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Standard XII

Chemistry

Common Ion Effect

If 0.1 M HC...

Question

If 0.1 M $H C I$ is mixed in the solution containing only $0.1 M$ $C d^{2 +}$ ions and saturated with $H_{2} S$ , then $[C d^{2 +}]$ remaining in the solution after $C d S$ stops to precipitate is

10^{- 8}

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8.2 \times 10^{- 9}

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5.6 \times 10^{- 6}

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5.6 \times 10^{- 10}

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Solution

The correct option is B $10^{- 8}$
$K_{1} K_{2} = \frac{[H^{\oplus}]^{2} [S^{2 -}]}{[H_{2} S]} \Rightarrow [S^{2 -}]_{m i n} = 10^{- 20} M$
$\Rightarrow [C d^{2 +}]_{a l l o w e d} = \frac{K_{s p} (C d S)}{[S^{2 -}]_{s a l t}} = 10^{- 8} M$

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Similar questions

Q. Find the concentration of

H^{+}

ion that must be maintained in a saturated solution of

H_{2} S (0.1 M)

so as to precipitate

C d S

not

Z n S

Given:

[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_{s p} (H_{2} S) = 1.1 \times 10^{- 21}

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?

Q. A solution containing

0.1 M

{Z n}^{2 +}

and

0.01 M

{C u}^{2 +}

saturated with

H_{2} S

. The

S^{2 -}

concentration is

8.1 \times 10^{- 31} M

. Will

Z n S

and

C u S

precipitate?

K_{s p} (Z n S) = 3.0 \times 10^{- 23}

and

K_{s p} (C u S) = 8.0 \times 10^{- 4}

Q. Find the

[H^{+}]

that must be maintained in a saturated

H_{2} S (0.1 M)

to precipitate

C d S

not

Z n S

[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_{s p} (H_{2} S) = 1.1 \times 10^{- 21}

Q. At

18^{o} C

, the solubility of

C d S

in water is

6.33 \times 10^{- 15}

M. What is the concentration of

C d^{2 +}

ion in a solution of pH = 1 saturated with

H_{2}

S gas, in which concentration of

H_{2} S = 0.1 M

? The product of the first and second ionization constants of

H_{2} S

1.1 \times 10^{- 22}

at this temperature.

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If 0.1 M HCI is mixed in the solution containing only 0.1MCd2+ ions and saturated with H2S, then [Cd2+] remaining in the solution after CdS stops to precipitate is

The correct option is B 10−8K1K2=[H⊕]2[S2−][H2S]⇒[S2−]min=10−20M⇒[Cd2+]allowed=Ksp(CdS)[S2−]salt=10−8M

If 0.1 M $H C I$ is mixed in the solution containing only $0.1 M$ $C d^{2 +}$ ions and saturated with $H_{2} S$ , then $[C d^{2 +}]$ remaining in the solution after $C d S$ stops to precipitate is

The correct option is B $10^{- 8}$
$K_{1} K_{2} = \frac{[H^{\oplus}]^{2} [S^{2 -}]}{[H_{2} S]} \Rightarrow [S^{2 -}]_{m i n} = 10^{- 20} M$
$\Rightarrow [C d^{2 +}]_{a l l o w e d} = \frac{K_{s p} (C d S)}{[S^{2 -}]_{s a l t}} = 10^{- 8} M$