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Calculate K for the reaction, A^(-)+H(3)...

Calculate K for the reaction, `A^(-)+H_(3)^(+)O hArr HA+H_(2)O`
if `K_(a)` value for the acid HA is `1.0xx10^(-6)`.

A

`1.0 xx 10^(6)`

B

`1.0 xx 10^(-8)`

C

`1.0 xx 10^(8)`

D

`1.0 xx 10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
1
The reaction is the reverse of the ionization reaction of HA, hence the equilibrium constant is the reciprocal of `K_(a)`.
`K = ([HA])/([H^(+)][A^(-)]) = 1/K_(a) = 1/(1.0 xx 10^(-6)) = 1.0 xx 10^(6)`
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