Chemistry - Ionic Equilibrium Question with Solution | TestHub

Solution (c) Let $x\mathrm{mL}$ of 0.10 M sodium formate is added.

Number of moles in xmL of 0.10 M sodium formate $=\frac{0.10}{1000}\times x$

Number of moles in 50 mL of 0.05 M formic acid $=\frac{0.05}{1000}\times 50$ $\therefore \frac{\text{ [Sodium formate] }}{\text{ [Formic acid] }}=\frac{\frac{0.10\times x}{1000}}{\frac{0.05\times 50}{1000}}=0.04\times$ From Henderson's equation,

$$\begin{gathered} \text{displaylines}\mathrm{pH}=\mathrm{p}{K}_{\mathrm{a}}+\log \frac{[salt]}{[acid]} \\ 4.0=3.80+{\log }_{}0.04x \\ {\log }_{}0.04x=0.2 \\ x=39.6ml \end{gathered}$$