Chapter 17.3, Problem 7RQ

Interpretation Introduction

Interpretation: The value of solubility product constant for the salt needs to be determined.

Concept introduction: When the rate at which forward and backward reaction occur become equal such that concentration of reactants and products remains to be constant, that state is described as equilibrium. It is indicated by double sided arrow $\rightleftharpoons$ . It can be physical or chemical in nature.

For an equilibrium reaction:

  $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

The expression for equilibrium constant is:

  $K=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Where,

  $K$ is equilibrium constant.

  $\left[\text{A}\right]$ and $\left[\text{B}\right]$ are reactant concentrations for A and B.

  $\left[\text{C}\right]$ and $\left[\text{D}\right]$ are product concentrations for C and D.

a, b, c and d are coefficients of reactants.

For sparingly soluble salts, $K$ replaced by $K_{\text{sp}}$ and concentration terms are replaced by solubility.

Answer to Problem 7RQ

  $K_{\text{sp}}$ for ${\text{PbCrO}}_4$ is $2.7667\times {10}^{-13}$ .

Explanation of Solution

Given information: Solubility of ${\text{PbCrO}}_4$ is $1.7\times {10}^{-4}\text{g/L}$ .

To covert unit of Solubility of ${\text{PbCrO}}_4$ from $\text{g/L}$ to $\text{mol/L}$ , we divide $1.7\times {10}^{-4}\text{g/L}$ by molar mass of ${\text{PbCrO}}_4$ .

Molar mass of ${\text{PbCrO}}_4$ is $323.1937 \text{g/mol}$ .

  $\begin{array}{c}\text{Solubility of}{\text{PbCrO}}_4=\frac{1.7\times {10}^{-4}\text{g/L}}{323.1937 \text{g/mol}}\\ =5.2600\times {10}^{-7}\text{mol/L}\end{array}$

Equilibrium reaction for ${\text{PbCrO}}_4$ is as follows:

  ${\text{PbCrO}}_4\left(s\right)\rightleftharpoons {\text{Pb}}^{2+}\left(aq\right)+{\text{CrO}}_4^{2-}\left(aq\right)$

The expression for $K_{\text{sp}}$ for above reaction is:

  $K_{\text{sp}}=\left[{\text{Pb}}^{2+}\right]\left[{\text{CrO}}_4^{2-}\right]$

Since concentration of $\left[{\text{Pb}}^{2+}\right]$ = $\left[{\text{CrO}}_4^{2-}\right]$ so, above expression can be reduced to

  $K_{\text{sp}}={\left[{\text{Pb}}^{2+}\right]}^2$

Substitute $5.2600\times {10}^{-7}\text{mol/L}$ for $\left[{\text{Pb}}^{2+}\right]$ in above equation

  $\begin{array}{c}{K}_{\text{sp}}={\left[{\text{Pb}}^{2+}\right]}^2\\ ={\left[5.2600\times {10}^{-7}\text{mol/L}\right]}^2\\ =2.7667\times {10}^{-13}\end{array}$

Conclusion

  $K_{\text{sp}}$ for ${\text{PbCrO}}_4$ is $2.7667\times {10}^{-13}$ .