Question

A solution which is 10^-3 M each in Mn^{2 +} , Fe^{2 +} , Zn^{2 +} and Hg^{2 +} is treated with 10 ^{- 16} M sulphide ion. If K_sp of MnS, FeS, ZnS and HgS are 10 ^{- 13} , 10^-18 , 10^-24 and 10^-53 respectively, which one will precipitate first ?

• A. FeS
• B. MgS
• C. HgS
• D. ZnS

Answer 1

Answer: (C)

[S^{2 -} ] need for precipitation of

$\text{FeS} = \frac{1 0^{- 1 8}}{1 0^{- 3}} = 1 0^{- 1 5} \text{M}$

$\text{MnS} = \frac{1 0^{- 1 3}}{1 0^{- 3}} = 1 0^{- 1 0} \text{M}$

$\text{HgS} = \frac{1 0^{- 5 3}}{1 0^{- 3}} = 1 0^{- 5 0 \, } \text{M}$

$\text{ZnS} = \frac{1 0^{- 2 4}}{1 0^{- 3}} = 1 0^{- 2 1} \text{M}$

Thus minimum [S^{2 -} ] for precipitation is for HgS it will be precipitated first.