The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons [tex]\text{H}^{+}[/tex] than hydroxide ions [tex]\text{OH}^{-}[/tex]. The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.
Assuming that [tex]x \; \text{L}[/tex] of the 0.307 [tex]\text{mol} \cdot \text{dm}^{-3}[/tex] sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, [tex]x[/tex] is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain [tex]0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x[/tex] moles of acetic acid and [tex]0.307 \; x[/tex] moles of acetate ions.
Let [tex]\text{HAc}[/tex] denotes an acetic acid molecule and [tex]\text{Ac}^{-}[/tex] denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.
[tex]\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}[/tex]
The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of [tex]10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}[/tex]. There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of [tex]10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} [/tex] in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by [tex]10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}[/tex] would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:
[tex]\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}[/tex]
By definition:
[tex]\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)[/tex]
The question states that
[tex]\text{K}_{a} = 1.75 \times 10^{-5}[/tex]
such that
[tex]10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241[/tex]
Thus it takes [tex]0.0241 \; \text{L}[/tex] of sodium hydroxide to produce this buffer solution.
