Calculate the solubility of MnS is pure water, `K_(sp) (MnS)=2.5 xx 10^(-10)`. Assume hydrolysis of `S^(2-)` ions, `K_(1) and K_(2)" for "H_(2)S" are "1 xx 10^(-7) and 1 xx 10^(-14)` respectively.

Explanation

Let the solubility of MnS be x mole/litre. On dissociation `[Mn^(2+)]="x but "[S^(2-)] ne x` because `S^(2-)` undergoes extensive hydrolysis. Let us consider the two stages of hydrolysis.
`S^(2-)+H_(2)O Leftrightarrow HS^(-)+OH^(-), K_(h)^(.)=(K_(w))/(K_(2))=(1 xx 10^(-14))/(1 xx 10^(-14))=1.0`
`HS^(-)+H_(2)O Leftrightarrow H_(2)S+OH^(-) , K_(h)^(..)=K_(w)/(K_(1))=(1 xx 10^(-14))/(1 xx 10^(-7))=1 xx 10^(-7)`
As `K_(h)^(..) gt gt K_(h)^(..)`, first stage of hydrolysis is almost complete.
`x=[Mn^(2+)]=[HS^(-)]=[OH^(-)]`
Consider first stage of hydrolysis
`K_(h)=([HS^(-)] [OH^(-)])/([S^(2-)])`
`or [S^(-2)]=([HS^(-)] [OH^(-)])/(K_(h.))=x^2/(10)`
At equilibrium,
`[Mn^(2+)] [S^(2-)]=K_(sp)=2.5 xx 10^(-10)`
`x=(2.5 xx 10^(-10))/(x^2)`
`x=6.3 xx 10^(-4)M`