Question

A 0.025 M solution of monobasic acid has a freezing point of $-{0.60}^{o}C$. $K_f$ of $H_{2}O=$ $1.86Kmolalit{y}^{-1}$. Assume molality equal to molarity. $K_a$ for the acid is:

• A. $2.96\times {10}^{-3}$
• B. $1.88\times {10}^{-3}$
• C. $4.76\times {10}^{-3}$
• D. None of these

Answer 1

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

For monobasic acid HA
$\underset{\underset{1-\alpha }{1}}{HA}\rightleftharpoons \underset{\underset{\alpha }{0}}{H^{+}}+\underset{\underset{\alpha }{0}}{A^{-}}$
According to Ostwald's dilution law;
$K_a=\frac{c{\alpha }^2}{1-\alpha }$ ....(i)
Also, $\Delta T=K_f^{\prime }\times molality$
$\because molality=molarity=0.025$ (As given)
$\therefore \Delta T_{exp}=1.86\times 0.025=0.0465$
$\because i=\frac{\Delta T_{exp}}{\Delta T_N}=1+\alpha$
$\therefore 1+\alpha =\frac{0.06}{0.0465}$
$\therefore \alpha =0.29$
By eq. (i), $K_a=\frac{0.025\times (0.29{)}^2}{1-0.29}=2.96\times {10}^{-3}$