Chapter 17, Problem 17.51QP

Interpretation Introduction

Interpretation: The value of $\text{pH}$ for $0.100\text{M}$ $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ 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 $0.100\text{M}$ $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$.

Answer to Problem 17.51QP

The value of $\text{pH}$ for $0.100\text{M}$ $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is $\underset{\_}{1.72}$.

Explanation of Solution

The concentration of $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is $0.100\text{M}$.

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

$\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}+3{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{Fe}{\left(\text{OH}\right)}_{\text{3}}+{\text{3HNO}}_{\text{3}}$

The net ionic equation is,

${\text{Fe}}^{3+}+3{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{Fe}{\left(\text{OH}\right)}_{\text{3}}+3{\text{H}}^{+}$

The $K_{\text{a}}$ of $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is $3.00\times {10}^{-3}$.

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

$\begin{array}{cccccc}& {\text{Fe}}^{3+}\left(aq\right)& \stackrel{}{\to }& \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(aq\right)& +& {\text{H}}^{+}\left(aq\right)\\ \text{I}& 0.100& & 0& & 0\\ \text{C}& -x& & x& & x\\ \text{E}& 0.100-x& & x& & x\end{array}$

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

$K_{\text{a}}=\frac{\left[\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{Fe}}^{3+}\right]}$

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

$\begin{array}{c}3.00\times {10}^{-3}=\frac{x^2}{0.100-x}\\ x^2=3.00\times {10}^{-3}\left(0.100-x\right)\\ x^2=3.00\times {10}^{-4}-3.00\times {10}^{-3}x\\ 0=x^2+3.00\times {10}^{-3}x-3.00\times {10}^{-4}\end{array}$

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 $x$.

$\begin{array}{c}x=\frac{-\left(3.00\times {10}^{-3}\right)\pm \sqrt{{\left(3.00\times {10}^{-3}\right)}^2-4\times 1\times \left(-3.00\times {10}^{-4}\right)}}{2\times 1}\\ =\frac{-3.00\times {10}^{-3}\pm \sqrt{9.0\times {10}^{-6}+1.2\times {10}^{-3}}}{2}\\ =\frac{-3.00\times {10}^{-3}+0.0347}{2}\\ =0.01885\text{M}\end{array}$

Hence, the concentration of ${\text{H}}^{+}$ is $0.01885\text{M}$.

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

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

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

$\begin{array}{c}\text{pH}=-\log \left(0.01885\right)\\ =\underset{\_}{1.72}\end{array}$

Hence, the $\text{pH}$ of a solution is $\underset{\_}{1.72}$.

Conclusion

The $\text{pH}$ of a $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ solution is $\underset{\_}{1.72}$.