Use values of Ksp for AgI and Kf for Ag(CN)2− to calculate the molar solubility of AgI in pure water, calculate the equilibrium constant for the reaction Agl(s)+2CN−(aq)⥫⥬==Ag(CN)−2(aq)+I−( aq), determine the molar solubility of AgI in a 0.100 M NaCN solution.

a)

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)

Since [Ag⁺] = [I⁻]:

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)

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:

  (6)

From equation (2):

  (7)

By entering equations (7) and (4) into equation (6) we have:

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

Solving the above equation for x:

Hence, the molar solubility of AgI in a NaCN solution is 0.05 M.

I hope it helps you!