Question

Calculate the $pH$ of a $0.05M$ weak monobasic acid solution. The value of dissociation constant is $8.0\times {10}^{-9}$.

• A. 2
• B. 4.69
• C. 6.6
• D. 6.37

Answer 1

The correct option is B 4.69

Given,
$K_a=8.0\times {10}^{-9}$, concentration (C) = $0.05M$
Let the weak monobasic acid is $HA$ . The dissociation of weak acid is given by :
$HA\rightleftharpoons H^{+}+A^{-}C00C-C\alpha C\alpha C\alpha$

$K_a=\frac{C{\alpha }^2}{C(1-\alpha )}As,\alpha <<1K_a=C{\alpha }^2$
$\alpha =\sqrt{\frac{K_a}{c}}\alpha =\sqrt{\frac{8\times {10}^{-9}}{0.05}}=\sqrt{16\times {10}^{-8}}=4\times {10}^{-4}$
Concentration of $\left[H^{+}\right]=C\alpha =(0.05\times 4\times {10}^{-4})=2\times {10}^{-5}mol{L}^{-1}$
$pH=-\log [2\times {10}^{-5}]=4.69$