Question

$0.6$mL of acetic acid having density $1.06gm{l}^{-1}$, is dissolved in $1$ litre of water. The depression of freezing point observed for the strength of acid was ${0.0205}^o$C. The dissociation constant of acid is:

• A. $1.86\times {10}^{-5}$
• B. $2.21\times {10}^{-7}$
• C. $3.56\times {10}^{-5}$
• D. $4.44\times {10}^{-7}$

Answer 1

The correct option is A $1.86\times {10}^{-5}$

number of moles of $C{H}_{3}COOH=\frac{\left(0.6\right)ml\times \left(1.06\right)g/ml}{\left(60\right)g/mol}=0.0106mol=n$

$Molarity=\frac{0.0106mol}{1000ml\times 1{gmol}^{-1}}=0.0106\frac{mol}{Kg}$

$\Delta T_f=\left(1.86KKg{mol}^{-1}\right)\left(0.0106mol{Kg}^{-1}\right)=0.0197K$

Von't Hoff Factor$=\frac{Observed}{Calculated}FreezingPoint=\frac{0.0205}{0.0197}=1.041$

$C{H}_{3}COOH\leftrightharpoons H^{+}+C{H}_{3}CO{O}^{-}$

$n$ $0$ $0$

$n(1-x)$ $nx$ $nx$

total moles$=n-nx+2nx=n-nx$

$i=\frac{n(1+x)}{n}=1+x=1.041$

Degree of Dissociation; $x=0.041$

$\left[{CH}_{3}COOH\right]=n(1-x)=\left(0.0106\right)(1-0.041)\left[{CH}_{3}CO{O}^{-}\right]=\left[H^{+}\right]=nx=\left(0.0106\right)\left(0.041\right)K_a=\frac{\left[{CH}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[{CH}_{3}COOH\right]}=\frac{{\left(0.0106\right)}^2{\left(0.041\right)}^2}{\left(0.0106\right)(1-0.041)}=1.86\times {10}^{-5}$