Chapter 17, Problem 118AP

${\text{AgNO}}_{\text{3}}$ is added slowly to a solution that contains 0.1 M each of ${\text{Br}}^{\text{-}}{\text{, CO}}_3^{2-},{\text{ and SO}}_4^{2-}$ ions. What compound will precipitate first and what compound will precipitate last?

Interpretation Introduction

Interpretation:

The compoundsthatprecipitate out firstand last areto be identified with given concentration of ions.

Concept introduction:

The amount of solute dissolved in a given volume of the solvent to form a saturated solution at a given temperature is termed as the solubility of the solute in the solvent at that temperature.

The solubility product of a sparingly soluble salt is given as the product of the concentration of the ions raised to the power equal to the respective stoichiometric coefficients after the dissociation of the electrolyte.

Number of moles of solute dissolved per litre of solution is called molar solubility.

The unit of molar solubility is $\text{mol}/\text{L}$.

Molar solubility can be evaluated from the solubility product constant $\left({\text{K}}_{\text{sp}}\right)$ and stoichiometry.

When equilibrium is reached between a solid and its constituent ions in a solution, it is known as the solubility product constant $\left({\text{K}}_{\text{sp}}\right)$.

At a given temperature, the product of molar concentrations of the ions of salt present in the solution is known as the solubility product of the salt. It is represented by $K_{sp}$.

Higher is the value of solubility product of a salt, more is its solubility.

The presence of common ions in the solution decreases the solubility of a given compound.

For a general reaction: $\text{AB}\left(s\right)\rightleftharpoons {\text{A}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)$

The solubility product can be calculated by the expression as: ${\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{][B}}^{-}\text{]}$

Here, ${\text{K}}_{\text{sp}}$ is the solubility product constant and $sp$ stands for solubility product.

The solubility product $\left({\text{K}}_{\text{sp}}\right)$ is considered as constant, which means that it will not be affected by the common ion. On the other hand, solubility ($\text{s}$) will be affected due to the presence of the common ion.

The more the molar concentration of the ions, the more time it’ll take to precipitate. The less the molar concentration of the ions, the lesser time it’ll take to precipitate.

Answer to Problem 118AP

Solution: No, comparison of two salts with similar formulas that have the same number of cations and anions is possible.

Explanation of Solution

Given information: The concentration of $\left[{\text{Br}}^{-}\right]$, $\left[{\text{CO}}_3^{2-}\right]$, and $\left[{\text{SO}}_4^{2-}\right]$ in the solution is $0.1\text{M}$.

The compound ${\text{AgNO}}_3$ is a strong electrolyte. It dissociates into ${\text{Ag}}^{+}$ and ${\text{NO}}_3^{-}$ in an aqueous solution. Adding ${\text{AgNO}}_3$ in a solution that contains $\left[{\text{Br}}^{-}\right]$, $\left[{\text{CO}}_3^{2-}\right]$ and $\left[{\text{SO}}_4^{2-}\right]$ will cause the slightly-soluble ionic compounds $\text{AgBr}\left(s\right)$, ${\text{Ag}}_2{\text{CO}}_3\left(s\right)$ and ${\text{Ag}}_2{\text{SO}}_4\left(s\right)$ to precipitate out from the solution.

The equation for the dissociation of $\text{AgBr}\left(s\right)$ in water is as follows:

$\text{AgBr}\left(s\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)$

Consider $s$ to be the molar solubility.

The concentration of $\left[{\text{Ag}}^{+}\right]$ is calculated as follows:

The ICE table for theionization of $\text{AgBr}\left(s\right)$ is as follows:

$\begin{array}{l}\text{AgBr}\left(s\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{ Br}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0.1\\ \text{Change}\left(\text{M}\right)& -s& +s& +s\\ \text{Equilibrium}\left(\text{M}\right)& -& s& 0.1+s\end{array}\end{array}$

The equilibrium expression for a reaction is written as follows:

${\text{K}}_{\text{sp}}{\text{= [Ag}}^{\text{+}}{\text{][Br}}^{-}\text{]}$

Here, ${\text{K}}_{\text{sp}}$ is the solubility product constant, $\left[{\text{Ag}}^{+}\right]$ is the concentration of the silver ions, and $\left[{\text{Br}}^{-}\right]$ is the concentration of the bromine ion.

Substitute the values of $\left[{\text{Ag}}^{+}\right]$, $\left[{\text{Br}}^{-}\right]$, and ${\text{K}}_{\text{sp}}$ of $\text{AgBr}\left(s\right)$ as $7.7\times {10}^{-13}$ in the above expression,

${\text{K}}_{\text{sp}}=\left[s\right]\left[0.1+s\right]$

The value of $s$ is small as compared to $0.1$. It can be neglected.

On solving further,

$\begin{array}{c}\text{7}\text{.7}\times {\text{10}}^{-13}\text{= }\left(s\right)\left(0.1\right)\\ s=7.7\times {10}^{-12}\text{M}\end{array}$

The concentration of $\left[{\text{Ag}}^{+}\right]=s$

Hence, the concentration of $\left[{\text{Ag}}^{+}\right]$ is $7.7\times {10}^{-12}\text{M}$.

