1.

The pH of 0.1 M solution of an organic acid is 4.0. Calculate its dissociation constant.

Answer» Correct Answer - `1xx 10^(-7)`
`pH =4 " or " - log [H_(3)O^(+)] =4`
`" or " " "log[(1)/(H_(3)O)] =4 " or " [H_(3) O^(+)] =10^(-4) m`
Let the acid HA ionise as :
`HA overset((aq))(hArr) H_(3)O^(+) + A^(-) (aq)`
`K_(a) =[[H_(3)O^(+)][A^(-)(aq)]]/[[HA]] =[[H_(3)O^(+)]^(2)]/[[HA]]`
`" Now "" "[H_(3)O^(+)] =10^(-4) M, HA =0.1 M`
`:." " K_(a) =((10^(-4))/((0.1)) =1xx 10^(-7)`


Discussion

No Comment Found

Related InterviewSolutions