Question

The measured freezing point depression for a $0.1m$ aqueous ${CH}_{3}COOH$ solution is ${0.19}^{o}C$. The acid dissociation constant $K_a$ at this concentration will be:

(Given $K_f$ the molal cryoscopic constant $=1.86K$ $kg{mol}^{-1}$)

• A. $4.76\times {10}^{-5}$
• B. $4\times {10}^{-5}$
• C. $8\times {10}^{-5}$
• D. $2\times {10}^{-5}$

Answer 1

Answer: (B)

$4\times {10}^{-5}$

$\because \Delta T_f=i\times K_f\times m$

$\therefore i=\frac{\Delta T_f}{K_f\times m}=\frac{0.19}{1.86\times 0.1}=1.02$

Again $\alpha =\frac{i-1}{n-1}=\frac{1.02-1}{2-1}$

$=0.02=2.0\times {10}^{-2}$

For ${CH}_{3}COOH\rightleftharpoons {CH}_{3}CO{O}^{-}+H^{+}$

$K_a={C\alpha }^2=0.1\times {(2.0\times {10}^{-2})}^2=4\times {10}^{-5}$