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Chemistry
If H2S gas is bubbled into 0.2 M NaCN solution which is 0.02 M in each [Cd(CN)4]2- and [Ag(CN)2]-.then which option will be correct if H2S produces 1 x 10-9 M sulphide ion in the solution ( Given Ksp Ag2S = 1 x 10-50 M3; Ksp CdS = 7.1 x 10-28 M2 Kinst [Ag(CN)2]- = 1 x 10-20M2 Kinst [Cd(CN)4]2- = 7.8 x 10-18 M4 )

Q. If H₂S gas is bubbled into 0.2 M NaCN solution which is 0.02 M in each [Cd(CN)₄]^2- and [Ag(CN)₂]^-.then which option will be correct if H₂S produces 1 x 10^-9 M sulphide ion in the solution
( Given K_sp Ag₂S = 1 x 10^-50 M³; K_sp CdS = 7.1 x 10^-28 M²
K_inst [Ag(CN)₂]^- = 1 x 10^-20M² K_inst [Cd(CN)₄]^2- = 7.8 x 10^-18 M⁴)

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Ag₂S precipitates first from the solution

CdS precipitates first from the solution

None of them precipitate under the given conditions

Ag₂S precipitates at a sulphide concentration of 1 x 10^-15 M

Solution:

From question
$\text{Ag} \left(\text{CN}\right)_{2}^{-} \rightleftharpoons \left(\text{Ag}\right)^{+} + 2 \left(\text{CN}\right)^{-} \left(\text{K}\right)_{\text{inst}} = 1 \times 1 0^{- 2 0} \left(\text{M}\right)^{2}$
$\text{Cd} \left(\text{CN}\right)_{4}^{2 -} \rightleftharpoons \left(\text{Cd}\right)^{2 +} + 4 \left(\text{CN}\right)^{-} \left(\text{K}\right)_{\text{inst}} = \text{7.8} \times 1 0^{- 1 8} \left(\text{M}\right)^{4}$
$∴ \, \, \left[\left(\text{Ag}\right)^{+}\right] = \frac{\left(\text{K}\right)_{\text{inst}} \times \left[\text{Ag} \left(\text{CN}\right)_{2}^{-}\right]}{\left(\left[\left(\text{CN}\right)^{-}\right]\right)^{2}}$
$= \frac{1 \times 1 0^{- 2 0} \times \text{0.02}}{\left(\text{0.2}\right)^{2}} = 5 \times 1 0^{- 2 1} \text{M}$
Similarly, $\left[\left(\text{Cd}\right)^{2 +}\right] = \frac{\text{7.8} \times 1 0^{- 1 8} \times \text{0.02}}{\left(\text{0.2}\right)^{4}} \text{M}$
$= \text{9.75} \times 1 0^{- 1 7} \text{M}$
[S^2-] needed to precipitate [Ag⁺]
$= \frac{\text{K}_{\text{sp}} \left[\text{Ag}_{2} \text{S}\right]}{\left[\text{Ag}^{+}\right]^{2}}$
$= \frac{1 \times 1 0^{- 5 0}}{\left(5 \times 1 0^{- 2 1}\right)^{2}} = 4 \times 1 0^{- 1 0} \text{M}$
[S²^-] needed to precipitate [Cd²⁺]
$\frac{\text{7.1} \times 1 0^{- 2 8}}{\text{9.75} \times 1 0^{- 1 7}} = \frac{\text{K}_{\text{sp}} \left[\text{CdS}\right]}{\left[\text{Cd}^{2 +}\right]} = \text{7.203} \times 1 0^{- 1 2} \text{M}$
Since [S²^-] needed to precipitate Cd²⁺ ion is smaller than that of Ag⁺ ion.

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Ag2S precipitates first from the solution