Question

Find the dissociation constant $K_a$ of a weak monobasic acid which is 3.5% dissociated in a $\frac{M}{10}$ solution at ${25}^{\circ }C$

• A. $0.5\times {10}^{-3}$
• B. $2.62\times {10}^{-6}$
• C. $4.2\times {10}^{-6}$
• D. $1.27\times {10}^{-4}$

Answer 1

The correct option is D $1.27\times {10}^{-4}$

$HA(aq.)\rightleftharpoons H^{+}(aq.)+A^{-}(aq.)Initial:C00Equilibrium:(C-C\alpha )C\alpha C\alpha C=\frac{1}{10}=0.1M\alpha =\frac{3.5}{100}=0.035K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}=\frac{C{\alpha }^2}{1-\alpha }{K}_a=\frac{0.1\times (0.035{)}^2}{1-0.035}{K}_a=1.269\times {10}^{-4}M$



Theory:

Relation between $K_{a}and{K}_b$:
Consider this weak acid reaction of $HA$
$HA\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+A^{-}\left(aq\right)$
$K_a=\frac{\left[H_3{O}^{+}\right]\left[A\right]}{\left[HA\right]}$
Consider this weak base reaction of the conjugate base of $HA$. I.e., $A^{-}$
$A^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons HA\left(aq\right)+O{H}^{-}\left(aq\right)$
$K_b=\frac{\left[O{H}^{-}\right]\left[HA\right]}{\left[A\right]}$
$K_a\times K_b=\frac{\left[H_3{O}^{+}\right]\left[A\right]}{\left[HA\right]}\times \frac{\left[O{H}^{-}\right]\left[HA\right]}{\left[A\right]}$
$K_a\times K_b=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]=K_w$ or $K_a\times K_b=\left[H^{+}\right]\left[O{H}^{-}\right]=K_w={10}^{-14}$
For any acid-base conjugate pair, $K_a\times K_b=K_w$