`Zn` salt is mixed with `(NH_(4))_(2)S` of `0.021M`. What amount of `Zn^(2+)` will remain uprecipitated in `12mL` of the solution? `K_(sp)` of `ZnS = 4.51 xx 10^(-24)`.

Zn salt is mixed with (NH_(4))_(2)S of molarity 0.021M . The amount of Zn^(2+) reamins unprecipitated in 12 mL of this solutikon : (K_(SP)for ZnS=4.51xx10^(-24))

Zn salt is mixed with (NH_4)_2S of molarity 0.021 M Calculate the amount ZN^(2+) remains unprecipitated in 12 ml of this solution.

0.2 millimoes of Zn^(2+) ion is mixed with (NH_(4))_(2) S of molarity 0.02 M. The amount of Zn^(2+) that remains uprecipitated in 20 mL of this solution would be : (Given : K_(sp) (ZnS)=4xx10^(-24))

20 ml of 1.5 xx10^(-5) M BaCI_2 solution is mixed with 40 ml of 0.9 xx 10^(-5) M sodium sulphate solution , will a precipitate of BaSO_4 get formed ? (K_(sp) "for" BaSO_4 =1.2 xx 10^(-10)) .

A solution containing both Zn^(2+) and Mn^(2+) ions at a concentration of 0.01M is saturated with H_(2)S . What is pH at which MnS will form a ppt ? Under these conditions what will be the concentration of Zn^(2+) ions remaining in the solution ? Given K_(sp) of ZnS is 10^(-22) and K_(sp) of MnS is 5.6 xx 10^(-16), K_(1) xx K_(2) of H_(2)S = 1.10 xx 10^(-21) .

A solution containing both Zn^(2+) and Mn^(2+) ions at a concentration of 0.01M is saturated with H_(2)S . What is pH at which MnS will form a ppt ? Under these conditions what will be the concentration of Zn^(2+) ions remaining in the solution ? Given K_(sp) of ZnS is 10^(-22) and K_(sp) of MnS is 5.6 xx 10^(-16), K_(1) xx K_(2) of H_(2)S = 1.10 xx 10^(-21) .

An acidified of 0.05 M Zn^(2+) is saturated with 0.1 M H_2S . What is the minimum molar concentration (M) or H^+ required to prevent the precipitation of ZnS ? Use K_(sp)(ZnS) = 1.25 xx10^(-22) and overall dissociation constant of H_2S, K_("NET") =K_1K_2=1xx10^(-21) .