ch 17: additional aspects of aqueous equilibria brown, lemay ch 17 ap chemistry

Ch 17: Additional Aspects of Aqueous Equilibria

Brown, LeMay Ch 17

AP Chemistry

2

17.1: Common Ion Effect• Addition of a “common ion”: solubility of solids

decrease because of Le Châtelier’s principle.

Ex: AgCl (s) ↔ Ag+ (aq) + Cl- (aq)

Addition of Cl- shifts equilibrium toward solid

3

17.4: Solubility Equilibria• Dissolving & precipitating of salts

Solubility rules discussed earlier are generalized qualitative observations of quantitative experiments.

Ex: PbCl2 (s) ↔ Pb2+ (aq) + 2 Cl- (aq)

Ksp = [Pb2+][Cl-]2 = 1.6 x 10-5

Ksp = solubility-product constant (found in App. D)

Recall that both aqueous ions and solid must be present in solution to achieve equilibriumChanges in pH will affect the solubility of salts composed of a weak acid or weak base ion.

4

• Molar solubility: moles of salt per liter of solution in a saturated solutionEx: What is the molar solubility of Ag2SO4 at 25ºC? What

mass of Ag2SO4 does that represent per liter? Ksp (Ag2SO4) = 1.4 x 10-5; MW Ag2SO4 = 311.80 g

Ag2SO4 (s) ↔ 2 Ag+ (aq) + SO4-2 (aq)

Ksp = [Ag+]2[SO4-2]

  Ag2SO4 (s) Ag+ (aq) SO42- (aq)

Initial

Change

Equilibrium

Ksp = [Ag+]2[SO4-2] = (2x)2(x) = 4x3

x = (1.4 x 10-5/4)1/3 = 0.015 M

(-x)

X+ 2x M + x M

2x M x M

X 0 M 0 M

= [SO42-] = [Ag2SO4]

5

Ex: What mass of Ag2SO4 does that represent per L?

(MW Ag2SO4 = 311.80 g, molar solubility, [Ag2SO4] = 0.015 M)

1mol

g 311.80

L 1

SOAg mol 0.015L 1 42 4.7 g Ag2SO4

Ex: 3.0 x 10-5 g Pb3(AsO4)2 will dissolve in 1.0 L water to make a saturated solution. What is Ksp of Pb3(AsO4)2? (MW of Pb3(AsO4)2 = 899.44 g/mol)

Pb3(AsO4)2 (s) ↔ 3 Pb+2 (aq) + 2 AsO4-3 (aq)

Ksp = [Pb2+]3[AsO4-3]2

[Pb3(AsO4)2] = (3.0 x 10-5 g)/(899.44 g/mol)/(1.0 L)

= 3.3 x 10-8 M Pb3(AsO4)2 = x

Ksp=[Pb2+]3[AsO4-3]2 = [3(3.3 x 10-8)]3 [2(3.3 x 10-8)]2 =

= 4.5 x 10-36

  Pb3(AsO4)2 (s) Pb2+ (aq) AsO43- (aq)

Initial

Change

Equilibrium

(-x)

X+ 3x M + 2x M

3x M 2x M

X 0 M 0 M

7

Ex: What is the molar solubility of CaF2 in a 1.0 L solution of 0.10 M Ca(NO3)2? Ksp (CaF2) = 3.9 x 10-11

CaF2 (s) ↔ Ca2+ (aq) + 2 F- (aq)

  CaF2 (s) Ca2+ (aq) F- (aq)

Initial

Change

Equilibrium

(-x)

X+ x M + 2x M

(0.10 + x) M 2x M22-11 ]][[109.3 FCaK sp

2)2)(10.0( xx211 )2)(10.0(109.3 x Assume x is negligible:

x = 9.9x10-6 moles of CaF2 are soluble per L.

X 0.10 M 0 M

8

Predicting whether a precipitate (ppt) will form depends on concentration of ions present and Ksp of all likely salts. Compare the ion-product, Q, with Ksp

Q has the same mathematical form as Ksp

Q must be calculated for each example

If:Q > Kspprecipitation occurs (continues until

Q = Ksp)

Q = Kspsolution is at equilibrium

Q < Kspsolid dissolves (continues until Q = Ksp)

9

Ex: Will a precipitate form when 0.10 L of 3.0 x 10-3 M Pb(NO3)2 is added to 0.400 L of 5.0 x 10-3 M Na2SO4?

