Chapter 17, Problem 54A

Interpretation Introduction

Interpretation:

The solubility of a given compound in gram per liter 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 54A

The solubility of lead chloride in g/L is $\text{918}\text{.266}\text{g/L}$ .

Explanation of Solution

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

  ${\text{PbCl}}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{2Cl}}^{\text{-1}}\left(\text{aq}\right)$

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

C $\text{= +x +2x }$ (change in concentration or change in an amount from an initial state to equilibrium)

E $\text{= x 2x }$ (equilibrium concentration)

Given information:

Solubility product $\text{=3}{\text{.6×10}}^{\text{-3}}$

The Molar mass of lead chloride = $\text{278}\text{.1}\text{g/mol}$

The solubility of the ions iscalculated as,

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\left[{\text{Pb}}^{\text{+2}}\right]\text{×}{\left[{\text{Cl}}^{\text{-1}}\right]}^{\text{2}}\\ {\text{K}}_{\text{sp}}\text{=}\left(\text{x}\right)\text{×}\left({\text{x}}^{\text{2}}\right)\\ {\text{K}}_{\text{sp}}\text{=}{\text{x}}^{\text{3}}\text{or}\\ \text{x}\text{=}\sqrt[\text{3}]{{\text{K}}_{\text{sp}}}\\ \text{x}\text{=}\sqrt[\text{3}]{\text{3}{\text{.6×10}}^{\text{-2}}}\\ \text{=}\sqrt[\text{3}]{{\text{36×10}}^{\text{-3}}}\\ \text{x}\text{=}\text{0}\text{.331}\end{array}$

The solubility of lead and chloride ions is $\text{0}\text{.331}$ .

The solubility in gram per liter is

  $\begin{array}{l}\text{S}\text{=}\text{0}\text{.331×278}\text{.1}\\ \text{=}\text{918}\text{.266}\text{g/L}\end{array}$

Conclusion

The solubility of lead and chloride ions in gram per liter is $\text{918}\text{.266}\text{g/L}$ .