An acid `HA` ionizes as `HAhArrH^(+)+A^(-)` The `pH` of `1.0 M` solution is `5`. Its dissociation constant would be

Options

A. `1xx10^(-10)`
B. 5
C. `5xx10^(-8)`
D. `1xx10^(-5)`

Answer

A

Explanation

pH of a solution = 5
`therefore[H^(+)]=10^(-ph)=10^(-5)`
HA (a weak acid) ionizes as
`Ha to H^(+) + A^(-)`
1.0 0 0
`(1-10^(-5)) 10^(-5) 10^(-5)`
`K_(a)=([H^(-)][A^(-)])/([HA])=(10^(-5)xx10^(-5))/((1-10^(-5))) = 10^(-10)/1 [1 gtgtgt 10^(-5)]=1xx10^(-10)M`