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…

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?

MARKS Content Team · Accepted Answer

Correct answer: (C) HgS. [S^(2-)] needed for precipitation of FeS=(10^(-18))/(10^(-3))=10^(-15)M MnS=(10^(-13))/(10^(-3))=10^(-10)M HgS=(10^(-53))/(10^(-3))=10^(-50)M ZnS=(10^(-24))/(10^(-3))=10^(-21)M Thus, minimum [S^(2-)] is for HgS and so it will be precipitated first.

A solution which is $1 0^{- 3}$ M each in $M n^{2 +}, F e^{2 +}, Z n^{2 +}$ and $H g^{2 +}$ is treated with $1 0^{- 16}$ M sulphide ion. If $K_{s p}$ of $M n S$ , FeS, ZnS and HgS are $1 0^{- 13}, 1 0^{- 18}, 1 0^{- 24}$ and $1 0^{- 53}$ respectively. Which one will precipitate first?

Solution

More Ionic Equilibrium Questions

A solution which is 10-3 M each in Mn2+,Fe2+,Zn2+ and Hg2+ is treated with 10-16 M sulphide ion. If Ksp of MnS , FeS, ZnS and HgS are 10-13,10-18,10-24 and 10-53 respectively. Which one will precipitate first?

Solution

More Ionic Equilibrium Questions

A solution which is $1 0^{- 3}$ M each in $M n^{2 +}, F e^{2 +}, Z n^{2 +}$ and $H g^{2 +}$ is treated with $1 0^{- 16}$ M sulphide ion. If $K_{s p}$ of $M n S$ , FeS, ZnS and HgS are $1 0^{- 13}, 1 0^{- 18}, 1 0^{- 24}$ and $1 0^{- 53}$ respectively. Which one will precipitate first?