A carefully prepared aqueous solution of an unknown weak monoprotic acid () exhibits an experimental — Solutions and Colligative Properties Chemistry Question
Question
A carefully prepared $0.20\text{ molal}$ aqueous solution of an unknown weak monoprotic acid ($HX$) exhibits an experimental degree of ionization ($\alpha$) of exactly $0.20$. Given that the cryoscopic constant ($K_f$) for pure water is strictly $1.86^\circ\text{C kg mol}^{-1}$, calculate the exact theoretical freezing point of this resulting solution.
💡 Solution & Explanation
The weak acid dissociates as $HX \rightleftharpoons H^+ + X^-$, so the maximum number of ions is $n=2$. With $\alpha = 0.20$, the van't Hoff factor is $i = 1 + \alpha(n-1) = 1 + 0.20(2-1) = 1.20$. The depression in freezing point is calculated as $\Delta T_f = i \times K_f \times m = 1.20 \times 1.86 \times 0.20 = 0.4464^\circ\text{C}$. Since pure water freezes at $0^\circ\text{C}$, the new freezing point is $0 - 0.4464 = -0.4464^\circ\text{C}$, which rounds to $-0.446^\circ\text{C}$.