Answer:
a) [tex]S_{AgI} = 9.11 \cdot 10^{-9} mol/L[/tex]
b) Keq = 8.3x10⁴
c) [tex] S_{[ AgI]} = 0.05 M [/tex]
Explanation:
a) To find the molar solubility of AgI in water we need to use the solubility product constant (Ksp) of the following reaction:
AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)
[tex] K_{sp} = [Ag^{+}][I^{-}] = 8.3\cdot 10^{-17} [/tex]
Since [Ag⁺] = [I⁻]:
[tex]K_{sp} = [Ag^{+}]^{2} \rightarrow S = \sqrt{K_{sp}} = \sqrt{8.3\cdot 10^{-17}} = 9.11 \cdot 10^{-9} mol/L[/tex]
Hence, the molar solubility of AgI in water is 9.11x10⁻⁹ mol/L.
b) The equilibrium constant of AgI in CN, first we need to evaluate the reactions involved:
AgI(s) ⇄ Ag⁺(aq) + I⁻(aq) (1)
[tex] K_{sp} = [Ag^{+}][I^{-}] [/tex] (2)
Ag⁺(aq) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) (3)
[tex] K_{f} = \frac{[Ag(CN)_{2}^{-}]}{[Ag^{+}][CN^{-}]^{2}} [/tex] (4)
The net equation is given by the sum of the reactions (1) and (3):
AgI(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + I⁻(aq) (5)
Hence, the equilibrium constant of the reaction (5) is:
[tex] K_{eq} = \frac{[Ag(CN)_{2}^{-}][I^{-}]}{[CN^{-}]^{2}} [/tex] (6)
From equation (2):
[tex] [I^{-}] = \frac{K_{sp}}{[Ag^{+}]} [/tex] (7)
By entering equations (7) and (4) into equation (6) we have:
[tex] K_{eq} = \frac{[Ag(CN)_{2}^{-}]}{[CN^{-}]^{2}}*\frac{K_{sp}}{[Ag^{+}]} [/tex]
[tex] K_{eq} = K_{f}*K_{sp} = 1.0 \cdot 10^{21}*8.3\cdot 10^{-17} = 8.3 \cdot 10^{4} [/tex]
Therefore, the equilibrium constant for the reaction (5) is 8.3x10⁴.
c) From reaction (5):
AgI(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + I⁻(aq)
0.1 - 2x x x
[tex] K_{eq} = \frac{[Ag(CN)_{2}^{-}][I^{-}]}{[CN^{-}]^{2}} = \frac{x*x}{(0.1 - 2x)^{2}} [/tex]
[tex] x^{2} - 8.3 \cdot 10^{4}*(0.1 - 2x)^{2} = 0 [/tex]
Solving the above equation for x:
[tex] x = 0.05 mol/L = S_{[ AgI]} [/tex]
Hence, the molar solubility of AgI in a NaCN solution is 0.05 M.
I hope it helps you!
