Chapter 22, Problem 22.23P

Interpretation Introduction

Interpretation:

The solubility in grams per liter of $AgBr\left(s\right)$ in $0.35MN{H}_3$ 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 22.23P

The solubility in grams per liter of $AgBr\left(s\right)$ is given as $0.2g{L}^{-1}$.

Explanation of Solution

The two applicable equations are given below:

  $\begin{array}{c}AgBr\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+B{r}_{}^{-}\left(aq\right)K_{sp}=5.4\times {10}^{-13}{M}^2\left(1\right)\\ \\ A{g}^{+}\left(aq\right)+2N{H}_3\rightleftharpoons {\left[Ag\left(N{H}_3\right)\right]}_2^{+}\left(aq\right)K_f=2.0\times {10}^7{M}^{-2}\left(2\right)\end{array}$

Addition of the both equation will give,

  $\begin{array}{c}AgBr\left(s\right)+2N{H}_3\rightleftharpoons {\left[Ag\left(N{H}_3\right)\right]}_2^{+}\left(aq\right)+B{r}_{}^{-}\left(aq\right)\left(3\right)\\ \\ K_c=K_{sp}.K_f\\ =5.4\times {10}^{-13}{M}^2\times 2.0\times {10}^7{M}^{-2}\\ =1.1\times {10}^{-5}\end{array}$

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

  $\begin{array}{l}AgBr\left(s\right)+2N{H}_3\rightleftharpoons {\left[Ag\left(N{H}_3\right)\right]}_2^{+}\left(aq\right)+B{r}_{}^{-}\left(aq\right)\\ \\ Initial0.35M0M\approx 0M\\ \\ Change-2x+x+x\\ \\ Equilibrium0.35M-2xxx\end{array}$

The expression for equilibrium constant is,

  $\begin{array}{c}{K}_c=\frac{x^2}{{\left(0.35M-2x\right)}^2}=1.1\times {10}^{-5}\\ \\ \frac{x}{\left(0.35M-2x\right)}=3.3\times {10}^{-3}\\ \\ x=0.00115\\ \\ {\left[Ag\left(N{H}_3\right)\right]}_2^{+}=x=0.00115\\ \\ \left[N{H}_3\right]=0.35M-2x=0.35M-\left(2\times 0.00115\right)\\ =0.3477\\ \end{array}$

The concentration of $A{g}^{+}$ can be calculated as follows:

  $\begin{array}{c}{K}_f=\frac{{\left[Ag\left(N{H}_3\right)\right]}_2^{+}}{\left[A{g}^{+}\right]{\left[N{H}_3\right]}^2}=2.0\times {10}^7{M}^{-2}\\ \\ \left[A{g}^{+}\right]=\frac{{\left[Ag\left(N{H}_3\right)\right]}_2^{+}}{2.0\times {10}^7{M}^{-2}{\left[N{H}_3\right]}^2}=\frac{0.00115M}{2.0\times {10}^7{M}^{-2}{\left[0.3477M\right]}^2}\\ \\ \left[A{g}^{+}\right]=4.8\times {10}^{-10}M\end{array}$

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

  $\begin{array}{c}s=\left[A{g}^{+}\right]+{\left[Ag\left(N{H}_3\right)\right]}_2^{+}\\ \\ =4.8\times {10}^{-10}M+0.00115M\\ \\ s\approx 1.15\times {10}^{-3}M\end{array}$

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

  $\begin{array}{c}s=1.15\times {10}^{-3}mol.L^{-1}\left(\frac{187.77g}{1molAgBr}\right)\\ \\ s=0.2g{L}^{-1}\end{array}$

The solubility in grams per liter of $AgBr\left(s\right)$ is calculated to be $0.2g{L}^{-1}$.