Answer:
0.05 M
Explanation:
The reactions are the following:
AgCl (s) ⇄ Ag⁺(aq) + Cl⁻(aq)
[tex] Ksp = [Ag^{+}][Cl^{-}] = 1.8 \cdot 10^{-10} [/tex] (1)
Ag⁺(aq) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq)
[tex] Kf = \frac{[Ag(CN)_{2}^{-}]}{[Ag^{+}][CN^{-}]^{2}} = 1.0 \cdot 10^{21} [/tex] (2)
The reaction of the solubility of AgCl in NaCN is:
AgCl (s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + Cl⁻(aq)
0.1M - 2x x x
[tex] Keq = \frac{[Ag(CN)_{2}^{-})][Cl^{-}]}{[CN^{-}]^{2}} [/tex] (3)
By introducing equation (2) and (1) into equation (3) we have:
[tex] Keq = \frac{[Ag(CN)_{2}^{-})]}{[CN^{-}]^{2}} \frac{Ksp}{[Ag^{+}]} = Kf \cdot Ksp = 1.0 \cdot 10^{21} \cdot 1.8 \cdot 10^{-10} = 1.8 \cdot 10^{11} [/tex]
This equilibrium constant is equal to:
[tex] Keq = \frac{x\cdot x}{(0.1 - 2x)^{2}} = \frac{x^{2}}{(0.1 - 2x)^{2}} [/tex] (4)
where x : is the molar solubility of the Ag(CN)₂⁻ and Cl⁻ and hence the molar solubility of AgCl in NaCN
By solving equation (4) for x we have:
S = 0.05 M = [Ag(CN)₂⁻] = [Cl⁻]
Therefore, the molar solubility of AgCl in 0.10 M NaCN is 0.05 M.
I hope it helps you!
