Chapter 15, Problem 15.BCP

(a)

Interpretation Introduction

Interpretation:

The solubility of barium ion in pure water has to be determined.

Concept Introduction:

The solubility product for an equilibrium reaction is defined as the product of ion concentration raised to power their stoichiometric coefficient.   The symbol for solubility product is $K_{\text{sp}}$.  If $K_{\text{sp}}$ value is higher, it means that compound is highly soluble.

Answer to Problem 15.BCP

The solubility of barium ion in pure water is $\underset{\_}{1.048\times 10^{-5}mol/L}$.

Explanation of Solution

The standard values of $K_{\text{sp}}$ for ${\text{BaSO}}_4$ is   $1.1\times {10}^{-10}$.

The dissociation reaction for ${\text{BaSO}}_4$ is shown as follows:

  $\begin{array}{l}{\text{BaSO}}_4\rightleftharpoons {\text{Ba}}^{2+}+{\text{SO}}_4{}^{2-}\\ \begin{array}{cc}\text{s}& \text{s}\end{array}\end{array}$

Where,

• $\text{s}$ is the solubility of  ${\text{Ba}}^{2+}$ ion and ${\text{SO}}_4{}^{2-}$ ion.

The expression for solubility product of the above reaction is shown below.

  $K_{\text{sp}}=\left[{\text{Ba}}^{2+}\right]\left[{\text{SO}}_4{}^{2-}\right]$        (1)

Where,

• $K_{\text{sp}}$ is the solubility product.
• $\left[{\text{Ba}}^{2+}\right]$ is the solubility of barium ion.
• $\left[{\text{SO}}_4{}^{2-}\right]$ is the solubility of suphate ion.

The value of $K_{\text{sp}}$ is $1.1\times {10}^{-10}$.

The value of $\left[{\text{Ba}}^{2+}\right]$ is $\text{s}$.

The value of $\left[{\text{SO}}_4{}^{2-}\right]$ is $\text{s}$.

Substitute the value of $K_{\text{sp}}$, $\left[{\text{Ba}}^{2+}\right]$ and $\left[{\text{SO}}_4{}^{2-}\right]$ in equation (1).

  $\begin{array}{c}1.1\times {10}^{-10}=\left(\text{s}\right)\left(\text{s}\right)\\ {\text{s}}^2=1.1\times {10}^{-10}\\ \text{s}=\sqrt{1.1\times {10}^{-10}}\\ =\underset{\_}{1.048\times 10^{-5}mol/L}\end{array}$

Thus, the solubility of barium ion in pure water is $\underset{\_}{1.048\times 10^{-5}mol/L}$.

(b)

Interpretation Introduction

Interpretation:

The solubility of barium ion in the solution of hydrochloric acid has to be calculated.

Concept Introduction:

Same as part (a).

Answer to Problem 15.BCP

The solubility of barium ion in the solution of hydrochloric acid is $\underset{\_}{1.057\times 10^{-4}mol/L}$.

Explanation of Solution

The standard values of $K_{\text{sp}}$ for ${\text{HSO}}_4{}^{-}$  is   $1.1\times {10}^{-2}$.

The ICE table for ${\text{HSO}}_4{}^{-}$ is shown below:

  $\begin{array}{l}{\text{HSO}}_4{}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{SO}}_4{}^{2-}\\ \begin{array}{cccc}\text{I:}& 0& 0.1& 1.048\times {10}^{-5}\\ \text{C:}& +x& -x& -x\\ \text{E:}& x& \left(0.1-x\right)& \left(1.048\times {10}^{-5}-x\right)\end{array}\end{array}$

Where,

• $\text{I}$ is the Initial concentration.
• $\text{C}$ is the Change concentration.
• $\text{E}$ is the Equilibrium concentration.
• $x$ is the change is concentration.

The expression for $K_{\text{a}}$ for the above reaction is shown below.

  $K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{SO}}_4{}^{2-}\right]}{\left[{\text{HSO}}_4{}^{-}\right]}$        (2)

Where,

• $K_{\text{a}}$ is the dissociation constant.
• $\left[{\text{HSO}}_4{}^{-}\right]$ is the concentration of barium ion.
• $\left[{\text{SO}}_4{}^{2-}\right]$ is the concentration of suphate ion.
• $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ion.

The value of $K_{\text{a}}$ is $1.1\times {10}^{-2}$.

The value of $\left[{\text{HSO}}_4{}^{-}\right]$ is $x$.

