A solution which is $10^{- 3} N$ each in $M n^{2 +}, F e^{2 +}, Z n^{2 +}$ and $H g^{2 +}$ is treated with $10^{- 16} M$ sulphide ion. If $K_{s p}$ values of $M n S, F e S, Z n S$ and $H g S$ are $10^{- 15}, 10^{- 23}, 10^{- 20}$ and $10^{- 54}$ respectively, which one will precipitate first ?

F e S

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M g S

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H g S

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Z n S

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $H g S$
Lesser the $K_{s p}$ , more is the precipitation.
So, $H g S$ will precipitate first.

Suggest Corrections

Similar questions

A solution which has $10^{- 3}$ M each of $M n^{2 +}, F e^{2 +}, Z n^{2 +}$ and $H g^{2 +}$ is treated with $10^{- 16}$ M sulphide ion. If $K_{s p}$ of $M n S$ , $F e S$ , $Z n S$ and $H g S$ are $10^{- 15}$ , $10^{- 23}$ , $10^{- 20}$ and $10^{- 54}$ respectively, which one will precipitate first?

A solution which is 10^-3 M each in Mn²⁺, Fe²⁺, Zn²⁺ and Hg²⁺ is treated with 10^-16 M sulphide ion. If K_sp of MnS, FeS, ZnS and HgS are 10^-15, 10^-23, 10^-20 and 10^-54 respectively, which one will precipitate first?