Chapter 17, Problem 78A

Interpretation Introduction

Interpretation:

The solubility of lead arsenate is to be determined when solubility product is given at a particular temperature.

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

The solubility product is given as

${\text{K}}_{\text{sp}}\text{ = }{\left[{\text{Pb}}^{\text{+3}}\right]}^{\text{3}}\text{×}{\left[{\text{ASO}}_{\text{4}}{}^{\text{-3}}\right]}^{\text{2}}$

Answer to Problem 78A

The molar solubility of lead arsenate is $\text{0}{\text{.8165× 10}}^{\text{-18}}$ mol/L

Explanation of Solution

Given information : The solubility product of lead arsenate is $\text{4}{\text{.0×10}}^{\text{-36}}$

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

${\text{ Pb}}_{\text{3}}{\left(\text{ASO4}\right)}_{\text{2}}\rightleftharpoons {\text{3Pb}}^{\text{2+}}{\text{+ 2ASO}}_{\text{4}}{}^{\text{-3}}$

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

$\text{C = +3x +2x }$ (Change in concentration or change in amount from initial state to equilibrium)

$\text{E= 3x 2x }$ (Equilibrium concentration)

The solubility product is given as

$\begin{array}{l}{\text{ K}}_{\text{sp}}\text{= }{\left[{\text{Pb}}^{\text{2+}}\right]}^{\text{3}}\text{×}{\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]}^{\text{2}}\hfill \\ \text{ = }{\left(\text{3x}\right)}^{\text{3}}\text{×}{\left(\text{2x}\right)}^{\text{2}}\hfill \end{array}$

${\text{= 108x}}^{\text{5}}$

$\sqrt[\text{5}]{\left(\text{Ksp/108}\right)}\text{ = x}$

Or, $\text{x =}\sqrt[\text{5}]{\left({\text{4x10}}^{\text{-36}}\text{/ 108}\right)}$

$\text{= }\sqrt[\text{5}]{\text{0}{\text{.03703×10}}^{\text{-36}}}$ mol/L

$\text{= }\sqrt[\text{5}]{\text{370}{\text{.37×10}}^{\text{-40}}}$ mol/L

$\text{=3}{\text{.264×10}}^{\text{-8}}$ mol/L

Or, The molar solubility is ( $\text{3}{\text{.264×10}}^{\text{-8}}$ mol/L)

Conclusion

The solubility of lead arsenate is $\text{0}{\text{.8165×10}}^{\text{-18}}$ mol/L.