Ionic EquilibriumHard
Question
0.2 millimoles of Zn2+ ion is mixed with (NH4)2 S of molarity 0.02 M. The amount of Zn2+ remains unprecipitated in 20 mL of this solution would be (Given : KSP ZnS = 4 × 10-24)
Options
A.5.2 × 10-22 g
B.2.6 × 10-22 g
C.2 × 10-23 g
D.none of these
Solution
S2- + Zn2+ → ZnS (s)
m.moles 0.4 0.2
0.2 -
so in solution,
[S2-] =
= 0.01 M
∴ [Zn2+] =
= 4 × 10-22 M
∴ mass of Zn2+ remain unprecipitated in 20 ml of solution =
× 20 × 65 = 5.2 × 10-22 gm.
m.moles 0.4 0.2
0.2 -
so in solution,
[S2-] =
= 0.01 M∴ [Zn2+] =
= 4 × 10-22 M∴ mass of Zn2+ remain unprecipitated in 20 ml of solution =
× 20 × 65 = 5.2 × 10-22 gm.Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Select the correct order :-...Given is a concentrated solution of a weak electrolyte $A_{x}B_{y}$ of concentration ' c ' and dissociation constant ' K...Ka for formic acid and acetic acid are 2.1 × 10-4 and 1.1 × 10-5 respectively. The relative strenth of acids i...Some chemists at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest...The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.2 M NaOH would be : (pKa for CH3COOH = ...