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Standard XII

Chemistry

Solubility Product

A solution wh...

Question

A solution which is $10^{- 3} M$ each in ${M n}^{2 +}, {F e}^{2 +}, {Z n}^{2 +}$ and ${H g}^{2 +}$ is treated with $10^{- 16} M$ sulphide ion. if $K_{s p}$ 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?

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Solution

Sobility product $(K s p)$ of $M n S = 10^{- 15}$
$F e S = 10^{- 23}$
$Z n S = 10^{- 20}$
$H g S = 10^{- 54}$
for precipitation first from the given data, $H g S$ has a $K s P$ of $10^{- 54}$ which is least. So, $H g S$ precipitates first.

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Similar questions

Q. A solution which is

10^{- 3} M

each in

M n^{2 +}, F e^{2 +}, Z n^{2 +}

, and

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

Sobility product (Ksp) of MnS=10−15 FeS=10−23 ZnS=10−20 HgS=10−54for precipitation first from the given data, HgS has a KsP of 10−54 which is least. So, HgS precipitates first.

A solution which is $10^{- 3} M$ each in ${M n}^{2 +}, {F e}^{2 +}, {Z n}^{2 +}$ and ${H g}^{2 +}$ is treated with $10^{- 16} M$ sulphide ion. if $K_{s p}$ 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?

Sobility product $(K s p)$ of $M n S = 10^{- 15}$
$F e S = 10^{- 23}$
$Z n S = 10^{- 20}$
$H g S = 10^{- 54}$
for precipitation first from the given data, $H g S$ has a $K s P$ of $10^{- 54}$ which is least. So, $H g S$ precipitates first.