Chapter 17, Problem 108AP

Interpretation Introduction

Interpretation:

The concentration of ${\text{Ba}}^{\text{2+}}$, ${\text{Sr}}^{\text{2+}}$, and ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ is to be calculated with given $K_{\text{sp}}$ value.

Concept introduction:

If the stoichiometry of the precipitated salt is same, then the salt with the minimum solubility product constant will be precipitated first, followed by higher solubility product constant.

If the ionic product is less than the solubility product constant, it means the solution is unsaturated, that is, more salt can be dissolved in it. If ionic product is greater than the solubility product constant.

The solubility product of the sparingly soluble salt is given as the product of the concentration of ions raise to the power equal to the number of times the ion occurs in the equation after the dissociation of electrolyte.

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

Answer to Problem 108AP

Solution: The concentration of $\left[{\text{Ba}}^{\text{2+}}\right]$ is $1.8\times {10}^{-7}\text{M}$, the concentration of $\left[{\text{Sr}}^{\text{2+}}\right]$ is $6.2\times {10}^{-4}\text{M}$, and the concentration of $\left[{\text{SO}}_4^{2-}\right]$ is $6.2\times {10}^{-4}\text{ M}$.

Explanation of Solution

Given information: The solubility product constant of ${\text{SrSO}}_4$ is $3.8\times {10}^{-7}$ and the solubility product constant of ${\text{BaSO}}_4$ is $1.1\times {10}^{-10}$.

The concentration of $\left[{\text{SO}}_4^{2-}\right]$ is calculated as follows:

If the stoichiometry of the precipitated salt is same, then the salt with the minimum solubility product constant will be precipitated first, followed by higher solubility product constant.

${\text{SrSO}}_4\left(s\right)\rightleftharpoons {\text{Sr}}^{2+}\left(aq\right)+{\text{ SO}}_4^{2-}\left(aq\right)$ $K_{\text{sp}}\left({\text{SrSO}}_4\right)=3.8\times {10}^{-7}$

${\text{BaSO}}_4\left(s\right)\rightleftharpoons {\text{Ba}}^{2+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)$ $K_{\text{sp}}\left({\text{BaSO}}_4\right)=1.1\times {10}^{-10}$

Comparing the $K_{\text{sp}}$ values of both the compounds. $K_{\text{sp}}{\text{ of SrSO}}_{\text{4}}$ is higher as compared to $K_{\text{sp}}$. ${\text{SrSO}}_{\text{4}}$ is more soluble as compared to ${\text{BaSO}}_{\text{4}}$.

Therefore, all the ${\text{SO}}_4^{2-}$ ions come from ${\text{SrSO}}_{\text{4}}$.

Summarise the concentration at equilibrium as:

Consider s to be the molar solubility.

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

The $K_{\text{sp}}$ value of ${\text{SrSO}}_{\text{4}}$ is $3.8\times {10}^{-7}$.

The equilibrium expression for a reaction is written as follows:

$K_{\text{sp}}{\text{= [Sr}}^{\text{2+}}{\text{][SO}}_4^{2-}\text{]}$

Here, $K_{\text{sp}}$ is the solubility product constant, ${\text{[Sr}}^{\text{2+}}\text{]}$ is the concentration of strontium ion, and ${\text{[SO}}_4^{2-}\text{]}$ is the concentration sulphate ion.

Substitute the value of $K_{\text{sp}}$, ${\text{[Sr}}^{\text{2+}}\text{]}$, and ${\text{[SO}}_4^{2-}\text{]}$ in the above expression,

$\begin{array}{l}{K}_{\text{sp}}=[s][s]\\ 3.8\times {10}^{-7}=s^2\\ s={\left(3.8\times {10}^{-7}\right)}^{\frac{1}{2}}\\ s=6.2\times {10}^{-4}\text{M}\end{array}$

Concentration of $\left[{\text{Sr}}^{2+}\right]$ is equal to the concentration of $\left[{\text{SO}}_4^{2-}\right]$, which is as follows:

$s=6.2\times {10}^{-4}\text{M}$

Now,

Calculating the concentration of $\left[{\text{Ba}}^{2+}\right]$:

For ${\text{BaSO}}_{\text{4}}$,

Summarize the concentration at equilibrium as follows:

Consider s to be the molar solubility.

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

$K_{\text{sp}}{\text{ of BaSO}}_4=1.1\times {10}^{-10}$

The equilibrium expression for a reaction is written as:

$K_{\text{sp}}{\text{= [Ba}}^{\text{2+}}{\text{][SO}}_{\text{4}}{}^{\text{2-}}\text{]}$.

Here, $K_{\text{sp}}$ is the solubility product constant, ${\text{[Ba}}^{\text{2+}}\text{]}$ is the concentration of barium ion, and ${\text{[SO}}_4^{2-}\text{]}$ is the concentration sulphate ion.

Substitute the value of $K_{\text{sp}}$, ${\text{[Ba}}^{\text{2+}}\text{]}$, and ${\text{[SO}}_4^{2-}\text{]}$ in the above expression:

$\begin{array}{l}\text{1}\text{.1}\times {\text{10}}^{-10}={\text{ [Ba}}^{\text{2}+}{\text{][SO}}_4^{2-}\text{]}\\ {\text{[Ba}}^{\text{2}+}\text{]}=\frac{\text{1}\text{.1}\times {\text{10}}^{-10}}{{\text{[SO}}_4^{2-}\text{]}}\\ =\frac{\text{1}\text{.1}\times {\text{10}}^{-10}}{6.2\times {10}^{-4}}\\ =1.8\times {10}^{-7}\text{M}\end{array}$

The concentration of $\left[{\text{Ba}}^{\text{2+}}\right]$ is $1.8\times {10}^{-7}\text{M}$.

Conclusion

The concentration of $\left[{\text{Ba}}^{\text{2+}}\right]$ is $1.8\times {10}^{-7}\text{M}$, the concentration of $\left[{\text{Sr}}^{\text{2+}}\right]$ is $6.2\times {10}^{-4}\text{M}$, and the concentration of $\left[{\text{SO}}_4^{2-}\right]$ is $6.2\times {10}^{-4}\text{ M}$.