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Question

Chemistry Question on Equilibrium

Dissociation constant of a weak acid is 10−510^{-5}. The pH of 0.1 M solution of this acid is

A

3

B

4

C

6

D

8

Answer

3

Explanation

Solution

{\underset{\text{weak acid}}{ {HA + H_2O}} <=> H_{3}O^{+} + A^{-} }
Ka=[H3O+][A−][HA]{ K_{a} = \frac{[H_3O^{+}][A^{-}]}{[HA]}}
But [H3O+][A−] {[H_3O^+][A^-]} [∵\because1 molecule of HAHA gives
1A+1 A^+ ion and 1H3O+1H_3O^+ ion]
Ka=[H3O+]2[HA]{ K_{a} = \frac{[H_3O^{+}]^{2}}{[HA]}}
10−5[H3O+]20.1 M10^{-5} \frac{[H_3O^{+}]^{2}}{0.1 \, M}
[H3O+]2=10−5×10 M=10−6M{[H_3O^{+}]^{2} = 10^{-5} \times 10 \, M = 10^{-6} M}
[H3O+]=10−3M{ [H_3O^{+}] = 10^{-3} M}
pH=−log[H3O+]=−log(10−3)=3{ pH = - log[H_3 O^{+}] = - log(10^{-3} ) = 3}