Question

The dissociation constant for a weak monobasic acid is $4\times {10}^{-10}.pH$ for this $0.01N$ weak monobasic acid is :

• A. 9.7
• B. 1.7
• C. 3.7
• D. 5.7

Answer 1

The correct option is D 5.7

Let, the monobasic acid is $HA$. Its dissociation is given below :
$HA\rightleftharpoons H^{+}+A^{-}$
Applying Ostwald's dilution law:$\alpha =\sqrt{\frac{K_a}{C}}$
$\alpha =\sqrt{\frac{4\times {10}^{-10}}{0.01}}=2\times {10}^{-4}$
Concentration of $H^{+}$
$\left[H^{+}\right]=C\alpha =0.01\times 2\times {10}^{-4}$$\left[H^{+}\right]=2\times {10}^{-6}mol{L}^{-1}pH=-\log \left[H^{+}\right]pH=-\log [2\times {10}^{-6}]pH=5.7$