Possible new salts: PbSO4 (Ksp = 1.6 x 10-8)

NaNO3 (Ksp = 112.12)

Q = [Pb2+][SO42-]

[Pb2+] = (3.0 x 10-3 M)(0.10 L) / (0.10 L + 0.400 L)= 3.0 x 10-4 mol / 0.50 L = 6.0 x 10-4 M Pb2+

[SO42-] = (5.0 x 10-3 M)(0.400 L) / (0.10 L + 0.400 L)

= 2.0 x 10-3 mol / 0.50 L = 4.0 x 10-3 M SO42-

Q = [Pb2+][SO42-] = (6.0x10-4)(4.0x10-3) = 2.4 x 10-6

٭ Since Q > Ksp, PbSO4 will precipitate.

10

Ex: Will a precipitate form if 20.0 mL of 3.50 x 10-3 M Hg2(NO3)2 solution are mixed with 40.0 mL of 2.00 x 10-3 M NaCl solution? Ksp

(Hg2Cl2) = 1.3 x 10-18

[Hg22+] = (3.50x10-3 M)(0.0200 L)/(0.0200 L + 0.0400 L)

= 0.00117 M

[Cl-] = (2.00 x 10-3 M)(0.0400 L)/(0.0200 L + 0.0400 L)= 0.00133 M

Q = [Hg22+][Cl-]2

Q = (.00117) (.00133)2 = 2.07 x 10-9

٭ Since Q > Ksp, there will be a precipitate (Hg2Cl2)

11

17.2: Buffers:٭ Solutions that resist drastic changes in pH upon

additions of small amounts of acid or base.Consist of a weak acid and its conjugate base (usually in salt form)Ex: acetic acid and sodium acetate:

HC2H3O2 + NaC2H3O2

Or consist of a weak base and its conjugate acid (usually in salt form)Ex: ammonia and ammonium chloride:

NH3 + NH4Cl

12

Consider a weak acid/salt buffer:HA (aq) ↔ H+ (aq) + A- (aq)

][

]][[

HA

AHKa

][

][][

A

HAKH a

٭ Thus, the pH of a buffer is determined by Ka of weak acid and the ratio of the concentrations of an acid & its conjugate base.

13

Henderson-Hasselbalch equation:

][

][][

A

HAKH a

][

][log]log[

A

HAKH a

][

][loglog]log[

A

HAKH a

][

][log

A

HApKpH a

][

][log

HA

ApKpH a

Use for all buffer

solutions!][

][log

acid

basepKpH a

Lawrence Henderson

(1878-1942)

Karl Hasselbalch(1874-1962)

][

][log

base

acidpKpOH b

14

Why/how do buffers work?٭ A small addition of [H+] will decrease the [A-] but

increase the [HA]:

H+ (aq) + A- (aq) ↔ HA (aq)pH decreases, but pH will

be small

٭ A small addition of [OH-] will decrease the [HA] but increase the [A-]:

OH- (aq) + HA (aq) ↔ H2O (l) + A- (aq) pH increases, but pH will

be small

15

٭ In general, when choosing a buffer, select one in which the acid form has a pKa close to the desired pH.

٭ If [base] > [acid], then pH > pKa

٭ If [base] < [acid], then pH < pKa

٭ If [base] = [acid], then pH = pKa and [H+] = Ka

][

][log

acid

basepKpH a

16

Buffer capacity:٭ Amount of acid or base a buffer can neutralize

before the pH begins to change “significantly”.The greater the concentrations of acid & base in the buffer, the greater the buffer capacity.

A “typical” buffer solution can hold the pH to ± 1.00

٭ Buffer solutions can also be considered common-ion solutions.

