1.

What is `[H^(+)]` of a solution that is 0.01 M in HCN and 0.02 M in NaCN (`K_(a)` for `HCN = 6.2 xx 10^(-10)`)A. `3.1 xx 10^(10)`B. `6.2 xx 10^(5)`C. `6.2 xx 10^(-10)`D. `3.1 xx 10^(-10)`

Answer» Correct Answer - D
`K_(a) = ([H^(+)][CN^(-)])/([HCN^(-)]) rArr 6.2 xx 10^(-10) = ([H^(+)] [0.02])/([0.01])`
`[H^(+)] = (6.2 xx 10^(-10) xx 0.01)/(0.02) = 3.1 xx 10^(-10)`.


Discussion

No Comment Found

Related InterviewSolutions