Question

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is the final ph? (the pka of acetic acid is 4.7.)

Answer 1

`pH = 13.5`

Step-by-step explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

`\text{HAc} + \text{OH}^(-) \to \text{Ac}^(-) + \text{H}_2\text{O}`

The mixture would contain

• 
  `0.4 * 0.5 - 0.1 * 0.5 = 0.15 \; \text{mol}` of
  `\text{OH}^(-)` and
• 
  `0.1 * 0.5 = 0.05 \; \text{mol}` of
  `\text{Ac}^(-)`

if
`\text{Ac}^(-)` undergoes no hydrolysis; the solution is of volume
`0.1 + 0.4 = 0.5 \; \text{L}` after the mixing. The two species would thus be of concentration
`0.30 \; \text{mol} \cdot \text{L}^(-1)` and
`0.10 \; \text{mol} \cdot \text{L}^(-1)`, respectively.

Construct a RICE table for the hydrolysis of
`\text{Ac}^(-)` under a basic aqueous environment (with a negligible hydronium concentration.)

The question supplied the acid dissociation constant
`pK_a`for acetic acid
`\text{HAc}`; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the base dissociation constant
`pK_b` for its conjugate base,
`\text{Ac}^(-)`. The following relationship relates the two quantities:

`pK_(b) (\text{Ac}^(-)) = pK_(w) - pK_(a)( \text{HAc})`

... where the water self-ionization constant
`pK_w \approx 14` under standard conditions. Thus
`pK_(b) (\text{Ac}^(-)) = 14 - 4.7 = 9.3`. By the definition of
`pK_b`:

`[\text{HAc} (aq)] \cdot [\text{OH}^(-) (aq)] / [\text{Ac}^(-) (aq) ] = K_b = 10^{-pK_(b)}`

`x \cdot (0.3 + x) / (0.1 - x) = 10^(-9.3)`

`x = 1.67 * 10^(-10) \; \text{M} \approx 0 \; \text{M}`

`[\text{OH}^(-)] = 0.30 +x \approx 0.30 \; \text{M}`

`pH = pK_(w) - pOH = 14 + \text{log}_(10)[\text{OH}^(-)] = 14 + \text{log}_(10){0.30} = 13.5`