17

Ex: Calculate the pH of a buffer made by adding 0.300 mol acetic acid and 0.300 mol sodium acetate to enough water to make 1.00 L. (Ka = 1.8 x 10-5)

CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq)

  CH3COOH (aq) CH3COO- (aq) H+ (aq)

I 0.300 M 0.300 M 0 M

C

E

-x M

(0.300 – x) M

+ x M + x M

(0.300 + x) M x M

][

]][[108.1

3

35-

COOHCH

HCOOCHKa

x

xx

300.0

))(300.0(

xx

300.0

))(300.0(108.1 5

74.4)108.1log( 5 pH

18

Or, use Henderson-Hasselbalch:

][

][log

acid

basepKpH a

]300.0[

]300.0[log)108.1log( 5 pH

74.4)108.1log( 5 pH

• Assumes that x is negligible

19

Ex: Calculate the pH of this buffer after the addition of 0.020 mol NaOH (assume volume remains constant.)

1. Neutralization: calculate the concentrations after the initial reaction of the strong base with the weak acid part of the buffer.

CH3COOH (aq) + OH- (aq) → CH3COO- (aq) + H2O (l)

  CH3COOH (aq) OH- (aq) CH3COO- (aq)

I 0.300 mol 0.020 mol 0.300 mol

C

F

-0.020 mol

0.280 mol

- 0.020 mol + 0.020 mol

0 mol 0.320 mol

Then, consider how these amounts behave in equilibrium!

20

2. Equilibrium:

CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq)

  CH3COOH (aq) CH3COO- (aq) H+ (aq)

I 0.280 mol 0.320 mol 0 mol

C

E (0.280 – x) mol (0.320 + x) mol x mol

-x mol + x mol + x mol

][

]][[108.1

3

35-

COOHCH

HCOOCHKa

x

xx

280.0

))(320.0(

280.0

))(320.0(108.1 5 x

80.4)106.1log( 5 pH

5106.1 x

21

Or, use Henderson-Hasselbalch:

][

][log

acid

basepKpH a

]280.0[

]320.0[log)108.1log( 5 pH

80.40580.074.4 pH

The pH increased, but not by much.

22

Ex: Calculate the pH of the original buffer after the addition of 0.020 mol HCl (assume volume remains constant.)

1. Neutralization: calculate the concentrations after the initial reaction of the strong acid with the weak conjugate base.

CH3COO- (aq) + H+ (aq) → CH3COOH (aq)

  CH3COO- (aq) H+ (aq) CH3COOH (aq)

I 0.300 mol 0.020 mol 0.300 mol

C

F

-0.020 mol

0.280 mol

- 0.020 mol + 0.020 mol

0 mol 0.320 mol

Then, consider how these amounts behave in equilibrium!

23

2. Equilibrium:

CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq)

  CH3COOH (aq) CH3COO- (aq) H+ (aq)

I 0.320 mol 0.280 mol 0 mol

C

E (0.320 – x) mol (0.280 + x) mol x mol

-x mol + x mol + x mol

][

]][[108.1

3

35-

COOHCH

HCOOCHKa

x

xx

320.0

))(280.0(

320.0

))(280.0(108.1 5 x

69.4)101.2log( 5 pH

5101.2 x

24

Or, use Henderson-Hasselbalch:

][

][log

acid

basepKpH a

]320.0[

]280.0[log)108.1log( 5 pH

69.40580.074.4 pH

The pH decreased, but not by much.

25

17.3: Acid-Base TitrationsTitration curve: graph of pH vs. volume of titrant added٭ Each type of titration has a uniquely shaped titration curve

Every titration consists of four regions:1. The initial pH2. Between the initial pH and the equivalence point: pH is

determined by amount of solution not yet neutralized3. The equivalence point: moles of acid = moles of base,

leaving a solution of the salt produced in the acid-base neutralization

4. After the equivalence point: pH is determined by amount of excess titrant

26

Strong Acid – Strong Base TitrationEx: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

14

pH 7

0

0 50Volume of NaOH added (mL)

Region 1: Initial pH• pH starts very low (here, pH = -log 0.100 = 1) because a

strong acid is being titrated.

