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Lewis Theory

H 2 S is bubb...

Question

$H_{2} S$ is bubbled into 0.2M NaCN solution which is 0.02 M in each $[C d (C N)_{4}]_{2 -} a n d [A g (C N)_{2}]^{-}$ . The $H_{2} S$ produces $1 \times 10^{- 9}$ M sulphide in the solution. (Given, $K_{s p} A g_{2} S = 1 \times 10^{- 50} M^{3}; K_{s p} C d S = 7.1 \times 10^{- 28} M^{2} K_{i n s t} [A g (C N)_{2}]^{- 1} = 1 \times 10^{- 20} M^{2} K_{i n s t} [C d (C N)_{4}]^{- 2} = 7.8 \times 10^{- 18} M^{4})$ Identify the correct statement

A

$A g_{2} S$ precipitates first from the solution

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B

CdS precipitates first from the solution

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C

None of them precipitate under the given conditions

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D

$A g_{2} S$ precipitates at a sulphide concentration of $1 \times 10 - 15 M$

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Solution

The correct option is B
CdS precipitates first from the solution

$A g (C N)_{2}^{-} ⇌ A g^{+} + 2 C N^{-} K_{i n s t} = 1 \times 10^{- 20} M^{2} C d (C N)_{4}^{- 2} ⇌ C d^{+ 2} + 4 C N^{-} K_{i n s t} = 7.8 \times 10^{- 18} M^{4} ∴ [A g^{+}] = \frac{K_{i n s t} \times ⌊ A g (C N_{2}^{-}) ⌋}{[C N^{-}]^{2}} = \frac{1 \times 10^{- 20} \times 0.02}{(0.2)^{2}} = 5 \times 10^{- 21} M$
Similarly, $[C d^{+ 2}] = \frac{7.8 \times 10^{- 18} \times 0.02}{(0.2)^{4}} M = 9.75 \times 10^{- 17} M$
$[S^{- 2}]$ needed to precipitate $[A g^{+}] = \frac{K_{S P} \times [A g_{2} S]}{[A g^{+}]^{2}} \frac{1 \times 10^{- 50}}{(5 \times 10^{- 21})^{2}} = 4 \times 10^{- 10} M$
Since $[S^{- 2}]$ needed to precipitate $C d^{+ 2}$ ion is smaller than that of $A g^{+}$ ion.
Therefore $C d^{+ 2}$ will precipitate first.

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Similar questions

H_{2} S

is bubbled into a 0.2 M NaCN solution which is, 0.02 M each in

A g (C N)_{2}^{⊝}

(C d (C N)_{4})^{2 -}

K_{s p}

A g_{2} S

C d S

10^{- 50}

7.1 \times 10^{- 28}

and K instability for

[A g (C N)_{2}^{⊝}]

[C d (C N)_{4}^{2 -}]

1.0 \times 10^{- 20}

7.8 \times 10^{- 18}

, which sulphide will precipitate first?

Q. A saturated solution of

H_{2} S

0.1

M

H C l

25^{o} C

S^{2 -}

ion concentration of

10^{- 23}

m o l

L^{- 1}

. The solubility product of some sulphides are

C u S = 10^{- 44}

F e S = 10^{- 14}

M n S = 10^{- 15}

C d S = 10^{- 25}

0.01 M

solution of these salts in

1 M

H C l

are saturated with

H_{2} S

, which of these will be precipitated?

p H

that must be maintained in a saturated

H_{2} S

(0.1 M) to precipitate

C d S

Z n S

x

and concentration of

C d^{2 +}

p H

y \times 10^{- 7}

x + y =

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

K_{s p} (C d S) = 8 \times 10^{- 27}

K_{s p} (Z n S) = 1 \times 10^{- 21}

K_{a 1} (H_{2} S) = 1.0 \times 10^{- 8}

K_{a 2} (H_{2} S) = 1.0 \times 10^{- 13}

K_{s p}

value for some sulphides are

K_{s p} (B i_{2} S_{3}) = 1.2 \times 10^{- 98};

K_{s p} (A g_{2} S) = 4 \times 10^{- 48};

K_{s p} (N i S) = 1.6 \times 10^{- 25}

. An unknown cation

X^{+}

was added to a solution of

N a_{2} S

. The precipitation of sulphide of the cation was found to occur before

N i S

A g_{2} S .

K_{s p}

value for the sulphide of cation

X^{+}

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}

Z n S

starts precipitating, what concentration of

C d^{2 +}

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