Question

Calculate the molar solubility of silver bromide. The Ksp of AgBr is 7.7 x 10^-13. The Kf of Ag(NH3)²⁺ is 1.5 x 10⁷.

a) in water

b) in a solution of 1.0M NH3.

ANS: a) 8.8 x 10-7 M b) 3.4 x 10-3 M

Answer 1

Now the molar solubility in a solution of 1.0 M NH3

AgBr = > Ag+ + Br- Ksp = [Ag+ ] [ Br-] = 7.7 * 10 ^ - 13

Ag+ + 2NH3 <=> Ag(NH3)2+

Kf = [Ag(NH3)2+] / [Ag+] [ NH3-] ^2 = 1.5 * 10 ^ 7

almost every Ag- ion formed from AgBr dissolution is transformed into the complex. ( Kf is very large and favours the RHS)

For each Ag + that dissolves 1 Br- ion is released

so the [Br-] ~ [Ag(NH3)2+] = c

Combining the equation and solving for [Ag+}

c^2 = Ksp * Kf * [NH3] ^2

c^2 = 7.7 * 10 ^-13 * 1.5* 10^7 * (1-2c)^2

0.9999538 c² + 0.0000462 c - 0.00001155 = 0

c = 3.4 * 10^-3 moles per liter

[Br-] which results from the dissolution of the

AgBr = 3.4* 10^ -3 moles/ liter

solubility of AgBr in 1.0 M NH3 is 3.4 * 10 ^ -3 M