The value of $\left[{\text{SO}}_4{}^{2-}\right]$ is $\left(1.048\times {10}^{-5}-x\right)$.

The value of $\left[{\text{H}}^{+}\right]$ is $\left(0.1-x\right)$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{HSO}}_4{}^{-}\right]$,  $\left[{\text{H}}^{+}\right]$ and $\left[{\text{SO}}_4{}^{2-}\right]$  in equation (2).

  $1.1\times {10}^{-2}=\frac{\left(0.10-x\right)\left(1.048\times {10}^{-5}-x\right)}{x}$

Assuming that $x$ is very small with respect to $0.10$.  Therefore, $\left(0.10-x\right)$ is approximately equal to $0.10$.

  $\begin{array}{c}1.1\times {10}^{-2}=\frac{\left(0.10\right)\left(1.048\times {10}^{-5}-x\right)}{x}\\ 1.1\times {10}^{-2}x=1.048\times {10}^{-6}-0.10x\\ 1.1\times {10}^{-2}x+0.10x=1.048\times {10}^{-6}\\ 0.111x=1.048\times {10}^{-6}\end{array}$

The above expression is simplified further as shown below.

  $\begin{array}{c}x=\frac{1.048\times {10}^{-6}}{0.111}\\ =9.44\times {10}^{-6}\text{mol/L}\end{array}$

Therefore, concentration of sulphate ion will be calculated as shown below.

  $\begin{array}{c}\left[{\text{SO}}_4{}^{2-}\right]=\left(1.048\times {10}^{-5}-9.44\times {10}^{-6}\right)\\ =1.04\times {10}^{-6}\text{mol/L}\end{array}$

The dissociation reaction for ${\text{BaSO}}_4$ is shown as follows:

  $\begin{array}{l}{\text{BaSO}}_4\rightleftharpoons {\text{Ba}}^{2+}+{\text{SO}}_4{}^{2-}\\ \begin{array}{cc}\text{s}& 1.04\times {10}^{-6}\end{array}\end{array}$

Where,

• $\text{s}$ is the solubility of ${\text{Ba}}^{2+}$ ion.

The value of $K_{\text{sp}}$ is $1.1\times {10}^{-10}$.

The value of $\left[{\text{Ba}}^{2+}\right]$ is $\text{s}$.

The value of $\left[{\text{SO}}_4{}^{2-}\right]$ is $1.04\times {10}^{-6}$.

Substitute the value of $K_{\text{sp}}$, $\left[{\text{Ba}}^{2+}\right]$ and $\left[{\text{SO}}_4{}^{2-}\right]$  in equation (1).

  $\begin{array}{c}1.1\times {10}^{-10}=\left(\text{s}\right)\left(1.04\times {10}^{-6}\right)\\ \text{s}=\frac{1.1\times {10}^{-10}}{1.04\times {10}^{-6}}\\ =1.057\times {10}^{-4}\text{mol/L}\end{array}$

Thus, the solubility of barium ion in the solution of hydrochloric acid is $\underset{\_}{1.057\times 10^{-4}mol/L}$.

(c)

Interpretation Introduction

Interpretation:

The solubility of barium ion in the solution of hydrochloric acid at $37°\text{C}$ has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 15.BCP

The solubility of barium ion in the solution of hydrochloric acid at $37°\text{C}$ is $\underset{\_}{1.85\times 10^{-4}mol/L}$.

Explanation of Solution

The standard values of $K_{\text{sp}}$ for ${\text{BaSO}}_4$ at $37°\text{C}$ is $1.5\times {10}^{-10}$.

The dissociation reaction for ${\text{BaSO}}_4$ is shown as follows:

  $\begin{array}{l}{\text{BaSO}}_4\rightleftharpoons {\text{Ba}}^{2+}+{\text{SO}}_4{}^{2-}\\ \begin{array}{cc}\text{s}& \text{s}\end{array}\end{array}$

The value of $K_{\text{sp}}$ is $1.5\times {10}^{-10}$.

The value of $\left[{\text{Ba}}^{2+}\right]$ is $\text{s}$.

The value of $\left[{\text{SO}}_4{}^{2-}\right]$ is $\text{s}$.

Substitute the value of $K_{\text{sp}}$, $\left[{\text{Ba}}^{2+}\right]$ and $\left[{\text{SO}}_4{}^{2-}\right]$  in equation (1).

