Chapter 17, Problem 17.52QP

Interpretation Introduction

Interpretation: The value of $\text{pH}$ for $1.00\text{M}$ $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2$ 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 $1.00\text{M}$ $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2$.

Answer to Problem 17.52QP

The value of $\text{pH}$ for $1.00\text{M}$ $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $\underset{\_}{3.76}$.

Explanation of Solution

The concentration of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $1.00\text{M}$.

The chemical equation for the dissociation of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2$ is,

$\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(aq\right)+2{\text{NO}}_3^{-}\left(aq\right)$

In general ${\text{Cu}}^{2+}\left(aq\right)$ ions is in hydrated form that is ${\left[\text{Cu}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{2+}$. Dissociation of this ion in water is,

${\left[\text{Cu}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\left[\text{Cu}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\left(\text{OH}\right)\right]}^{+}\left(aq\right)$

The $K_{\text{a}}$ of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $3\times {10}^{-8}$.

The ICE table for the dissociation of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2$ is,

$\begin{array}{cccccc}& {\left[\text{Cu}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{2+}\left(aq\right)& \stackrel{}{\to }& {\left[\text{Cu}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\left(\text{OH}\right)\right]}^{+}\left(aq\right)& +& {\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)\\ \text{I}& 1.00& & 0& & 0\\ \text{C}& -x& & x& & x\\ \text{E}& 1.00-x& & x& & x\end{array}$

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

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

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

$\begin{array}{c}3\times {10}^{-8}=\frac{x^2}{1.00-x}\\ x^2=3\times {10}^{-8}\left(1.00-x\right)\\ x^2=3\times {10}^{-8}-3\times {10}^{-8}x\\ 0=x^2+3\times {10}^{-8}x-3\times {10}^{-8}\end{array}$

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 $x$.

$\begin{array}{c}x=\frac{-\left(3\times {10}^{-8}\right)\pm \sqrt{{\left(3\times {10}^{-8}\right)}^2-4\times 1\times \left(-3\times {10}^{-8}\right)}}{2\times 1}\\ =\frac{-3\times {10}^{-8}\pm \sqrt{9\times {10}^{-16}+1.2\times {10}^{-7}}}{2}\\ =\frac{-3\times {10}^{-8}+3.464\times {10}^{-4}}{2}\\ =1.73\times {10}^{-4}\text{M}\end{array}$

Hence, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is $1.73\times {10}^{-4}\text{M}$.

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 $\text{pH}$.

$\begin{array}{c}\text{pH}=-\log \left(1.73\times {10}^{-4}\right)\\ =\underset{\_}{3.76}\end{array}$

Hence, the $\text{pH}$ of a solution is $\underset{\_}{3.76}$.

Conclusion

The $\text{pH}$ of a $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2$ solution is $\underset{\_}{3.76}$.