Answer:
a) ![S_{AgI} = 9.11 \cdot 10^{-9} mol/L](https://tex.z-dn.net/?f=S_%7BAgI%7D%20%3D%209.11%20%5Ccdot%2010%5E%7B-9%7D%20mol%2FL)
b) Keq = 8.3x10⁴
c)
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)
<u>Since [Ag⁺] = [I⁻]:</u>
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)
(2)
Ag⁺(aq) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) (3)
(4)
<u>The net equation is given by the sum of the reactions (1) and (3):</u>
AgI(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + I⁻(aq) (5)
<u>Hence, the equilibrium constant of the reaction (5) is:</u>
(6)
<u>From equation (2):</u>
(7)
<u>By entering equations (7) and (4) into equation (6) we have</u>:
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
![K_{eq} = \frac{[Ag(CN)_{2}^{-}][I^{-}]}{[CN^{-}]^{2}} = \frac{x*x}{(0.1 - 2x)^{2}}](https://tex.z-dn.net/?f=%20K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BAg%28CN%29_%7B2%7D%5E%7B-%7D%5D%5BI%5E%7B-%7D%5D%7D%7B%5BCN%5E%7B-%7D%5D%5E%7B2%7D%7D%20%3D%20%5Cfrac%7Bx%2Ax%7D%7B%280.1%20-%202x%29%5E%7B2%7D%7D%20)
<u>Solving the above equation for x:</u>
![x = 0.05 mol/L = S_{[ AgI]}](https://tex.z-dn.net/?f=%20x%20%3D%200.05%20mol%2FL%20%3D%20S_%7B%5B%20AgI%5D%7D%20)
Hence, the molar solubility of AgI in a NaCN solution is 0.05 M.
I hope it helps you!