The -Ag- in a solution prepared by shaking Ag2S with saturated H2S 0-01M in a solution of 0-15M H3O- is X-10-15M- value of X is- Ksp of Ag2S-1-6-10-49 and K1- K2 for H2S are 1-0-10-7 and 1-0-10-14 respectively-

For the dissociation H_2S ⇌ 2H^+ +S^2 , K_1 × K_2 = 1.0 × 10^ 7× 1.0 × 10^ 14 = 1.0 × 10^ 21 . H_2S H^+ S^2 Equilibrium concentrations ...

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Byju's Answer

Standard XII

Chemistry

Solubility Product

The [Ag+] i...

Question

The $[{A g}^{+}]$ in a solution prepared by shaking ${A g}_{2} S$ with saturated $H_{2} S$ ( $0.01 M$ ) in a solution of $0.15 M$ $H_{3} O^{+}$ is $X \times 10^{- 15} M$ , value of $X$ is:
$K_{s p}$ of ${A g}_{2} S = 1.6 \times 10^{- 49}$ and $K_{1}$ , $K_{2}$ for $H_{2} S$ are $1.0 \times 10^{- 7}$ and $1.0 \times 10^{- 14}$ respectively.

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Solution

For the dissociation $H_{2} S ⇌ 2 H^{+} + S^{2 -}$ , $K_{1} \times K_{2} = 1.0 \times 10^{- 7} \times 1.0 \times 10^{- 14} = 1.0 \times 10^{- 21}$ .

$H_{2} S$
$H^{+}$
$S^{2 -}$
Equilibrium concentrations (M)
0.01
0.15
x
The equilibrium constant expression is
$K = \frac{[H^{+}]^{2} [S^{2 -}]}{[H_{2} S]}$
$1.0 \times 10^{- 21} = \frac{x (0.15)^{2}}{0.01}$
$x = 4.44 \times 10^{- 22}$ .
$[A g^{+}]^{2} = \frac{K_{s p}}{[S^{2}]} = \frac{1.6 \times 10^{- 49}}{4.44 \times 10^{- 22}} = 3.6 \times 10^{- 27}$
$[A g^{+}] = 6 \times 10^{- 14} = 60 \times 10^{- 15}$
Thus, $X = 60$

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	$H_{2} S$	$H^{+}$	$S^{2 -}$
Equilibrium concentrations (M)	0.01	0.15	x

The [Ag+] in a solution prepared by shaking Ag2S with saturated H2S (0.01M) in a solution of 0.15M H3O+ is X×10−15M, value of X is: Ksp of Ag2S=1.6×10−49 and K1, K2 for H2S are 1.0×10−7 and 1.0×10−14 respectively.