A solution of `Cu^(2+)` `Zn^(2+)` `Bi^(3+)`. `Mn^(2+)` and `Co^(2+)` at pH = 1 is reacted with `H_(2)S`. Which ions will be precipitated as sulphide?

Upon treatment with ammonical H_(2)S solution, the metal ion precipitating as sulphide is:

Cu^(2+) ions give white precipitate with :

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) .

It is given that 0.001 mol each of Cd^(2+) and Fe^(2+) ions are contained in 1.0L of 0.02M HC1 solution. This solutions is now saturated with H_(2)S gas at 25^(@)C . a. Determine whether or not each of these ions will be precipitated as sulphide? b. How much Cd^(2+) ions remains in the solution at equilibrium? K_(1)(H_(2)S) = 1.0 xx 10^(-7), K_(2) (H_(2)S) = 1.0 xx 10^(-14) : ltbRgt K_(sp) (CdS) = 8 xx 10^(-27): K_(sp) (FeS) = 3.7 xx 10^(-19) .

An acidic solution contains both Zn^(2+) and Hg^(2+) ions. Which ion will get precipitated passing H_2S into it ?

0.5 M HCI solution has ions Hg^(+), Cd^(2), Sr^(+2), Fe^(+2), Cu^(+2) . If H_(2)S gas in passed through this solution, the ions which are precipitated out are

When H_(2)S gas is passed through ZnCl_(2) solution. ZnS is not precipitated, why?

Actylene reacts with ammonical Cu_(2)Cl_(2) to give precipitate of

Aqueous solution contains Zn(CH_(2)COO)_(2),Cd(CH_(3)COO)_(2) and Cu(CH_(3)COO)_(2) on passing H_(2)S gas, therte is a precipitate of ……. As sulphide

Assertion (A): A solution contains 0.1M each of pB^(2+), Zn^(2+),Ni^(2+) , ions. If H_(2)S is passed into this solution at 25^(@)C . Pb^(2+), Ni^(2+), Zn^(2+) will get precpitated simultanously. Reason (R): Pb^(2+) and Zn^(2+) will get precipitated if the solution contains 0.1M HCI . [K_(1) H_(2)S = 10^(-7), K_(2)H_(2)S = 10^(-14), K_(sp) PbS =3xx 10^(-29) K_(sp) NiS = 3 xx 10^(-19). K_(sp) ZnS = 10^(-25)]