Question

The percentage dissociation $\left(\alpha \right)$ of $C{H}_{3}COOH$ having dissociation constant $K_a=1.8\times {10}^{-5}M$ is given as $5$%.
Find the inital concentration of $C{H}_{3}COOH$.

• A. $2.44\times {10}^{-2}M$
• B. $1.98\times {10}^{-2}M$
• C. $1.28\times {10}^{-3}M$
• D. $6.84\times {10}^{-3}M$

Answer 1

The correct option is D $6.84\times {10}^{-3}M$

$C{H}_{3}COOH\left(l\right)\rightleftharpoons C{H}_{3}CO{O}^{-}(aq.)+H^{+}(aq.)Initial:C00Equilibrium:C(1-\alpha )c\alpha c\alpha$

$K_a=1.8\times {10}^{-5}=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$
$K_a=\frac{C\alpha \times C\alpha }{C(1-\alpha )}=\frac{C{\alpha }^2}{1-\alpha }=\frac{C\times (0.05{)}^2}{1-0.05}$

$\therefore C=\frac{1.8\times {10}^{-5}\times 0.95}{0.05\times 0.05}=6.84\times {10}^{-3}M$
Initial Concentration of $C{H}_{3}COOH$ $C=6.84\times {10}^{-3}M$