color will be mix of the colors of Ind

when Ind:HInd < 0.1, the color will be mix of the colors of HInd

81

Phenolphthalein

Methyl Red

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N NH

NaOOC

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N N

NaOOC

H3O+ OH-

Monitoring a Titration with an Indicator

• for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point

• an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pHpKa of HInd ≈ pH at equivalence point

Acid-Base Indicators

Solubility Equilibria

• all ionic compounds dissolve in water to some degree however, many compounds have such low solubility in water that

we classify them as insoluble

• we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water

Solubility Product

• the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp

• for an ionic solid MnXm, the dissociation reaction is:

MnXm(s) ⇌ nMm+(aq) + mXn−(aq)

• the solubility product would be

Ksp = [Mm+]n[Xn−]m

Solubility Product

• e.g., the dissociation reaction for PbCl2 is

PbCl2(s) Pb⇌ 2+(aq) + 2 Cl−(aq)

• and its equilibrium constant (solubility product) is

Ksp = [Pb2+][Cl−]2

Molar Solubility

• solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature

• the molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated solution

• for the general reaction MnXm(s) ⇌ nMm+(aq) + mXn−(aq)

mnmn

sp

mn

K

solubilitymolar

Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C

Write the dissociation reaction and Ksp expression

Create an ICE table defining the change in terms of the solubility of the solid

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S


Ksp = [Pb2+][Cl−]2

Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C

Substitute into the Ksp expression

Find the value of Ksp from Table 16.2, plug into the equation and solve for S

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S

Ksp = [Pb2+][Cl−]2

Ksp = (S)(2S)2

M 1043.14

1017.1

4

4

2

35

3

3

S

SK

SK

sp

sp

Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M


Write the dissociation reaction and Ksp expression

Create an ICE table defining the change in terms of the solubility of the solid

[Pb2+] [Br−]

Initial 0 0

Change +(1.05 x 10-2) +2(1.05 x 10-2)

Equilibrium (1.05 x 10-2) (2.10 x 10-2)

PbBr2(s) Pb⇌ 2+(aq) + 2 Br−(aq)

Ksp = [Pb2+][Br−]2


Substitute into the Ksp expression

plug into the equation and solve

Ksp = [Pb2+][Br−]2

Ksp = (1.05 x 10-2)(2.10 x 10-2)2

[Pb2+] [Br−]

Initial 0 0

Change +(1.05 x 10-2) +2(1.05 x 10-2)

Equilibrium (1.05 x 10-2) (2.10 x 10-2)

6

222

1063.4

1010.21005.1

sp

sp

K

K

Ksp and Relative Solubility

• molar solubility is related to Ksp

• but you cannot always compare solubilities of 2 compounds by comparing their Ksps

• in order to compare Ksps, the compounds must have the same dissociation stoichiometry (same values of n and m)

The Effect of Common Ion on Solubility

• addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt

• for example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2


addition of Cl− shifts the equilibrium to the left

The Effect of pH on Solubility

• for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide and the lower the pH, the higher the solubility higher pH = increased [OH−]

M(OH)n(s) M⇌ n+(aq) + nOH−(aq)

e.g. Mg(OH)2(s) Mg⇌ 2+(aq) + 2OH−(aq)

Shifts left with inc. OH-

The Effect of pH on Solubility

• for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility

M2(CO3)n(s) 2 M⇌ n+(aq) + nCO32−(aq)

H3O+(aq) + CO32− (aq) ⇌ HCO3

− (aq) + H2O(l)

Precipitation• precipitation will occur when the concentrations of the ions exceed the

solubility of the ionic compound

• if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur Q = Ksp, the solution is saturated, no precipitation Q < Ksp, the solution is unsaturated, no precipitation Q > Ksp, the solution would be above saturation, the salt above

saturation will precipitate

• some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions

precipitation occurs if Q > Ksp

a supersaturated solution will precipitate if a seed crystal is added

Selective Precipitation

• a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others

• a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different

Ex 16.13 What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater?

precipitating may just occur when Q = Ksp

22 ]OH][Mg[ Q

6

13

132

109.1059.0

1006.2]OH[

1006.2]OH][059.0[

spKQ

Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M

22 ]OH][Ca[ Q

2

6

62

1060.2011.0

1068.4]OH[

1068.4]OH][011.0[

spKQ

Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M

M 108.4

1060.2

1006.2]Mg[

1006.2]1060.2][Mg[

when

1022

132

13222

spKQ

precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M

22 ]OH][Mg[ Q when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10-10 M

Complex Ion Formation

• transition metals tend to be good Lewis acids

• they often bond to one or more H2O molecules to form a hydrated ion H2O is the Lewis base, donating electron pairs to form coordinate

covalent bonds

Ag+(aq) + 2 H2O(l) Ag(H⇌ 2O)2+(aq)

• ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H2O)2

+

• the attached ions or molecules are called ligands e.g., H2O

Complex Ion Equilibria

• if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand

Ag(H2O)2+

(aq) + 2 NH3(aq) Ag(NH⇌ 3)2+

(aq) + 2 H2O(l)

generally H2O is not included, since its complex ion is always present in aqueous solution

Ag+(aq) + 2 NH3(aq) Ag(NH⇌ 3)2

+(aq)

Formation Constant• the reaction between an ion and ligands to form a complex

ion is called a complex ion formation reaction


+(aq)

• the equilibrium constant for the formation reaction is called the formation constant, Kf

23

23

]NH][[Ag

])[Ag(NH

fK

Formation Constants

Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?

Write the formation reaction and Kf expression.

Look up Kf value

Determine the concentration of ions in the diluted solutions

Cu2+(aq) + 4 NH3(aq) Cu(NH⇌ 3)22+(aq)

134

32

243 107.1

]NH][Cu[

])Cu(NH[

fK

M 107.6L 0.250 L 200.0

L 1mol 101.5

L 200.0]Cu[ 4

-3

2

M 101.1L 0.250 L 200.0

L 1mol 100.2

L 250.0]NH[ 1

-1

3

Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What

is the [Cu2+] at equilibrium?

Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium

[Cu2+] [NH3] [Cu(NH3)22+]

Initial 6.7E-4 0.11 0

Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4

Equilibrium x 0.11 6.7E-4

Cu2+(aq) + 4 NH3(aq) Cu(NH⇌ 3)22+(aq)

132

43

43

2

107.1])Cu(NH[

]NH][Cu[

fK

Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?

Cu2+(aq) + 4 NH3(aq) Cu(NH⇌ 3)22+(aq)

134

32

243 107.1

]NH][Cu[

])Cu(NH[

fK

Substitute in and solve for x

confirm the “x is small” approximation [Cu2+] [NH3] [Cu(NH3)2

2+]

Initial 6.7E-4 0.11 0

Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4

Equilibrium x 0.11 6.7E-4

13413

4

4

413

107.211.0107.1

107.6

11.0

107.6107.1

x

x

since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid

The Effect of Complex Ion Formation on Solubility

• the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands

AgCl(s) Ag⇌ +(aq) + Cl−(aq) Ksp = 1.77 x 10-10


+(aq) Kf = 1.7 x 107

• adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+

Solubility of Amphoteric Metal Hydroxides

• many metal hydroxides are insoluble

• all metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH−

• Act as a base and react with H3O+

e.g. Al(OH)3 dissolved in acid

Al(OH)3(s) + 3H3O+ →Al3+(aq) + 6 H2O(l)

Solubility of Amphoteric Metal Hydroxides

• some metal hydroxides also become more soluble in basic solution acting as a acid

Al(OH)3(s) + OH- →Al(OH4)-(aq)

• substances that behave as both an acid and base are said to be amphoteric (e.g. Cr3+, Zn2+, Pb2+, and Sn2+)

• Therefore Al(OH)3 is soluble at high pH and at low pH but insoluble in a pH-neutral solution

Al3+

• Al3+ is hydrated in water to form an acidic solution

Al(H2O)63+

(aq) + H2O(l) Al(H⇌ 2O)5(OH)2+(aq) + H3O+

(aq)

• addition of OH− drives the equilibrium to the right and continues to remove H from the molecules

Al(H2O)5(OH)2+(aq) + OH−

(aq) Al(H⇌ 2O)4(OH)2+

(aq) + H2O (l)

Al(H2O)4(OH)2+

(aq) + OH−(aq) Al(H⇌ 2O)3(OH)3(s) + H2O

(l)

Al(H2O)3(OH)3(s) + OH−(aq) Al(H⇌ 2O)2(OH)4(s) + H2O

(l)

Where Al(H2O)3(OH)3(s) = Al(OH)3(s)

Chapter 16 Aqueous Ionic Equilibria

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Transcript of Chapter 16 Aqueous Ionic Equilibria