Question

Degree of dissociation of $1MC{H}_{3}COOH$ in presence of $1M(C{H}_{3}COO{)}_{2}Pb$ having degree of dissociation $80\%$ and $3M(C{H}_{3}CCO{)}_{2}Ca$ is : ($K_a$ for acetic acid is $1.8\times {10}^{-5}$)

• A. $2.3\times {10}^{-6}$
• B. $1.36\times {10}^{-6}$
• C. $1.34\times {10}^{-3}$
• D. $4.24\times {10}^{-3}$

Answer 1

The correct option is A $2.3\times {10}^{-6}$

$C{H}_{3}COO{)}_{2}Ca$ is strong electrolyle

$\therefore$ it will dissociate completely

$\Rightarrow 3H(C{H}_{3}COO{)}_{2}Ca$ gives $6HC{H}_{3}CO{O}^{-}$

$(C{H}_{3}COO{)}_{2}Pb$ dissociate $80\%$

$\left(C{H}_{3}COO\right)2Pb\rightleftharpoons P{b}^{2+}+2C{H}_{3}CO{O}^{-}$

$\therefore 1MPb$ will yield $2\times \left(80\%of1M\right)C{H}_{3}CO{O}^{-}$

$=2\times 0.8<C{H}_{3}CO{O}^{-}$

$=1.6M$

$\therefore$ Total $C{H}_{3}CO{O}^{-}=7.6M$

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

$t=0$ $1$ $7.6$ $0$

$t=t_{eq}$ $1-\alpha$ $7.6+\alpha$ $\alpha$

$\alpha \ll 7.6$

$\therefore 7.6+\alpha \approx 7.6$

$\therefore$ Now,

$\frac{\alpha \times 7.6}{1-\alpha }=K_a=1.8\times {10}^{-5}$

$\alpha \ll 1\therefore 1-\alpha \approx 1$

$\alpha \times 7.6=1.8\times {10}^{-5}$

$\Rightarrow \alpha =\frac{1.8}{7.6}\times {10}^{-5}$

$\approx 2.86\times {10}^{-6}$