Question

The equivalent conductance of an aqueous solution of $1.0283\times {10}^{-3}$ g equivalent acetic acid per litre is $48.15{\pi }^{-1}$ cm${}^2$ equiv${}^{-1}$ at ${25}^{o}C$. At infinite dilution value is $390.7{\pi }^{-1}$ cm${}^2$ equiv${}^{-1}$. Calculate the degree of ionisation and ionisation constant of acetic acid is:

• A. $0.1232,1.78\times {10}^{-5}$
• B. $0.223,102\times {10}^{-5}$
• C. $0.229,1.78\times {10}^{-5}$
• D. $0.531,2.85\times {10}^{-5}$

Answer 1

The correct option is A $0.1232,1.78\times {10}^{-5}$

The degree of ionisation $\alpha =\frac{\Lambda }{{\lambda }_0}=\frac{48.15{\pi }^{-1}c{m}^{2}equi{v}^{-1}}{390.7{\pi }^{-1}c{m}^{2}equi{v}^{-1}}=0.1232$
The ionisation constant of acetic acid :

$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}=\frac{\left(C\alpha \right)\times \left(C\alpha \right)}{\left(C\right(1-\alpha \left)\right)}=C\frac{{\alpha }^2}{1-\alpha }$

$K_a=1.0283\times {10}^{-3}\times \frac{(0.1232{)}^2}{1-0.1232}=1.78\times {10}^{-5}$