Question

The concentration of sulphide ion in $0.1M{\text{ HCl}}$ solution saturated with hydrogen sulphide is $1.0 \times {10^{ - 19}}M$ . If $10ml$ of this is added to $5ml$ of $0.04M$ solution of ${\text{FeS}}{{\text{O}}_{\text{4}}},{\text{MnC}}{{\text{l}}_{\text{2}}}{\text{,ZnC}}{{\text{l}}_{\text{2}}}$ ${\text{and CdC}}{{\text{l}}_{\text{2}}}$. In which of these solutions precipitation will take place?

Answer 1

Hint: The precipitation of solution depends upon two factors or we can the property of solutions, solubility products and ionic products. If the ionic product is greater than the solubility product then the solution will precipitate.

Complete step by step answer:
In a chemical reaction, the product which settles down at the bottom of the vessel due to the low solubility in the solution is known as precipitate. This precipitation occurs due to increase in the temperature which decreases the solubility of solids and hence precipitation occurs.
 The precipitation occurs when the ionic product is greater than that of solubility product.
Ionic product is the product of concentration of ions with their stoichiometric coefficients raised to the powers and solubility product indicates the solubility of the solid in water, more is the value of solubility product constant more the complex is soluble. So to settle down the solids as precipitate we have to maintain the solubility product lower than the ionic products.
As provided in the problem that $10ml$ of this sulphide solution is added to $5ml$ of the other solution. Hence the total volume of solution becomes $15ml$ .
Therefore the concentration of sulphide ion is; $\left[ {{{\text{S}}^{{\text{ - 2}}}}} \right] = \dfrac{{1.0 \times {{10}^{ - 19}} \times 10}}{{15}} = 6.67 \times {10^{ - 20}}M$
And the concentration of metal ions are:
 \[[F{e^{2 + }}] = [M{n^{2 + }}] = [Z{n^{2 + }}] = [C{u^{2 + }}] = \dfrac{5}{{15}} \times 0.04 = 1.33 \times {10^{ - 2}}\]
Hence the ionic product of all of these will be same which is:
$\left[ {{{\text{M}}^{{\text{ + 2}}}}} \right]\left[ {{S^{ - 2}}} \right] = 1.33 \times {10^{ - 2}} \times 6.67 \times {10^{ - 20}} = 8.77 \times {10^{ - 22}}$
But the solubility products will be different. The ionic product will be more in the case of zinc sulphide and cadmium sulphide so the ${\text{ZnC}}{{\text{l}}_{\text{2}}}{\text{ and CdC}}{{\text{l}}_{\text{2}}}$solutions will precipitate.

Note:
The solubility and the ionic product, both represent the product of concentrations of ions in the solution. The ionic product is used for the saturated and unsaturated both types of solutions but the solubility product is used for only saturated solutions.