Question

A $0.025M$ aqueous solution of a monobasic acid has a freezing point of $-{0.06}^{\circ }C$ .
Calculate acid dissociation constant $\left(K_a\right)$
Given:
$(K_f{)}_{H_{2}O}=1.86Kkgmo{l}^{-1}$
Assume molality = molarity for a dilute solution

• A. $2.96\times {10}^{-3}$
• B. $3.75\times {10}^{-4}$
• C. $4.66\times {10}^{-4}$
• D. $5.1\times {10}^{-3}$

Answer 1

The correct option is A $2.96\times {10}^{-3}$

Since for pure water :
$\Delta T_f^{\circ }=0^{\circ }C$
$(\Delta T_f{)}_{\text{observed}}=(0-(-0.06){)}^{\circ }C={0.06}^{\circ }C$
$(\Delta T_f{)}_{\text{calculated}}=K_f\times m$

$(\Delta T_f{)}_{\text{calculated}}=1.86\times 0.025={0.0465}^{\circ }C.$

$\therefore i=\frac{\text{Observed depression in f.p.}}{\text{Calculated depression in f.p.}}=\frac{0.06}{0.0465}=1.29$

Since $HA$ dissociates into two ions,
$n=2$
$i=1+(n-1)\alpha 1.29=1+(2-1)\times \alpha \alpha =0.29$

$HA\rightleftharpoons H^{+}+A^{-}Initially0.02500Eqb.0.025(1-0.29)(0.025\times 0.29)(0.025\times 0.29)$

$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}{K}_a=\frac{(0.025\times 0.29{)}^2}{0.025\times (1-0.29)}{K}_a=2.96\times {10}^{-3}$