Question

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

• A. $1.269\times {10}^{-4}$
• B. $1.269\times {10}^{-5}$
• C. $2.269\times {10}^{-4}$
• D. $2.269\times {10}^{-5}$

Answer 1

The correct option is A $1.269\times {10}^{-4}$

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