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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

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 the $K_{s p}$ 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?

F e S

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M g S

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H g S

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Z n S

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Solution

The correct option is C $H g S$
Compound with lowest $K_{s p}$ will precipitate first. So $H g S$ will precipitate first as it has lowest $K_{s p}$ among all.

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Similar questions

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?

Q. A solution 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}

values of MnS, Fes, ZnS and HgS are

10^{- 15}, 10^{- 23}, 10^{- 20}

and

10^{- 54}

respectively, which one will precipitate first?

Q. 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}

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?

Q. 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 MnS,FeS,ZnS andHgS are

10^{- 15}, 10^{- 23}, 10^{- 20}

and

10^{- 54}

respectively, which one will precipitate first?

Q. 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 ?

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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 the Ksp of MnS,FeS,ZnS and HgS are 10−15,10−23,10−20 and 10−54 respectively, which one will precipitate first?

The correct option is C HgSCompound with lowest Ksp will precipitate first. So HgS will precipitate first as it has lowest Ksp among all.

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 the $K_{s p}$ 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?

The correct option is C $H g S$
Compound with lowest $K_{s p}$ will precipitate first. So $H g S$ will precipitate first as it has lowest $K_{s p}$ among all.