Chapter 17, Problem 17.50QP

Interpretation Introduction

Interpretation: The value of $\text{pH}$ for $0.50\text{M}$ ${\text{CrCl}}_{\text{3}}$ is to be calculated.

Concept introduction: The potential of ${\text{H}}^{\text{+}}$ ions is called as $\text{pH}$. The value of $\text{pH}$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $\text{pH}$ is $7$.

To determine: The value of $\text{pH}$ for $0.50\text{M}$ ${\text{CrCl}}_{\text{3}}$.

Answer to Problem 17.50QP

The value of $\text{pH}$ for $0.50\text{M}$ ${\text{CrCl}}_{\text{3}}$ is $\underset{\_}{2.15}$.

Explanation of Solution

The concentration of ${\text{CrCl}}_{\text{3}}$ is $0.50\text{M}$.

The chemical equation for the dissociation of ${\text{CrCl}}_{\text{3}}$ in water is,

${\text{CrCl}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{3+}$

At equilibrium,

${\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{3+}+{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{OH}\right]}^{2+}+{\text{H}}_{\text{3}}{\text{O}}^{+}$

The $K_{\text{a}}$ of ${\text{CrCl}}_{\text{3}}$ is $1.0\times {10}^{-4}$.

The ICE table for the dissociation of $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is,

$\begin{array}{cccccc}& {\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{3+}& \stackrel{}{\to }& {\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{OH}\right]}^{2+}& +& {\text{H}}_{\text{3}}{\text{O}}^{+}\\ \text{I}& 0.50& & 0& & 0\\ \text{C}& -x& & x& & x\\ \text{E}& 0.50-x& & x& & x\end{array}$

The concentrations of both ${\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{OH}\right]}^{2+}$ and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ are equal and assumed to be $x$ and is calculated by the equilibrium constant expression,

$K_{\text{a}}=\frac{\left[{\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{OH}\right]}^{2+}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{3+}\right]}$

Substitute the values of dissociation constant and concentration of base in the above expression to calculate the concentration of ${\text{H}}^{+}$ ions.

$\begin{array}{c}1\times {10}^{-4}=\frac{x^2}{0.50-x}\\ x^2=1\times {10}^{-4}\left(0.50-x\right)\\ x^2=5\times {10}^{-5}-1\times {10}^{-4}x\\ 0=x^2+1\times {10}^{-4}x-5\times {10}^{-5}\end{array}$

Simplify the above equation by the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute the values of $a$ and $b$ in the above formula to calculate the value of $x$.

$\begin{array}{c}x=\frac{-\left(1\times {10}^{-4}\right)\pm \sqrt{{\left(1\times {10}^{-4}\right)}^2-4\times 1\times \left(-5\times {10}^{-5}\right)}}{2\times 1}\\ =\frac{-1\times {10}^{-4}\pm \sqrt{1\times {10}^{-8}+2\times {10}^{-4}}}{2}\\ =\frac{-1\times {10}^{-4}+0.0141}{2}\\ =7\times {10}^{-3}\text{M}\end{array}$

Hence, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is $7\times {10}^{-3}\text{M}$.

The $\text{pH}$ of a solution is calculated by the formula,

$\text{pH}=-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$

Substitute the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ in the above formula to calculate the value of $\text{pH}$.

$\begin{array}{c}\text{pH}=-\log \left(7\times {10}^{-3}\text{M}\right)\\ =\underset{\_}{2.15}\end{array}$

Hence, the $\text{pH}$ of a solution is $\underset{\_}{2.15}$.

Conclusion

The $\text{pH}$ of a ${\text{CrCl}}_{\text{3}}$ solution is $\underset{\_}{2.15}$