InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Calculate the approximate pH of a 0.100 M aqueous `H_(2)S` solution. `K_(1) and K_(2)` for `H_(2)S` are `1.00xx10^(-7) and 1.30xx10^(-13)` respectively at `25^(@)C`. |
|
Answer» `K_(2) lt lt K_(1)`. Hence `H^(+)` ions are mainly from 1st dissociation, i.e., `H_(2)S hArr H^(+)+HS^(-)` `K_(1)=([H^(+)][HS^(-)])/([H_(2)S])=([H^(+)]^(2))/([H_(2)S]) or [H^(+)]=sqrt(K_(1)[H_(2)S]):. [H^(+)]=sqrt(10^(-7)xx10^(-1))=10^(-4)`. Hence, pH = 4 |
|