Chapter 16 Aqueous Ionic Equilibrium 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community

Chapter 16Aqueous

IonicEquilibrium

2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

2

The Danger of Antifreeze• each year, thousands of pets and wildlife die

from consuming antifreeze• most brands of antifreeze contain ethylene

glycol sweet taste initial effect drunkenness

• metabolized in the liver to glycolic acid HOCH2COOH

• if present in high enough concentration in the bloodstream, it overwhelms the buffering ability of HCO3

−, causing the blood pH to drop• when the blood pH is low, it ability to carry O2

is compromised acidosis

• the treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol

ethylene glycol(aka 1,2–ethandiol)

Tro, Chemistry: A Molecular Approach 3

Buffers

• 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 the weak acid molecules, HA – these molecules react with added base to neutralize ityou can also think of the H3O+ combining with the OH− to

make H2O; the H3O+ is then replaced by the shifting equilibrium

• the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more 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−


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 the position of equilibrium to the left

• this causes the pH to be higher than the pH of the acid solutionlowering the H3O+ ion concentration


Common Ion Effect


Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?

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

HC2H3O2 + H2O C2H3O2 + 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

OHHC

]OH][OH[C

232

3-232



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 HC2H3O2 = 1.8 x 10-5



Ka for HC2H3O2 = 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



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




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 0.100 0.100 ≈ 0

change -x +x +x


5

5232

3-232

108.1100.0

108.1100.0

OHHC

]OH][OH[C

aK

Ka for HC2H3O2 = 1.8 x 10-5


Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?





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

HF + H2O F + H3O+

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change

equilibrium


[HA] [A-] [H3O+]

initial 0.14 0.071 0

change

equilibrium





+x+xx

0.14 x 0.071 + x x

x

xxKa

14.0

071.0

HF

]OH][[F 3-

HF + H2O F + H3O+


x

xxKa

14.0

071.0

HF

]OH][[F 3-




x 3104.1

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x



0.071 +x

14.0

071.0100.7 4 x

Ka

Ka for HF = 7.0 x 10-4

4

15.3

100.7

1010

a

pKa

K

K a



Ka for HF = 7.0 x 10-4


%5%1%100104.1

104.11

3


x = 1.4 x 10-3

[HA] [A2-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x




x = 1.4 x 10-3


M 14.0104.114.014.0HF 3 x

M 104.1]OH[ 33

x

[HA] [A2-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x


M 072.0104.1071.0071.0]OHC[ 3232 x

0.071 + x x 0.14 x




85.2104.1log

OH-logpH3

3

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x





the values are close enough

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x


4

3

3-

102.714.0

104.1072.0

HF

]OH][[F

aK

Ka for HF = 7.0 x 10-4


Henderson-Hasselbalch Equation• calculating the pH of a buffer solution can be

simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation

• the equation calculates the pH of a buffer from the Ka and initial concentrations of the weak acid and salt of the conjugate baseas long as the “x is small” approximation is valid

initial

initiala acid][weak

anion] base conjugate[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



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

Substitute into the Henderson-Hasselbalch Equation

Check the “x is small” approximation


][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





find the pKa from the given Ka




HF + H2O F + H3O+

][HA

][AlogppH

-

aK

86.2

14.0

0.071log15.3pH

32.86-3

-pH3

104.110]OH[

10]OH[

%5%1%10014.0

104.1 3


Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?

• the Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable

• generally, the “x is small” approximation will work when both of the following are true:

a) the initial concentrations of acid and salt are not very dilute

b) the Ka is fairly small• for most problems, this means that the initial acid and

salt concentrations should be over 1000x larger than the value of Ka


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 parts1. 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 HC2H3O2 and 0.100 mol NaC2H3O2 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

HC2H3O2 + OH− C2H3O2 + H2O

HA A- OH−

mols Before 0.100 0.100 0

mols added - - 0.010

mols After




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

HC2H3O2 + OH− C2H3O2 + H2O

HA A- OH−

mols Before 0.100 0.100 ≈ 0


mols After 0.090 0.110 ≈ 0






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


[HA] [A-] [H3O+]

initial 0.090 0.110 0

change

equilibrium






+x+xx

0.090 x 0.110 + x x

x

xxKa

090.0

110.0

OHHC

]OH][OH[C

232

3-232



x

xxKa

090.0

110.0

OHHC

]OH][OH[C

232

3-232




x 51074.1

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +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



0.010 mol NaOH added to it?Ka for HC2H3O2 = 1.8 x 10-5


%5%016.0%100100.9

1074.12

5


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


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





83.41074.1log

OH-logpH5

3

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x






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 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that

has 0.010 mol NaOH added to it?

find the pKa from the given Ka



547.4

108.1log

logp5

aa KK

Ka for HC2H3O2 = 1.8 x 10-5

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x



Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2

in 1.00 L that has 0.010 mol NaOH added to it?Substitute into the Henderson-Hasselbalch Equation



][HA

][AlogppH

-

aK

83.4

090.0

0.110log547.4pH

%5%016.0%100090.0

1074.1 5

pKa for HC2H3O2 = 4.745

54.83-3

-pH3

1074.110]OH[

10]OH[


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?HC2H3O2 + H2O C2H3O2

+ H3O+

][HA

][AlogppH

-

aK

83.4

090.0

0.110log547.4pH

pKa for HC2H3O2 = 4.745

M 010.0L 1.00

mol 010.0]OH[

00.2100.1log

]OHlog[pOH2

12.00

2.00-14.00

pOH -14.00pH

00.14pOHpH


Basic BuffersB:(aq) + H2O(l) H:B+

(aq) + OH−(aq)

• buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−

H2O(l) + NH3 (aq) NH4+

(aq) + OH−(aq)


Ex 16.4 - What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?

find the pKa of the conjugate acid (NH4

+) from the given Kb

Assume the [B] and [HB+] equilibrium concentrations are the same as the initial



NH3 + H2O NH4+ + OH−

][HB

[B]logppH aK

65.9

20.0

0.50log25.9pH

109.65-3

-pH3

1032.210]OH[

10]OH[

%5%10020.0

1032.2 10

9.254.75-14p -14p

14pp

ba

ba

KK

KK


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

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

53

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 ± 1when 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

2 24.3][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]


Buffering Capacity

a concentrated buffer can neutralize more added acid or base than a dilute buffer


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 completewhen the reaction is complete we have reached the endpoint

of the titration

• an indicator may be added to determine the endpointan 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 solutionequivalence 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

72

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


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 solutioncalculate like a weak acid equilibrium problem

e.g., 15.5 and 15.6

• before the equivalence point, the solution becomes a buffercalculate 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


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


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:HIndwhen 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

91

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

94

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

• for example, the dissociation reaction for PbCl2 isPbCl2(s) Pb2+(aq) + 2 Cl−(aq)

• and its equilibrium constant is Ksp = [Pb2+][Cl−]2



Molar Solubility• solubility is the amount of solute that will dissolve in a

given amount of solutionat 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

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)

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





[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) Pb2+(aq) + 2 Br−(aq)

Ksp = [Pb2+][Br−]2




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 compounds by comparing their Ksps

• in order to compare Ksps, the compounds must have the same dissociation stoichiometry


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

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)addition of Cl− shifts the equilibrium to the left


Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C



[Ca2+] [F−]

Initial 0 0.100

Change +S +2S

Equilibrium S 0.100 + 2S

CaF2(s) Ca2+(aq) + 2 F−(aq)

Ksp = [Ca2+][F−]2


Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C


assume S is small

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

[Ca2+] [F−]

Initial 0 0.100

Change +S +2S

Equilibrium S 0.100 + 2S

Ksp = [Ca2+][F−]2

Ksp = (S)(0.100 + 2S)2

Ksp = (S)(0.100)2

M 1046.1

100.0

1046.1

100.0

8

2

10

2

S

S

SKsp


The Effect of pH on Solubility• for insoluble ionic hydroxides, the higher the pH, the

lower the solubility of the ionic hydroxideand the lower the pH, the higher the solubilityhigher pH = increased [OH−]

M(OH)n(s) Mn+(aq) + nOH−(aq)

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

M2(CO3)n(s) 2 Mn+(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 occurQ = Ksp, the solution is saturated, no precipitationQ < Ksp, the solution is unsaturated, no precipitationQ > 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


Qualitative Analysis

• an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis schemewet chemistry

• a sample containing several ions is subjected to the addition of several precipitating agents

• addition of each reagent causes one of the ions present to precipitate out


Qualitative Analysis

117


Group 1

• group one cations are Ag+, Pb2+, and Hg22+

• all these cations form compounds with Cl− that are insoluble in wateras long as the concentration is large enoughPbCl2 may be borderline

molar solubility of PbCl2 = 1.43 x 10-2 M

• precipitated by the addition of HCl

119

Group 2

• group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+

• all these cations form compounds with HS− and S2− that are insoluble in water at low pH

• precipitated by the addition of H2S in HCl

120

Group 3

• group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides

• all these cations form compounds with S2− that are insoluble in water at high pH

• precipitated by the addition of H2S in NaOH


Group 4

• group four cations are Mg2+, Ca2+, Ba2+

• all these cations form compounds with PO43−

that are insoluble in water at high pH

• precipitated by the addition of (NH4)2HPO4

122

Group 5• group five cations are Na+, K+, NH4

+

• all these cations form compounds that are soluble in water – they do not precipitate

• identified by the color of their flame


Complex Ion Formation• transition metals tend to be good Lewis acids• they often bond to one or more H2O molecules to form

a hydrated ionH2O is the Lewis base, donating electron pairs to form

coordinate covalent bonds

Ag+(aq) + 2 H2O(l) Ag(H2O)2+(aq)

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

+

• the attached ions or molecules are called ligandse.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(NH3)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(NH3)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(NH3)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

128



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


132

43

43

2

107.1])Cu(NH[

]NH][Cu[

fK





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+

131


Solubility of Amphoteric Metal Hydroxides

• many metal hydroxides are insoluble• all metal hydroxides become more soluble in acidic

solutionshifting the equilibrium to the right by removing OH−

• some metal hydroxides also become more soluble in basic solutionacting as a Lewis base forming a complex ion

• substances that behave as both an acid and base are said to be amphoteric

• some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+


Al3+

• Al3+ is hydrated in water to form an acidic solutionAl(H2O)6

3+(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+(aq)

• addition of OH− drives the equilibrium to the right and continues to remove H from the moleculesAl(H2O)5(OH)2+

(aq) + OH−(aq) Al(H2O)4(OH)2

+(aq) + H2O

(l)

Al(H2O)4(OH)2+

(aq) + OH−(aq) Al(H2O)3(OH)3(s) + H2O

(l)


Documents

Chapter 16 Aqueous Ionic Equilibrium 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community