Ch. 17 – Other Aspects of Equilibrium

The concept of equilibrium may be used to describe the solubility of salts and the buffering action of a

solution. A buffer is a solution that resists changes in pH. Buffers are found in living organisms to

pharmaceuticals. Some salts are very soluble while some are not. Both of these concepts revolve around

the central ideal of a common-ion.

The Common-Ion Effect

Imagine the you made the following solution and you allowed it to reach equilibrium –

CH3COOH(aq) H+(aq) + CH3COO-

(aq)

Now imagine you added sodium acetate, NaCH3COO, to this system at equilibrium. What would happen to the position of the equilibrium?

Buffers• A buffer is a solution that resists

changes in pH when either an acid or a base is added.

• Buffers consist of either a weak acids with one of its salts, or a weak base with one of its salts.

Buffers• For example, if a solution is make by

dissolving carbonic acid (weak acid) and sodium bicarbonate (salt of the acid) we get the following;

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3

-1(aq)

Buffers

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3

-1(aq)

• If we add a base to this buffered solution, the H3O+ will scoop it up.

H3O+(aq) + OH(aq)

- 2H2O(l)

Buffers

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3

-1(aq)

• If we add an acid to this buffered solution, the HCO3

-1 will scoop it up forming a weak acid.

H+(aq) + HCO3

-1 (aq)

H2CO3(aq)

Buffers

• Write the chemical reaction for the phosphoric acid – dihydrogen phosphate buffer reaction.

Buffers• What is the pH of a buffer that is 0.12 M in lactic

acid (HC3H5O3), and 0.10 M in sodium lactate, NaC3H5O3? Ka for lactic acid = 1.4 x 10-4.

Buffers• Henderson-Hasselbalch equation –

pH = pKa + log [base] [acid]

Buffers• Calculate the pH of a buffer composed of 0.12 M

benzoic acid (HC7H5O2) and 0.20 M sodium benzoate (NaC7H5O2). Ka of benzoic acid = 6.3 x 10-5.

Buffers• Calculate the pH of a buffered solution that was

made by mixing 65.0 mL of 0.20 M NaHCO3 with 75.0 mL of 0.15 M Na2CO3.

Buffers• How many moles of NH4Cl must be added to 2.0 L of

a 0.10 M NH3 to form a buffer whose pH is 9.00? Kb of NH3 = 1.8 x 10-5.

Buffers• Buffer Capacity – The amount of acid or base that

can be added before the pH begins to change considerably.

• The optimal pH of any buffer is when the pH = pKa. It is acceptable to make buffers that have a pH = pKa +/- 1.

• What is the optimal pH buffered by a solution that contains CH3COOH and NaCH3COO (pKa for CH3COOH = 1.8 x 10-5)?

Addition of Strong Acids and Strong Bases to Buffers

• A buffer is made by adding 0.300 mol of CH3COOH and 0.300 mole of NaCH3COO in enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 moles of NaOH is added. Ka of acetic acid = 1.8 x 10-5.

Addition of Strong Acids and Strong Bases to Buffers

• For giggles, what would the the pH if 0.020 moles of NaOH is added to pure water ( a non-buffered solution)?

Titrations

• A titration is the process by which a known concentration of a base is added to an acid (or vice versa) until the equivalence point is reached.

• The equivalence point is when an equal number of moles of the base and acid has reacted.

Titrations

• An indicator is used to signal when the equivalence point has been reached.

• Indicators are selected depending on the pH range of the equivalence point.

Titrations

• Calculate the pH when the following quantities of 0.100 M NaOH is added to 50.0 mL of a 0.100 M HCl solution.• A.) 49.0 mL

• B.) 51.0 mL

Titrations

• Thing become more complicated if a weak acid or a base it titrated with a strong acid or base. Why?

• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of a 0.100 M solution of CH3COOH (Ka = 1.8 x 10-5).

Titrations

• Calculate the pH of a the solution formed when 10.0 mL of 0.050 M NaOH is added to 40.0 mL of 0.0250 M benzoic acid (C6H5COOH), Ka = 6.3 x 10-5.

Titrations

• Titration Curves of Weak Acid / Base with a Strong Acid / Base.

Titrations

• Titration Curves of Polyprotic Acids with a Strong Base.

Solubility Equilibria

• Imagine a saturated solution of CuS;o There is a dynamic equilibrium at the interphase where

dissolved Cu+2 and S-2 ions come in contact with undissolved CuS.

CuS(s) Cu+2(aq) + S-2

(aq)

Ksp = [Cu+2] [S-2]

Solubility Equilibria

• Imagine a saturated solution of CuSO4;

CuS(s) Cu+2(aq) + S-2

(aq)

Ksp = [Cu+2] [S-2] = 6.0 x 10-37

Since the Ksp of CuS is very small, we

can conclude that CuS is not very soluble at all.

Solubility Equilibria

• Write the Solubility Product expression for silver sulfate. Based in its Ksp value, is silver sulfate soluble in water? Ksp Ag2SO4 = 1.5 x 10-5.

Solubility Equilibria

• A saturated solution of silver chromate is made and the concentration of silver ion is determined to be [Ag+1] = 1.3 x 10-4. Calculate the value of Ksp for silver chromate.