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K(a) of HA is 10^(-4). The equilibrium c...

`K_(a)` of `HA` is `10^(-4)`. The equilibrium constant for its reaction with a strong base is

A

`1.0xx10^(-4)`,

B

`1.0xx10^(-10)`,

C

`1.0xx10^(10)`,

D

`1.0xx10^(14)`,

Text Solution

Verified by Experts

The correct Answer is:
C
`K=(K_(a))/(K_(w))=10^(10)`
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