Question

What is the molar solubility of AgCl in 0.10 M NaCN if the colorless complex ion Ag(CN)2- forms? Ksp for AgCl is 1.8 × 10-10 and Kf for Ag(CN)2- is 1.0 × 1021

Answer 1

Answer:

0.05 M

Explanation:

The reactions are the following:

AgCl (s) ⇄ Ag⁺(aq) + Cl⁻(aq)        

$Ksp = [Ag^{+}][Cl^{-}] = 1.8 \cdot 10^{-10}$       (1)  

Ag⁺(aq) + 2CN⁻(aq)  ⇄  Ag(CN)₂⁻(aq)          

$Kf = \frac{[Ag(CN)_{2}^{-}]}{[Ag^{+}][CN^{-}]^{2}} = 1.0 \cdot 10^{21}$   (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    

$Keq = \frac{[Ag(CN)_{2}^{-})][Cl^{-}]}{[CN^{-}]^{2}}$     (3)

By introducing equation (2) and (1) into equation (3) we have:  

$Keq = \frac{[Ag(CN)_{2}^{-})]}{[CN^{-}]^{2}} \frac{Ksp}{[Ag^{+}]} = Kf \cdot Ksp = 1.0 \cdot 10^{21} \cdot 1.8 \cdot 10^{-10} = 1.8 \cdot 10^{11}$        

This equilibrium constant is equal to:

$Keq = \frac{x\cdot x}{(0.1 - 2x)^{2}} = \frac{x^{2}}{(0.1 - 2x)^{2}}$     (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!