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0.01M solution of H(2)A has pH equal to ...

0.01M solution of `H_(2)A` has pH equal to 4. If `k_(a1)` for the acid is `4.45 xx 10^(-7)`, the concentration of `HA^(-)` ion in solution would be

A

0.01 M

B

`4.45 xx 10^(-5)`

C

`8.0 xx 10^(-5)`

D

unpredictable

Text Solution

Verified by Experts

The correct Answer is:
B
`H_(2)A hArr HA^(-)+H^(+)`
Now, `[HA^(-)]=(K_(a)xx[H_(2)A])/([H^(+)])`
`=(4.45 xx 10^(-7)xx 1xx 10^(-2))/(1xx10^(-4))`
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