Chapter 22, Problem 22.25P

Interpretation Introduction

Interpretation:

The solubility in grams per liter of $CuBr\left(s\right)$ in $0.50MN{H}_3$ should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

  ${\text{M}}_{\text{n}}{\text{X}}_{\text{m}}\text{(s)}\stackrel{}{\rightleftharpoons }\text{n}{\text{M}}^{\text{m+}}\text{(aq)+m}{\text{X}}^{\text{n+}}\text{(aq)}$

The relation between solubility product and molar solubility is as follows:

  ${\text{K}}_{\text{sp}}\text{=}{\left[{\text{M}}^{\text{m+}}\right]}^{\text{n}}{\left[{\text{X}}^{\text{n-}}\right]}^{\text{m}}$

Here

The solubility product of salt is ${\text{K}}_{\text{sp}}$.

The molar solubility of ${\text{M}}^{\text{m+}}$ ion is $\left[{\text{M}}^{\text{m+}}\right]$

The molar solubility of ${\text{X}}^{\text{n-}}$ ion is $\left[{\text{X}}^{\text{n-}}\right]$

Answer to Problem 22.25P

The solubility in grams per liter of $CuBr\left(s\right)$ is given as $35g{L}^{-1}$.

Explanation of Solution

The two applicable equations are given below:

  $\begin{array}{c}CuBr\left(s\right)\rightleftharpoons C{u}^{+}\left(aq\right)+B{r}_{}^{-}\left(aq\right)K_{sp}=6.3\times {10}^{-9}{M}^2\left(1\right)\\ \\ C{u}^{+}\left(aq\right)+2N{H}_3\rightleftharpoons {\left[Cu\left(N{H}_3\right)\right]}_2^{+}\left(aq\right)K_f=6.3\times {10}^{10}{M}^{-2}\left(2\right)\end{array}$

Addition of the both equation will give,

  $\begin{array}{c}CuBr\left(s\right)+2N{H}_3\rightleftharpoons {\left[Cu\left(N{H}_3\right)\right]}_2^{+}\left(aq\right)+B{r}_{}^{-}\left(aq\right)\left(3\right)\\ \\ K_c=K_{sp}.K_f\\ =6.3\times {10}^{-9}{M}^2\times 6.3\times {10}^{10}{M}^{-2}\\ =396.9\end{array}$

Since, $K_c>>K_{sp}$, equation (1) can be neglected and (3) can be used to form the concentration table.

  $\begin{array}{l}CuBr\left(s\right)+2N{H}_3\rightleftharpoons {\left[Cu\left(N{H}_3\right)\right]}_2^{+}\left(aq\right)+B{r}_{}^{-}\left(aq\right)\\ \\ Initial0.50M0M\approx 0M\\ \\ Change-2x+x+x\\ \\ Equilibrium0.50M-2xxx\end{array}$

The expression for equilibrium constant is,

  $\begin{array}{c}{K}_c=\frac{x^2}{{\left(0.50M-2x\right)}^2}=396.9\\ \\ \frac{x}{\left(0.50M-2x\right)}=19.9\\ \\ x=0.244\\ \\ {\left[Cu\left(N{H}_3\right)\right]}_2^{+}=x=0.244\\ \\ \left[N{H}_3\right]=0.50M-2x=0.50M-\left(2\times 0.244\right)\\ =0.012\end{array}$

The concentration of $C{u}^{+}$ can be calculated as follows:

  $\begin{array}{c}{K}_f=\frac{{\left[Cu\left(N{H}_3\right)\right]}_2^{+}}{\left[C{u}^{+}\right]{\left[N{H}_3\right]}^2}=6.3\times {10}^{10}{M}^{-2}\\ \\ \left[C{u}^{+}\right]=\frac{{\left[Cu\left(N{H}_3\right)\right]}_2^{+}}{6.3\times {10}^{10}{M}^{-2}{\left[N{H}_3\right]}^2}=\frac{0.244M}{6.3\times {10}^{10}{M}^{-2}{\left[0.012M\right]}^2}\\ \\ \left[C{u}^{+}\right]=2.7\times {10}^{-8}M\end{array}$

The total solubility of $CuBr\left(s\right)$ can be calculated as follows:

  $\begin{array}{c}s=\left[C{u}^{+}\right]+{\left[Cu\left(N{H}_3\right)\right]}_2^{+}\\ \\ =2.7\times {10}^{-8}M+0.244M\\ \\ s\approx 0.244M\end{array}$

Therefore, the solubility of $CuBr\left(s\right)$ in grams per liter is given below:

  $\begin{array}{c}s=0.244mol.L^{-1}\left(\frac{143.45g}{1molCuBr}\right)\\ \\ s=35g{L}^{-1}\end{array}$

The solubility in grams per liter of $CuBr\left(s\right)$ is calculated to be $35g{L}^{-1}$.