Chapter 15

Applications of Aqueous Equilibria

The Common-Ion Effect

Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium CH 3 CO 2 H(aq) + H 2 O(l) H 3 O + (aq) + CH 3 CO 2 – (aq)

What are the ions present in the solution

?

The Common-Ion Effect

Le Châtelier’s Principle

CH 3 CO 2 H(aq) + H 2 O(l) H 3 O + (aq) + CH 3 CO 2 – (aq) The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left.

15.3 Buffer Solutions Buffer Solution: A solution which contains a weak acid and its conjugate base and

resists drastic changes in pH

.

Weak acid +

For Example: CH 3 CO 2 H HF + F 1 + CH 3 NH 4 1+ + NH 3 H 2 PO 4 1 + HPO 4 2 CO 2 1 Conjugate base

Example

 Calculate the pH of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution. Is this a buffer solution?

Ka = 1.8 x 10 -5

Example

 Calculate the pH of 0.15M HF and 0.25NaF mixture. Is this a buffer solution?

Example

 Calculate the pH of 100.0 mL DI water  Calculate the new pH after adding 1.0 mL of 0.10M HCl to the above water solution.

Buffer Solutions

 HA(aq) Weak acid + H 2 O(l) H 3 O 1+ (aq) + A 1 (aq) Conjugate base (M + A ) Add a small amount of base ( OH) to a buffer solution ◦ Acid component of solution neutralizes the added base Addition of OH 1 to a buffer: HA(aq) + OH 1 (aq) 100% H 2 O(l) + A 1 (aq)

Buffer Solutions

HA(aq) Weak acid + H 2 O(l) H 3 O 1+ (aq) + A 1 (aq) Conjugate base (M + A )  Add a small amount of acid (H 3 O + ) to a buffer solution ◦ Base component of solution neutralizes the added acid Addition of H 3 O 1+ to a buffer: A 1 (aq) + H 3 O 1+ (aq) 100% H 2 O(l) + HA (aq)

Buffer Solutions

 The addition of – OH or H 3 O + the pH of the solution, but to a buffer solution will change

not as drastically

as the addition of – OH or H 3 O + to a non-buffered solution

Buffer Solutions

Example

 pH of human blood (pH = 7.4) controlled by conjugated acid-base pairs (H 2 CO 3 /HCO 3 ). Write an equation for this buffer mixture then neutralization equation for the following effects ◦ With addition of HCl ◦ With addition of NaOH

Example

 50.0 mL of 0.100 M HCl was added to a .100L buffer consisting of 0.025 moles of sodium acetate and 0.030 moles of acetic acid. What is the pH of the buffer before and after the addition of the acid? Ka of acetic acid is 1.7 x 10 -5 . Assume the volume is constant

Example

 Calculate the pH of 0.100L of a buffer solution that is 0.25M in HF and 0.50 M in NaF. ◦ What is the change in pH on addition of 0.010 moles KOH ◦ Calculate the pH after addition of 0.080 moles HBr •

Assume the volume remains constant

•

Ka = 3.5 x 10 -4

Example

 Calculate the pH of the buffer that results from mixing 60.0mL of 0.250M HCHO 2 and 15.0 mL of 0.500M NaCHO 2 K a = 1.7 x 10 -4 ◦ Calculate the pH after addition of 10.0 mL of 0.150 MHBr. Assume volume is additive

Buffer Capacity

    A measure of amount of acid or base that the solution can absorb without a significant change in pH.

Depends on how many moles of weak acid and conjugated base are present.

For an equal volume of solution:

the more concentrated the solution, the greater buffer capacity

For an equal concentration:

the greater the volume, the greater the buffer capacity

Example

 The following pictures represent solutions that contain a weak acid HA and/ or its sodium salt NaA. (Na + ions and solvent water molecules have been omitted for clarity   Which of the solutions are buffer solution?

Which solution has the greatest buffer capacity?

Example

 What is the maximum amount of acid that can be added to a buffer made by the mixing of 0.35 moles of sodium hydrogen carbonate with 0.50 moles of sodium carbonate? How much base can be added before the pH will begin to show a significant change?

15.4 The Henderson-Hasselbalch Equation

Weak acid Acid(aq) + H 2 O(l)

K

a

= [H 3 O 1+ ][ Base ] [ Acid ] Conjugate base H 3 O 1+ (aq) + Base(aq) [H 3 O 1+ ] = K a [ Acid ] [ Base ] pH = pK a + log [Base] [ Acid ]

Examples

 Calculate the pH of a buffer solution that is 0.50 M in benzoic acid (HC 7 H 5 O 2 ) and 0.150 M in sodium benzoate (NaC 7 H 5 O 2 ). Ka = 6.5 x 10 -5

Example

 How would you prepare a NaHCO 3 -Na 2 CO 3 has pH = 10.40 Ka 2 = 4.7 x 10 -11 buffer solution that

Example

 You prepare a buffer solution of .323 M NH 3 What molarity of (NH 4 ) 2 SO 4 and (NH 4 ) 2 SO 4 . is necessary to have a pH of 8.6? (pK b NH 3 = 4.74)