Chapter 15, Problem 15.118SP

Interpretation Introduction

Interpretation:

Whether $\text{FeS}$ gets precipitate or not if the pH of the given solution is adjusted to $\text{pH 8}$ with a ${\text{NH}}_{\text{4}}{}^{\text{+}}{\text{-NH}}_{\text{3}}$ buffer has to be explained.

Concept Introduction:

Solubility product in acid:

The mixture of metal cations ${\text{M}}^{\text{2+}}$ can be separated by precipitation of metal sulphides.

In this precipitation, the metal cation are classified into two types they are very insoluble sulphides and more soluble sulphides. The more soluble sulphides are ${\text{Mn}}^{\text{2+}}{\text{, Fe}}^{\text{2+}}{\text{, Co}}^{\text{2+}}{\text{, Ni}}^{\text{2+}}{\text{, and Zn}}^{\text{2+}}$. The very insoluble sulphides are ${\text{Pb}}^{\text{2+}}{\text{, Cu}}^{\text{2+}}{\text{, andHg}}^{\text{2+}}$

These solubility difference is used for separation of metal cation in the mixture in acidic condition. The equilibrium reaction of metal sulphide in acid is,

${\text{MS}}_{\text{(s)}}{\text{ + 2 H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{M}}^{\text{2+}}{}_{\text{(aq) }}{\text{+ H}}_{\text{2}}{\text{S}}_{\text{(aq)}}{\text{ + 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The equilibrium constant of above reaction is called the solubility product ${\text{K}}_{\text{spa}}$ in acid, it is given by,

${\text{K}}_{\text{spa}}\text{=}\frac{{\text{[M}}^{\text{2+}}{\text{][H}}_{\text{2}}\text{S]}}{{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^{\text{2}}}$

The reaction quotient ${\text{Q}}_{\text{c}}$ for the above reaction at given time is,

${\text{Q}}_{\text{c}}\text{=}\frac{{{\text{[M}}^{\text{2+}}\text{]}}_{\text{t}}{{\text{[H}}_{\text{2}}\text{S]}}_{\text{t}}}{{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^{\text{2}}{}_{\text{t}}}$

Precipitation will occurs if ${\text{Q}}_{\text{c}}{\text{ > K}}_{\text{spa}}$ and precipitation will not occurs if ${\text{Q}}_{\text{c}}{\text{ < K}}_{\text{spa}}$.