Question

The dissociation constant for an acid HA is `1.6 xx 10^(-5)`. Calculate its `H_(3)O^(+)` in concentration in 0.01 M solution .

Answer 1

Correct Answer - `4.0 xx 10^(-4) " mol " L^(-1)`
the dissociation of the acid HA may be represented as :
`HA (aq) hArr H^(+) (aq) +A^(-) (aq)`
`K_(a) =[[H^(+)(aq)][A^(-)(aq)]]/[[HA(aq)]]=[[H^(+)(aq)]^(2)]/[[HA(aq)]]`
`K_(a) =1.6 xx 10^(-5) , [HA] =0.01 M`
`1.6xx 10^(-5) =[[H^(+)(aq)]^(2)]/(0.01) " or " [H^(+) (aq) ] =sqrt(1.6 xx 10^(-5) xx0.01)=4.0 xx 10^(-4) "mol " L^(-1)`