Chapter 17, Problem 51A

Interpretation Introduction

Interpretation:

The solubility product is to be calculated.

Concept Introduction :

The solubility product is the product of the concentration of its constituent ions with each ion concentration term raised to the power equal to the number of ions produced or dissociated.

Answer to Problem 51A

The solubility product of $\text{NiS}$ is $\text{0}{\text{.1573×10}}^{\text{-10}}\text{mol/L}$ .

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

  $\text{NiS}\left(\text{s}\right)\rightleftharpoons {\text{Ni}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{S}}^{\text{-2}}\left(\text{aq}\right)$

I $\text{= 0 0 }$ (initial concentration)

C $\text{= +x +x }$ (change in concentration or change in an amount from an initial state to equilibrium)

E $\text{= x x }$ (equilibrium concentration)

Given information:

Solubility $\text{=3}{\text{.6×10}}^{\text{-4}}\text{g/L}$

  $\begin{array}{l}\text{=}\frac{\text{3}{\text{.6×10}}^{\text{-4}}\text{g/L}}{\text{90}\text{.76}\text{g/mol}}\\ \text{=0}{\text{.3966×10}}^{\text{-5}}\text{mol/L}\end{array}$

Thus, solubility will be:

  $\text{0}{\text{.3996×10}}^{\text{-5}}\text{mol/L}$

The Molar mass of nickel sulphide = $\text{90}\text{.76}\text{g/mol}$

So, the solubility of nickel ion and sulphide ion is equal to $\text{0}{\text{.3996×10}}^{\text{-5}}\text{mol/L}$ .

The solubility product is calculated as,

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\left[{\text{Ni}}^{\text{+2}}\right]\text{×}\left[{\text{S}}^{\text{-2}}\right]\\ {\text{K}}_{\text{sp}}\text{=}\left(\text{0}{\text{.03966×10}}^{\text{-4}}\right)\text{×}\left(\text{0}{\text{.03966×10}}^{\text{-4}}\right)\\ {\text{K}}_{\text{sp}}\text{=0}{\text{.15729×10}}^{\text{-10}}\text{mol/L}\\ {\text{K}}_{\text{sp}}\text{=}\text{0}{\text{.1573×10}}^{\text{-10}}\text{mol/L}\end{array}$

The solubility product of $\text{NiS}$ is $\text{0}{\text{.1573×10}}^{\text{-10}}\text{mol/L}$ .

Conclusion

The solubility product of $\text{NiS}$ is $\text{0}{\text{.1573×10}}^{\text{-10}}\text{mol/L}$ .