In the reaction: `H_(2)S hArr 2H^(+) + S^(-)`, when `NH_(4)OH` is added, then

Options

A. `S^(--)` is precipitate
B. No action takes places
C. Concentration of `S^(--)` decreases
D. Concentration of `S^(--)` increases

Answer

D

Explanation

`H_(2)S hArr 2H^(+) +S^(--)`
`NH_(4)OH hArr NH_(4)^(+) +OH^(-)`
The `OH^(-)` ion of `NH_(4)OH` will react with `H^(+)` ion from `H_(2)S` .
`H^(+) +OH^(-) hArr H_(2)O `
So concentration of `H^(+)` ion will decrease and by Le–chatelier principle, the reaction will move in the forward direction and more `S^(--)` ions will form.