The equation for the dissociation of ${\text{Ag}}_2{\text{CO}}_3\left(s\right)$ in water is as follows:

${\text{Ag}}_2{\text{CO}}_3\left(s\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)$

The concentration of $\left[{\text{Ag}}^{+}\right]$ is calculated as follows:

The ICE tablefor the ionisation of ${\text{Ag}}_2{\text{CO}}_3\left(s\right)$ is as follows:

Consider $s$ to be the molar solubility.

$\begin{array}{l}{\text{Ag}}_2{\text{CO}}_3\left(s\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(aq\right)+{\text{ CO}}_3^{2-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0.1\\ \text{Change}\left(\text{M}\right)& -s& +2s& +s\\ \text{Equilibrium}\left(\text{M}\right)& -& 2s& 0.1+s\end{array}\end{array}$

The equilibrium expression for a reaction is written as follows:

${\text{K}}_{\text{sp}}{\text{= [Ag}}^{\text{+}}{\text{]}}^2{\text{[CO}}_3^{2-}\text{]}$

Here, ${\text{K}}_{\text{sp}}$ is the solubility product constant, $\left[{\text{Ag}}^{+}\right]$ is the concentration of the silver ion and $\left[{\text{CO}}_3^{2-}\right]$ is the concentration of the carbonate ion.

Substitute the values of $\left[{\text{Ag}}^{+}\right]$, $\left[{\text{CO}}_3^{2-}\right]$, and the ${\text{K}}_{\text{sp}}$ of ${\text{Ag}}_2{\text{CO}}_3\left(s\right)$ as $8.1\times {10}^{-12}$ in the above expression,

${\text{K}}_{\text{sp}}={\left[2s\right]}^2\left[0.1+s\right]$

The value of $s$ is small as compared to $0.1$. It can be neglected.

$\begin{array}{c}\text{8}\text{.1}\times {\text{10}}^{-12}\text{= }{\left(2s\right)}^2\left(0.1\right)\\ 0.4s^2=8.1\times {10}^{-12}\text{M}\\ \text{s}\text{=}\text{9}\text{.0}\times {\text{10}}^{-6}\text{M}\end{array}$

The concentration of $\left[{\text{Ag}}^{+}\right]=2s$

Hence, the concentration of $\left[{\text{Ag}}^{+}\right]$ is $9.0\times {10}^{-6}\text{M}$.

The equation for the dissociation of ${\text{Ag}}_2{\text{SO}}_4\left(s\right)$ in water is as follows:

${\text{Ag}}_2{\text{SO}}_4\left(s\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)$

The concentration of $\left[{\text{Ag}}^{+}\right]$ is calculated as follows:

The ICE table for the ionisation of ${\text{Ag}}_2{\text{SO}}_4\left(s\right)$ is as follows:

Consider $s$ to be the molar solubility.

$\begin{array}{l}{\text{Ag}}_2{\text{SO}}_4\left(s\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(aq\right)+{\text{ SO}}_4^{2-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0.1\\ \text{Change}\left(\text{M}\right)& -s& +2s& +s\\ \text{Equilibrium}\left(\text{M}\right)& -& 2s& 0.1+s\end{array}\end{array}$

The equilibrium expression for a reaction is written as follows:

${\text{K}}_{\text{sp}}{\text{= [Ag}}^{\text{+}}{\text{][SO}}_4^{2-}\text{]}$

Here, ${\text{K}}_{\text{sp}}$ is the solubility product constant, $\left[{\text{Ag}}^{+}\right]$ is the concentration of the silver ions, and $\left[{\text{SO}}_4^{2-}\right]$ is the concentration of the sulphate ion.

Substitute the values of $\left[{\text{Ag}}^{+}\right]$, $\left[{\text{SO}}_4^{2-}\right]$, and ${\text{K}}_{\text{sp}}$ of ${\text{Ag}}_2{\text{SO}}_4\left(s\right)$ as $1.5\times {10}^{-5}$ in the above expression.

${\text{K}}_{\text{sp}}={\left[2s\right]}^2\left[0.1+s\right]$

The value of $s$ is small as compared to $0.1$. It can be neglected.

$\begin{array}{c}\text{1}\text{.5}\times {\text{10}}^{-5}\text{= }{\left(2s\right)}^2\left(0.1\right)\\ 0.4s^2=1.5\times {10}^{-5}\\ s=6.1\times {10}^{-3}\text{M}\end{array}$

The concentration of $\left[{\text{Ag}}^{+}\right]=2s$

Hence, the concentration of $\left[{\text{Ag}}^{+}\right]$ is $1.2\times {10}^{-2}\text{M}$.

The order of precipitation depends upon the molar concentration of ${\text{Ag}}^{+}$ in the solution. The more the molar concentration of the ions, the more time it’ll take to precipitate. The less the molar concentration of the ions, the lesser time it’ll take to precipitate.

$\text{AgBr}\left(s\right)$ will precipitate first because the molar concentration of $\left[{\text{Ag}}^{+}\right]$ is less than the molar concentration of $\left[{\text{Ag}}^{+}\right]$ in the other compound. ${\text{Ag}}_2{\text{SO}}_4\left(s\right)$ will precipitate last because the molar concentration of $\left[{\text{Ag}}^{+}\right]$ is more than the molar concentration of $\left[{\text{Ag}}^{+}\right]$ in the other compound.

Conclusion

$\text{AgBr}\left(s\right)$ will precipitate first, then ${\text{Ag}}_2{\text{CO}}_3$, and finally ${\text{Ag}}_2{\text{SO}}_4\left(s\right)$.