Question

Problem 2.11 : 0.01m aqueous formic acid solution freezes at -0.021 °C. Calculate its degree of dissociation. K$_{f}$ = 1.86 K kg mol$^{-1}$ $\Delta$T$_{f}$ = i K$_{f}$m

Answer 1

0.129

Answer 2

1. Calculate the freezing point depression $\Delta T_f = T_f^\circ - T_f = 0 - (-0.021) = 0.021$ °C = 0.021 K.
2. Use the formula $\Delta T_f = i K_f m$ to find the van't Hoff factor $i$. Substitute the known values: $0.021 = i \times 1.86 \times 0.01$.
3. Solve for $i$: $i = \frac{0.021}{1.86 \times 0.01} = \frac{0.021}{0.0186} = \frac{35}{31}$.
4. For the dissociation of formic acid ($HCOOH \rightleftharpoons HCOO^- + H^+$), the van't Hoff factor is related to the degree of dissociation ($\alpha$) by $i = 1 + \alpha$.
5. Substitute the value of $i$ and solve for $\alpha$: $\frac{35}{31} = 1 + \alpha \implies \alpha = \frac{35}{31} - 1 = \frac{4}{31}$.
6. Convert the fraction to a decimal and round to an appropriate number of significant figures: $\alpha = \frac{4}{31} \approx 0.129$.