  $\begin{array}{c}1.5\times {10}^{-10}=\left(\text{s}\right)\left(\text{s}\right)\\ {\text{s}}^2=1.5\times {10}^{-10}\\ \text{s}=\sqrt{1.5\times {10}^{-10}}\\ =1.22\times {10}^{-5}\text{mol/L}\end{array}$

Thus, the solubility of barium ion in pure water is $\underset{\_}{1.22\times 10^{-5}mol/L}$.

The standard values of $K_{\text{sp}}$ for ${\text{HSO}}_4{}^{-}$ at $37°\text{C}$ is $7.1\times {10}^{-3}$.

The ICE table for ${\text{HSO}}_4{}^{-}$ is shown below:

  $\begin{array}{l}{\text{HSO}}_4{}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{SO}}_4{}^{2-}\\ \begin{array}{cccc}\text{I:}& 0& 0.1& 1.22\times {10}^{-5}\\ \text{C:}& +x& -x& -x\\ \text{E:}& x& \left(0.1-x\right)& \left(1.22\times {10}^{-5}-x\right)\end{array}\end{array}$

Where,

• $\text{I}$ is the Initial concentration.
• $\text{C}$ is the Change concentration.
• $\text{E}$ is the Equilibrium concentration.
• $x$ is the change is concentration.

The value of $K_{\text{a}}$ is $7.1\times {10}^{-3}$.

The value of $\left[{\text{HSO}}_4{}^{-}\right]$ is $x$.

The value of $\left[{\text{SO}}_4{}^{2-}\right]$ is $\left(1.22\times {10}^{-5}-x\right)$.

The value of $\left[{\text{H}}^{+}\right]$ is $\left(0.1-x\right)$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{HSO}}_4{}^{-}\right]$, $\left[{\text{H}}^{+}\right]$  and $\left[{\text{SO}}_4{}^{2-}\right]$  in equation (2).

  $7.1\times {10}^{-3}=\frac{\left(0.10-x\right)\left(1.22\times {10}^{-5}-x\right)}{x}$

Assuming that x is very small with respect to $0.10$.  Therefore, $\left(0.10-x\right)$ is approximately equal to $0.10$.

  $\begin{array}{c}7.1\times {10}^{-3}=\frac{\left(0.10\right)\left(1.22\times {10}^{-5}-x\right)}{x}\\ 7.1\times {10}^{-3}x=1.22\times {10}^{-6}-0.10x\\ 7.1\times {10}^{-3}x+0.10x=1.22\times {10}^{-6}\\ 0.1071x=1.22\times {10}^{-6}\end{array}$

The above expression is simplified further as shown below.

  $\begin{array}{c}x=\frac{1.22\times {10}^{-6}}{0.1071}\\ =1.139\times {10}^{-5}\text{mol/L}\end{array}$

Therefore, concentration of suphate ion will be calculated as shown below.

  $\begin{array}{c}\left[{\text{SO}}_4{}^{2-}\right]=\left(1.22\times {10}^{-5}-1.139\times {10}^{-5}\right)\text{mol/L}\\ =8.1\times {10}^{-7}\text{mol/L}\end{array}$

The dissociation reaction for ${\text{BaSO}}_4$ is shown as follows:

  $\begin{array}{l}{\text{BaSO}}_4\rightleftharpoons {\text{Ba}}^{2+}+{\text{SO}}_4{}^{2-}\\ \begin{array}{cc}\text{s}& 8.1\times {10}^{-7}\end{array}\end{array}$

Where,

• $\text{s}$ is the solubility of ${\text{Ba}}^{2+}$ ion.

The value of  $K_{\text{sp}}$ is  $1.5\times {10}^{-10}$.

The value of $\left[{\text{Ba}}^{2+}\right]$ is $\text{s}$.

The value of $\left[{\text{SO}}_4{}^{2-}\right]$ is $8.1\times {10}^{-7}$.

Substitute the value of $K_{\text{sp}}$, $\left[{\text{Ba}}^{2+}\right]$ and $\left[{\text{SO}}_4{}^{2-}\right]$  in equation (1).

  $\begin{array}{c}1.5\times {10}^{-10}=\left(\text{s}\right)\left(8.1\times {10}^{-7}\right)\\ \text{s}=\frac{1.5\times {10}^{-10}}{8.1\times {10}^{-7}}\\ =\underset{\_}{1.85\times 10^{-4}mol/L}\end{array}$

Thus, the solubility of barium ion in the solution of hydrochloric acid at $37°\text{C}$ is $\underset{\_}{1.85\times 10^{-4}mol/L}$.