27

Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl

14

pH 7

0

0 50Volume of NaOH added (mL)

Region 2: Between initial pH and equivalence point• As base is added, pH increases slowly, then rapidly as it

approaches the equivalence point.• Before equivalence point, moles H+ > moles OH- added; pH

is determined by amount of solution not yet neutralized

Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl

14

pH 7

0

0 50Volume of NaOH added (mL)

moles H+ from HCl = moles OH-

added

Region 3: Equivalence point

• Reached when moles H+ = moles OH- added; here, at pH = 7 because only species present are H2O and the salt (Na+1 and Cl-1), which do not hydrolyze to produce an acidic or basic solution, and at V = 50 mL because MAVA

= MBVB = (0.100)(.0500) = (0.100)VB

• Endpoint: point at which an indicator changes color; an approximation of the actual equivalence point

Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl

14

pH 7

0

0 50Volume of NaOH added (mL)

Region 4: After equivalence point

As base is added, pH rises rapidly, then continues to increase slowly.

Moles H+ < moles OH- added

pH eventually approaches an upper limit as base is added; here at pH = 13 (-log [OH-] = 1).

30

pH Indicators:٭ Chemicals with acid and base

forms with significant color differences

For a titration, choose an indicator with a range near the equivalence point

* Methyl red: red (HMer, acid form) to yellow (Mer-, base form) at pH ≈ 4.2 – 6.3

Phenolphthalein: colorless (HPh: acid form above) to pink (Ph-, base form) at pH ≈ 8.5-9.5

31

pH Indicators and Titrations

Phenophthalein is٭often used as an indicator in the titration of a strong acid with a strong base since it turns pink around pH 9, only a small addition of titrant beyond the equivalence point at pH 7.

32

Weak Acid – Strong BaseEx: 0.100 M NaOH (aq) is added to 50.0 mL of 0.100 M

CH3COOH (aq)

Equivalence point: MAVA = MB VB

(0.100 M)(0.0500 L) = (0.100 M)(VB)VB = 0.0500 L NaOH

Equation 1: Neutralization: reaction goes to completionCH3COOH (aq) + OH-(aq) → H2O (l) + CH3COO-(aq)

Equation 2: Equilibrium: extent of reaction based on Ka

CH3COOH (aq) ↔ H+ (aq) + CH3COO- (aq)

33

Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH

Region 1: Initial pH• pH starts below pH 7 because a weak acid is being titrated.

14

pH 7

0

0 50Volume of NaOH added (mL)

34

Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH

Region 2: Between initial pH and equivalence point• Moles CH3COOH > moles OH- added

• pH is determined by considering extent of neutralization and equilibrium reactions

14

pH 7

0

0 50Volume of NaOH added (mL)

35

Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH

Region 3: Equivalence point

• Equivalence point at pH > 7 because only species present are H2O and the salt (Na+ and CH3COO-); acetate ion is a weak base and hydrolyzes to produce a basic solution.

14

pH 7

0

0 50Volume of NaOH added (mL)

moles H+ from CH3COOH = moles OH- added

36

Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH

Region 4: After equivalence point

Moles acid < moles base added

pH is determined by concentration of excess base in solution (as in strong acid/strong base); equilibrium can be ignored

14

pH 7

0

0 50Volume of NaOH added (mL)

37

Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH; calculate pH at regular intervals of titration (Ka = 1.8 x 10-5): (similar to Exercise 17.7, p. 637)

After 30.0 mL of 0.100 M NaOH has been added:mol OH- = (0.100 M)(0.0300 L) = 3.00 x 10-3 mol OH-

mol CH3COOH = (0.100 M)(0.0500 L) = 5.00 x 10-3 mol

CH3COOH (aq) + OH- (aq) → CH3COO- (aq) +H2O

I 5.00 x 10-3 mol 3.00 x 10-3 mol 0

C

F

- 3.00 x 10-3 mol

2.00 x 10-3 mol

- 3.00 x 10-3 mol + 3.00 x 10-3 mol

0 3.00 x 10-3 mol

[CH3COOH] = (2.00 x 10-3 mol)/(0.0800 L) = 0.0250 M

[CH3COO-] = (3.00 x 10-3 mol)/(0.0800 L) = 0.0375 M

38

After 30.0 mL of 0.100 M NaOH has been added:CH3COOH (aq) ↔ H+ (aq) + CH3COO- (aq)

