Equilibrium of a Sparingly Soluble Ionic Compound

daphnelee-mh · Nov 5, 2019

Printed chemistry problem: 18. 100.0 mL of the mixed solution of NH3-NH4Cl containing 0.10 mol L-1 NH3 (Kb(NH3 H2O) = 1.78 x 10-5) is added to 100 mL of 0.20 mol L-1 MnCl2. If there is no precipitation of Mn(OH)2 (Ksp = 2.06 x 10-13), what is the mass of NH4Cl in the mixed solution?

Answer given is 6.61g.

Borek · Nov 5, 2019

You are looking for the highest concentration of OH^- that doesn't produce precipitation, don't you?

daphnelee-mh · Nov 8, 2019

Handwritten chemistry working on lined paper: Ksp of Mn(OH)2 calculation, [OH-] from Ksp and [Mn2+] = 0.10 mol/L, then NH3 and NH4+ equilibrium with Kb, giving [NH4+] = 0.64 and a mass calculation m = 34.24.

May I know which step is wrong? thanks

mjc123 · Nov 8, 2019

You have worked out the concentration in g/L. What is the mass in the actual solution?

daphnelee-mh · Nov 8, 2019

Handwritten chemistry working on lined paper: Ksp of Mn(OH)2 calculation giving [OH-] = 1.4 x 10^-6, then Kb expression for NH3 giving [NH4+] = 0.64 mol/L, followed by mass calculation n = 0.128 mol and m = 6.85 g.

Is this correct ? thanks.

mjc123 · Nov 11, 2019

It is not quite equal to the given answer; I think that's because you rounded off [OH^-] to 1.4e-6 M instead of 1.435e-6 M. Keep the extra digits through the calculation, and round off to the appropriate sig figs at the end.

Equilibrium of a Sparingly Soluble Ionic Compound

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