Question

Calculate the $H^{+}$ ion concentration of a $0.01M$ weak monobasic acid. The value of dissociation constant is $4.0\times {10}^{-10}$.

• A. $2\times {10}^{-8}$ $mol{L}^{-1}$
• B. $2\times {10}^{-2}$ $mol{L}^{-1}$
• C. $2\times {10}^{-4}$$mol{L}^{-1}$
• D. $2\times {10}^{-6}$ $mol{L}^{-1}$

Answer 1

The correct option is D $2\times {10}^{-6}$ $mol{L}^{-1}$

Given,
$K_a=4.0\times {10}^{-10}$, concentration = 0.01 M
Let the weak monobasic acid is $HA$ . The dissociation of weak acid is given by :
$HA\rightleftharpoons H^{+}+A^{-}C00C-C\alpha C\alpha C\alpha$

Applying Ostwald's dilution law
$\alpha =\sqrt{\frac{K_a}{c}}\alpha =\sqrt{\frac{4\times {10}^{-10}}{0.01}}=\sqrt{4\times {10}^{-8}}=2\times {10}^{-4}$
Concentration of $\left[H^{+}\right]=C\alpha =(0.01\times 2\times {10}^{-4})=2\times {10}^{-6}mol{L}^{-1}$
Hence, the concentration of $\left[H^{+}\right]=2\times {10}^{-6}mol{L}^{-1}$