I 0.0250 M 0 0.0375 M

C

E

- x

0.0250 - x

+ x + x

+ x 0.0375 + x

][

]][[108.1

3

35-

COOHCH

HCOOCHKa

x

xx

0250.0

))(0375.0(

0250.0

))(0375.0( x

92.4)102.1log( 5 pH

5102.1 x

92.4)0250.0(

)0375.0(log)108.1log( 5 pH

Or use H-H since its a buffer solution:

39

After 50.0 mL of 0.100 M NaOH has been added:٭ Initial:

mol OH- = (0.100 M)(0.0500 L)= 5.00 x 10-3 mol OH-

mol CH3COOH = (0.100 M)(0.0500 L)= 5.00 x 10-3 mol

CH3COOH (aq) + OH- (aq) → CH3COO- (aq) +H2O

I 5.00 x 10-3 mol 5.00 x 10-3 mol 0

C

F

[CH3COO-] = (5.00 x 10-3 mol)/(0.100 L) = 0.0500 M

- 5.00 x 10-3 mol

0

- 5.00 x 10-3 mol + 5.00 x 10-3 mol

0 5.00 x 10-3 mol

40

After 50.0 mL of 0.100 M NaOH has been added:CH3COO- (aq) + H2O (l) ↔ OH- (aq) + CH3COOH (aq)

I 0.0500 M 0 0

C

E

- x

0.0500 - x

+ x + x

+ x x

][

]][[

3

3

COOCH

OHCOOHCHKb

)0500.0(

))((

x

xx

0500.0106.5

210 x

72.8;28.5 pHpOH

][OHM 102.5 -6 x

5

14

108.1

100.1

a

w

K

K

41

After 70.0 mL of 0.100 M NaOH is added:mol OH- = (0.100 M)(0.0700 L) = 7.00 x 10-3 mol OH-

mol CH3COOH = (0.100 M)(0.0500 L) = 5.00 x 10-3 mol

mol OH- in excess = (7.00 x 10-3 – 5.00 x 10-3)= 2.00 x 10-3 mol OH-

[OH-] = (2.00 x 10-3 mol)/(0.1200 L) = 0.0167 M OH-

pOH = 1.778 and pH = 12.22 (pH + pOH = 14.00)

42

Weak Base – Strong AcidEx: 0.100 M HCl (aq) is added to 40.0 mL of 0.0750 M NH3

(aq)

Equivalence point: MAVA = MB VB

(0.100 M)(VA) = (0.0750 M)(0.0400 L)

VA = 0.0300 L HCl

Equation 1: Neutralization: reaction goes to completion

NH3 (aq) + H+ (aq) → NH4+ (aq)

Equation 2: Equilibrium: extent of reaction based on Kb

NH3 (aq) + H2O (l) ↔ OH- (aq) + NH4+ (aq)

43

Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq)

Region 1: Initial pH• pH starts above pH 7 because a weak base is being titrated.

14

pH 7

0

0 30Volume of HCl added (mL)

44

Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq)

Region 2: Between initial pH and equivalence point

• Moles base > moles acid added

• pH is determined by considering extent of neutralization and equilibrium reactions.

14

pH 7

0

0 30Volume of HCl added (mL)

45

Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq)

Region 3: Equivalence point

• Equivalence point at pH < 7 because only species present are H2O and the salt (NH4

+ and Cl-); ammonium ion is a weak acid and hydrolyzes to produce an acidic solution.

14

pH 7

0

0 30Volume of HCl added (mL)

moles NH3 = moles H+ added

46

Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq)

Region 4: After equivalence point

Moles base < moles acid added

pH is determined by concentration of excess acid in solution (as in strong acid/strong base); equilibrium can be ignored

14

pH 7

0

0 30Volume of HCl added (mL)

47

Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq); calculate pH at regular intervals of titration. (Kb = 1.8 x 10-5)

After 15.0 mL of 0.100 M HCl has been added:mol H+ = (0.100 M)(0.0150 L) = 1.50 x 10-3 mol H+

mol NH3 (aq) = (0.0750 M)(0.0400 L) = 3.00 x 10-3 mol

NH3 (aq) + H+ (aq) → NH4+ (aq)

