What mass of AgBr will dissolve in 500.0 mL of 2.50 M NH₃? Ksp for AgBr is 5.0 × 10⁻¹³. For the reaction Ag⁺(aq) + 2NH₃(aq) ⇌ Ag(NH₃)₂⁺(aq) Kf = 1.7 × 10⁷.

Answer 1

AgBr(s) <==> Ag⁺(aq) + Br^-(aq) ... Ksp = 5.0x10^-13

Ag⁺(aq) + 2NH₃(aq) <==> Ag(NH₃)₂⁺(aq) ... Kf = 1.7x10⁷

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AgBr(s) + 2NH₃(aq) <==> Ag(NH₃)₂⁺ + Br^-(aq) ... K = 8.5x10^-6

K = [Ag(NH₃)₂⁺][Br^-] / [NH₃]² = 8.5x10^-6

AgBr(s) + 2NH₃(aq) <==> Ag(NH₃)₂⁺ + Br^-(aq)

................2.50.....................0.................0........Initial

...............-2x.......................+x...............+x.......Change

..............2.50-2x..................x.................x........Equilibrium

8.5x10^-6 = (x)(x) / 2.50 - 2x and assuming x is small relative to 2.50 we can neglect it in the denominator

8.5x10^-6 = x²/(2.50)^2

x² = 53.125x10^-6

x = 7.29x10^-3 M

Molar mass AgBr = 188 g/mol

Mass of AgBr that will dissolve in 500.0 ml of 2.50 M NH₃ will be....

7.296x10^-3 mol/L x 0.500 L x 188 g/mol = 0.685 g AgBr