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- 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 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 )
Solution:
From question
Ag(CN)−2⇌(Ag)++2(CN)−(K)inst=1×10−20(M)2
Cd(CN)2−4⇌(Cd)2++4(CN)−(K)inst=7.8×10−18(M)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}
[S2-] 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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