Chapter 8, Problem 8.35P

Interpretation Introduction

Interpretation:

The pH of $\text{Cr}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{3}}$ and Fraction of chromium is in the form $\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{w-1}}{\left(\text{OH}\right)}^{\text{2+}}$ have to be calculated.

Explanation of Solution

The given compound $\text{Cr}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{3}}$ is dissociates into ${\text{Cr}}^{\text{3+}}\text{and}{\text{ClO}}_{\text{4}}^{\text{-}}.$ ${\text{Cr}}^{\text{3+}}$ ion is acidic in nature and when it reacts with water gives $\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}^{\text{3+}}.$

The reaction as follows,

  $\begin{array}{l}\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}^{\text{3+}}\rightleftharpoons \text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{\text{2+}}\text{+}{\text{H}}^{\text{+}}\\ \text{Initial}\left(\text{M}\right):0.01000\\ \text{Change:}-x+x+x\\ \text{Equilibrium:}0.010-xxx\end{array}$

Acid dissociation constant ${\text{K}}_{\text{a}}$ is represented as follows,

  ${\text{K}}_{\text{a}}=\frac{\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{\text{2+}}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}^{\text{3+}}\right]}$

As known that,

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{=}{\text{10}}^{-{\text{pK}}_{\text{a}}},\\ \\ \text{substitute}{\text{pK}}_{\text{a}}\text{=}\text{3}\text{.80,}\\ \\ K_a={10}^{-3.80}=1.6\times {10}^{-4}\\ \\ 1.6\times {10}^{-4}=\frac{x^2}{0.010-x}\end{array}$

By solving the above expression, the value of ‘x’ is $\mathrm{0.00119.}$

Hence,

  $\begin{array}{l}\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{\text{2+}}\right]\text{=}\left[{\text{H}}^{\text{+}}\right]\text{=}\text{x}\text{=}\text{0}\text{.00119}\text{M}\\ \\ \text{pH}\text{=}-\text{log}\left[{\text{H}}^{\text{+}}\right]=-\log \left(0.00119\right)=2.92\end{array}$

Therefore, the value of $\text{pH}$ is $2.92.$

The fraction of chromium in the form $\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{\text{2+}}$ as follows,

  $\text{=}\frac{\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{\text{2+}}\right]}{{\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}^{\text{3+}}\right]}_{\text{initial}}}=\frac{0.00119}{0.010}=0.12.$

Therefore, the fraction of chromium in the form $\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}{\left(\text{OH}\right)}^{\text{2+}}$ is $0.12.$