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## Additional Aqueous Equilibria

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### Additional Aqueous Equilibria

Lawrence J. Henderson

1878-1942.

Discovered how acid-base

equilibria are maintained

in nature by carbonic acid/

bicarbonate system in the

blood. Developed buffer

equation.

Karl A. Hasselbalch 1874-1962

Developed logarithmic form of buffer equation.

Use this “decision tree” to calculate pH

values of solutions of specific solutions.

- Is it pure water? If yes, pH = 7.00.
- Is it a strong acid? If yes, pH = -log[HZ]
- Is it a strong base? If yes, pOH = -log[MOH]
- or pOH = -log (2 x [M(OH)2])
- Is it a weak acid? If yes, use the relationship
- Ka = x2/(HZ – x), where x = [H+]
- Is it a weak base? If yes, use the relationship
- Kb = x2/(base – x), where x = [OH-]
- Is it a salt (MZ)? If yes, then decide if it is neutral, acid,
- or base; calculate its K value by the relationship
- KaKb = Kw, where Ka and Kb are for a conjugate system;
- then treat it as a weak acid or base.
- Is it a mixture of a weak acid and its weak conjugate base?
- It is a buffer; use the Buffer Equation.

So far, we’ve looked at solutions of weak acids

and solutions of weak bases.

HX H+ + X-

(described by Ka)

Weak acid equilibrium:

(described by Kb)

Weak base equilibrium:

X- + H2O HX+ OH-

What if you had both HX and X- in the same solution?

This could be obtained by adding some Na+X- salt to a

solution of HX.

The result is called a buffered solution.

A buffered solution resists changes to pH when an acid or base is added.

A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X-).

An example of preparing a buffered solution: Add 0.10 mole of lactic acid (H-Z) and 0.14 mole of sodium lactate (Na-Z) to a liter of solution.

A buffer resists a change in pH when a small amount of OH- or H+ is added.

The reason the buffered solution resists pH change becomes clear when we remember LeChâtlier’s principle.

LeChâtlier says the above equation will (1) shift to the left if we add HCl or

(2) shift to the right if we add NaOH -- thus resisting a change in [H+]

[H+][X-]

+

=

[

H

K

]

=

Ka

a

[X-]

[HX]

Buffered SolutionsWe can predict the pH of a buffered solution:

Rearranging,

Taking –logs of both sides and rearranging:

or

Henderson-Hasselbalch equation

or Buffer equation

What is the pH of a buffered solution of 0.2 M HF and 0.1 M NaF?

(buffer is weak acid HF and its conj. base (salt) F-)

Use:

pKa = -log Ka = -log(6.8 x 10-4) = 3.16

For HF,

So,

One often wants a buffer to have a specific [base]/[acid] ratio.

The H2CO3/HCO3- buffer in blood should be at a pH=7.40.

What must the [base]/[acid] ratio be to obtain this?

pKa = 6.37

For H2CO3 H+ + HCO3-

Ka = 4.3x10-7

Henderson-Hasselbalch:

Rearrange and solve for [base]/[acid]:

Adding strong acids and strong bases to buffers

Assume 1 L of a lactic acid/lactate buffer,

[HLac] = 0.061 M; and [Lac-1] = 0.079 M.

The pKa for lactic acid is 3.85.

The pH is pKa + log[(0.079)/(0.061)] = 3.96

Suppose you add 0.020 mol H+ (strong acid) to this buffer with no change in volume. Then, there will be a neutralization reaction between H+ and Lac-

H+ + Lac- HLac

init.

0.020M

0.061M

0.079M

change

-0.020M -0.020M +0.020 M

0 0.059 M 0.081 M

after rxn

Henderson-Hasselbalch:

Adding strong acids and strong bases to buffers

Assume 1 L of the lactic acid/lactate buffer from earlier slide.

[HLac] = 0.061 M; and [Lac-1] = 0.079 M

The pKa for lactic acid is 3.85.

The pH is pKa + log[(0.079)/(0.061)] = 3.96

Suppose you add 0.020 mol OH- (strong base) to this buffer with no change in volume. Then, there will be a neutralization reaction between OH- and HLac

OH- + HLac Lac-

init.

0.020M

0.079M

0.061M

change

-0.020M -0.020M +0.020 M

0 0.041 M 0.099 M

after rxn

Henderson-Hasselbalch:

0

.

099

=

+

=

+

=

3

.

85

log

3

.

85

(

+0.38

)

4

.

23

0

.

