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Standard XII

Chemistry

Solubility Product

On passing ...

Question

On passing $H_{2} S$ into a solution containing both $Z n^{2 -}$ and $C u^{2 -}$ in acidic medium, only CuS gets precipitated. This is because :

K_{s p} o f C u S

K_{s p}

of ZnS

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K_{s p} o f C u S

K_{s p}

of ZnS

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K_{s p} o f C u S

K_{s p}

of ZnS

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CuS is more stable then ZnS

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Solution

The correct option is A $K_{s p} o f C u S$ < $K_{s p}$ of ZnS
Since the solubility product of CuS is less than that of ZnS, CuS gets precipitated.

$K_{s p} (C u S) < K_{s p} (Z n S)$

$K_{s p} (C u S) = 8 \times 10^{- 37}$

$K_{s p} (Z n S) = 1.2 \times 10^{- 23}$

Hence the correct option is A.

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Similar questions

Q. A solution, containing

0.01 M Z n^{+ 2}

and

0.01 M C u^{2 +}

is saturated by passing

H_{2} S

gas. The

S^{- 2}

concentration is

8.1 \times 10^{- 21} M

K s p

for

Z n S

and

C u s

are

3.0 \times 10^{- 22}

and

8.0 \times 10^{- 36}

respectively. Which of the following will occur in the solution :-

K_{s p}

C d S = 8 \times 10^{- 27}

K_{s p}

Z n S = 1 \times 10^{- 21}

K_{a}

H_{2} S = 1 \times 10^{- 21}

.
When

Z n S

starts precipitating, what concentration of

C d^{2 +}

is left?

K_{s p}

C d S = 8 \times 10^{- 27}

K_{s p}

Z n S = 1 \times 10^{- 21}

K_{a}

H_{2} S = 1 \times 10^{- 21}

.
What

[H^{+}]

must be maintained in saturated

H_{2} S (0.1 M)

to precipitate

C d S

but not

Z n S

, if

[C d^{2 +}] = [Z n^{2 +}] = 0.1 M

initially?

K_{s p}

C d S = 8 \times 10^{- 27}

K_{s p}

Z n S = 1 \times 10^{- 21}

K_{a}

H_{2} S = 1 \times 10^{- 21}

.
What

[H^{+}]

must be maintained in saturated

H_{2} S (0.1 M)

to precipitate

C d S

but not

Z n S

, if

[C d^{2 +}] = [Z n^{2 +}] = 0.1 M

initially?

Z n S

is not precipitated by passing

H_{2} S

in acidic medium but

C u S

precipitated. The reason for it is:

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On passing H2S into a solution containing both Zn2− and Cu2− in acidic medium, only CuS gets precipitated. This is because :

The correct option is A Ksp of CuS<Ksp of ZnSSince the solubility product of CuS is less than that of ZnS, CuS gets precipitated. Ksp(CuS)<Ksp(ZnS) Ksp(CuS)=8×10−37 Ksp(ZnS)=1.2×10−23Hence the correct option is A.

On passing $H_{2} S$ into a solution containing both $Z n^{2 -}$ and $C u^{2 -}$ in acidic medium, only CuS gets precipitated. This is because :

The correct option is A $K_{s p} o f C u S$ < $K_{s p}$ of ZnS
Since the solubility product of CuS is less than that of ZnS, CuS gets precipitated.

$K_{s p} (C u S) < K_{s p} (Z n S)$

$K_{s p} (C u S) = 8 \times 10^{- 37}$

$K_{s p} (Z n S) = 1.2 \times 10^{- 23}$

Hence the correct option is A.