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Tasya [4]
3 years ago
12

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 th

e final ph? (the pka of acetic acid is 4.7.)
Chemistry
1 answer:
Tpy6a [65]3 years ago
6 0

pH = 13.5

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 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol} of \text{OH}^{-} and
  • 0.1 \times 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.)

\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}

The question supplied the <em>acid</em> dissociation constant pK_afor acetic acid \text{HAc}; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> 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 \times 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

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