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The pH of a 0.1 molar solution of the ac...

The `pH` of a `0.1` molar solution of the acid `HQ` is `3`. The value of the ionisation constant, `K_(a)` of the acid is

A

`3x10^(-1)`

B

`1xx10^9-3)`

C

`1xx10^(-5)`

D

`1xx10^(-7)`

Text Solution

Verified by Experts

The correct Answer is:
C
`HQ=H^(+) + Q^(-)`
`[H^(+)]=sqrt(K_(a)C)` by Ostwald’s dilution law
`[H^(+)]=10^(-ph)=10^(-3)M`
C=0.1M
Thus, `10^(-3)=sqrt(K_(a)xx0.1)`
`10^(-6)=K_(a)xx0.1`
`therefore K_(a)=10^(-5)`
Alternation Method
`HQ to H^(+) + Q^(-)`
Initial concentration 0.1 M 0 0
Given, pH = 3. This suggests `H^(+)]10^(-3)M` at equilibrium `=10^(-3)M`.
Hence `[HQ]=0.1M-10^(-3)M =0.1M[10^(-3)M ltltlt 0.1M]`
`K_(a)` for the above reaction is given by:`K_(a)=([H^(+)][Q^(-)])/([HQ])= ([10^(-3)][10^(-3)])/[0.1]`
`K_(a)=1xx10^(-5)`
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