041

Strong Acid-Base Titrations

HCl + NaOH NaCl

Titration curve

buret

(with 0.10 M OH- (aq)

beaker

(with 0.10 M H+ (aq)

& indicator)

V=50.0 mL

HAc + NaOH NaAc

Strong Base – Weak Acid Titration

At the end of the titration,

the pH is determined by

the concentration of

[NaAc] = 0.050 M.

The pH = 8.72

Half-way through the titration,

the pH = pKa = 4.74

Titrations of Polyprotic Acids

- In polyprotic acids, each ionizable proton dissociates in steps.
- In the titration of H2CO3 with NaOH there are two equivalence points:

- one for the formation of HCO3-
- H2CO3 + OH-→ HCO3- + H2O

- one for the formation of CO32-
- HCO3- + OH-→ CO32- + H2O

Solubility-Product Constant, Ksp

- Consider

- for which

= Ksp

- Ksp is the solubility product. (Remember, BaSO4 is ignored because it is a pure solid so its concentration is constant.) The larger the solubility product, the more soluble the salt.

Solubility-Product Constant, Ksp

- In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers.

Examples: for Ag2CO3 Ksp = [Ag+]2[[CO3-2]

for Al2(SO4)3 is Ksp = [Al+3]2[SO4-2]3

- Solubility is the amount (grams) of substance that dissolves to form a saturated solution.
- Molar solubility (s) is the number of moles of solute dissolving to form a liter of saturated solution.

- Problem solving – always “let s = molar solubility”
- Two types of problems:
- Calculate Ksp from solubility
- Calculate solubility from Ksp
- Problem examples:
- (a) Calculate Ksp for PbS, if the solubility = 1.73 x 10-14 M
- Let s = molar solubility = [PbS(aq)] = 1.73 x 10-14
- [Pb+2] = s = 1.73 x 10-14
- [S-2] = s = 1.73 x 10-14
- Ksp = [Pb+2][S-2] = s2 = (1.73 x 10-14)2 = 2.95 x 10-28

(b) Calculate Ksp for Co(OH)2 ; solubility = 6.88 x 10-6 M

Let s = molar solubility = [Co(OH)2(aq)] = 6.88 x 10-6

[Co+2] = s = 6.88 x 10-6

[OH-] = 2s = 2(6.88 x 10-6) = 1.38 x 10-5

Ksp = [Co+2][OH-]2 = (6.88 x 10-6)(1.38 x 10-5)2

= 1.31 x 10-15

[Alternatively, Ksp = 4s3 = 4(6.88 x 10-6)3 = 1.30 x 10-15]

(c) Calculate solubility of CdS; Ksp = 8 x 10-28

Let s = molar solubility (unknown) = [CdS(aq)]

Ksp = [Cd+2][S-2] = (s)(s) = s2 = 8 x 10-28

s = 3 x 10-14

(d) Calculate solubility of CaF2; Ksp = 3.9 x 10-11

Let s = molar solubility (unknown) = [CaF2(aq)]

Ksp = [Ca+2][F-]2 = (s)(2s)2 = 4s3 = 3.9 x 10-11

s = 2.1 x 10-4

Factors That Affect Solubility

Common-Ion Effect

- Solubility is decreased when a common ion is added.
- This is an application of Le Châtelier’s principle:

- As F- (from NaF, say) is added, the equilibrium shifts to the left.
- Therefore, CaF2(s) is formed and precipitation occurs.
- As NaF is added to the system, the solubility of CaF2decreases.

Example problem. What is solubility of AgCl with and without NaCl?

Given: Ksp = 1.8 x 10-10 for AgCl

Let s = solubility of AgCl (unknown) = [AgCl(aq)]

Ksp = 1.8 x 10-10 = s2

Solubility = s = (1.8 x 10-10 )1/2 = 1.35 x 10-5 M (without NaCl)

What is solubility of AgCl in a 0.1 M solution of NaCl (which contains Cl-, a common ion)?

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

x x+0.1

Ksp = [Ag+][Cl-]

1.8 x 10-10

(s)(0.1) =

= (s)(s+0.1)

s = 1.8 x 10-9 M (with NaCl)

How would we increase the solubility of a sparingly soluble salt such as CaF2?

In water, the equilibrium is: CaF2(s) Ca2+(aq) + 2 F- (aq)

LeChatelier says that to increase solubility, we must remove F- ions.

If we add a strong acid, H+, this will cause the reaction:

H+(aq) + F- (aq)

HF (aq)

(Remember, HF is a weak acid)

This reaction removes F- ions thus driving equilibrium in the top equation to the right which means more CaF2 dissolves.

As acid is added, pH decreases

and solubility of CaF2 increases.

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