I 3.00 x 10-3 mol 1.50 x 10-3 mol 0

C

F

- 1.50 x 10-3 mol

1.50 x 10-3 mol

- 1.50 x 10-3 mol + 1.50 x 10-3 mol

0 1.50 x 10-3 mol

[NH3] = (1.50 x 10-3 mol)/(0.0550 L) = 0.0273 M

[NH4+] = (1.50 x 10-3 mol)/(0.0550 L) = 0.0273 M

48

After 15.0 mL of 0.100 M HCl has been added:NH3 (aq) + H2O (l) ↔ OH- (aq) + NH4

+ (aq)

I 0.0273 M 0 0.0273 M

C

E

- x

0.0273 - x

+ x + x

+ x 0.0273 + x

][

]][[108.1

3

45-

NH

OHNHKb

x

xx

0273.0

))(0273.0(

0273.0

))(0273.0( x

74.4)108.1log( 5 pOH5108.1 x

26.9)0273.0(

)0273.0(log)

108.1

100.1log(

5

14

pH

Or use H-H since its a buffer solution:

26.9pH

• Half-equivalence point: where half of base has been neutralized, producing ammonium; can be used to determine Kb (see above)

49

After 30.0 mL of 0.100 M HCl has been added:٭ Initial:

mol H+ = (0.100 M)(0.0300 L) = 3.00 x 10-3 mol H+

mol NH3 = (0.0750 M)(0.0400 L) = 3.00 x 10-3 mol

NH3 (aq) + H+ (aq) → NH4+ (aq)

I 3.00 x 10-3 mol 3.00 x 10-3 mol 0

C

F

[NH4+] = (3.00 x 10-3 mol)/(0.0700 L) = 0.0429 M

- 3.00 x 10-3 mol

0

- 3.00 x 10-3 mol + 3.00 x 10-3 mol

0 3.00 x 10-3 mol

50

After 30.0 mL of 0.100 M HCl has been added:NH4

+ (aq) ↔ H+ (aq) + NH3 (aq)

I 0.0429 M 0 0

C

E

- x

0.0429 - x

+ x + x

+ x x

][

]][[

4

3

NH

HNHKa

)0.0429(

))((

x

xx

0429.0106.5

210 x

31.5pH

][HM 109.4 6 x

5

14

108.1

100.1

b

w

K

K

51

After 60.0 mL of 0.100 M HCl is added:mol H+ = (0.100 M)(0.0600 L) = 6.00 x 10-3 mol H+

mol NH3 = (0.0750 M)(0.0400 L) = 3.00 x 10-3 mol

mol H+ in excess = (6.00 x 10-3 – 3.00 x 10-3)= 3.00 x 10-3 mol H+

[H+] = (3.00 x 10-3 mol)/(0.1000 L) = 0.0300 M H+

pH = 1.523

52

Titration of a Polyprotic Acid:Ex: 0.100 M NaOH (aq) is added to 50.0 mL of 0.100 M H2CO3 (aq)Neutralization #1:

H2CO3(aq) + OH-(aq) → HCO3-(aq) + H2O (l)

Equivalence point #1: MAVA = MB VB

(0.100 M H2CO3)(0.0500 L) = (0.100 M)(VB)

VB = 0.0500 L NaOHNeutralization #2:

HCO3- (aq) + OH- (aq) → CO3

2- (aq) + H2O (l)

Equivalence point #2: MAVA = MB VB

(0.100 M HCO3-)(0.0500 L) = (0.100 M)(VB)

VB = 0.0500 L NaOH (additionally)

53

0.100 M NaOH (aq) is added to 50.0 mL of 0.100 M H2CO3 (aq)

14

pH 7

0

0 50 100Volume of NaOH added (mL)

Eq pt #1: primarily HCO3-; can be

used to calculate orig. [H2CO3] Half-eq. pt #1: mol H2CO3

= mol HCO3-;

can be used to calculate Ka1

Half-eq. pt #2: mol HCO3- = mol CO3

2-; can be used to calculate Ka2

Eq pt #2: primarily CO3

2-; can be used to calculate orig. [H2CO3]