Question

${\mathbf{Cr}}^{\mathbf{3}\mathbf{+}}$is acidic by virtue of the hydrolysis reaction ${\mathbf{Cr}}^{\mathbf{3}\mathbf{+}}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\stackrel{{\mathbf{K}}_{\mathbf{aI}}}{\mathbf{\rightleftharpoons }}\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathbf{H}}^{\mathbf{+}}$. [Further reactions produce $\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}\mathbf{}\mathbf{,}\mathbf{}}^{\mathbf{+}}\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}$and.] Find the value role="math" localid="1654864894349" ${\mathbf{K}}_{\mathbf{aI}}$of in Figure 6-9. Considering only the ${\mathbf{K}}_{\mathbf{aI}}$reaction, find pH value ofrole="math" localid="1654865110221" $\mathbf{0}\mathbf{.}\mathbf{010}\mathbf{MCr}{\mathbf{\left(}{\mathbf{CIO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{3}}$What fraction of chromium is in the formrole="math" localid="1654865163335" $\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}^{\mathbf{2}\mathbf{+}}$?

Short answer

The pH value of $0.010\mathrm{MCr}{\left({\mathrm{CIO}}_4\right)}_3=2.86$

The fraction of chromium which is in the form of$\mathrm{Cr}{\left(\mathrm{OH}\right)}^{2+}=14\%$

Step-by-step solution

Define pH and fraction of dissociation

pH stands for Potential hydrogen. It is the symbol for the logarithm of the reciprocal of hydrogen ion concentration in gram atoms per litre, used to express the acidity and alkalinity of a solution on a scale of 0 to 14

The dissociation degree a is the fraction of original solute molecules that have dissociated. $\mathrm{a}=\mathrm{x}/\mathrm{F}$

Hydrolysis reaction of Cr3+

${\mathrm{Cr}}^{3+}{\mathrm{H}}_2\mathrm{O}\stackrel{}{\rightleftharpoons }\mathrm{Cr}{\left(\mathrm{OH}\right)}^{2+}+{\mathrm{H}}^{+}$

Calculating the value of x

$$\begin{gathered} \left[{\mathrm{Cr}}^{3+}\right]=0.01-\mathrm{x} \\ \left[\mathrm{Cr}\left({\mathrm{OH}}^{2+}\right)\right]=\left[{\mathrm{H}}^{+}\right]=\mathrm{x} \end{gathered}$$

By using the formula${\mathrm{K}}_{\mathrm{a}}=\begin{array}{c}{\mathrm{x}}^2\\ \mathrm{F}-\mathrm{x}\end{array}$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{x}}^2}{0.01-\mathrm{x}} \\ {10}^{-3.66}=\frac{{\mathrm{x}}^2}{0.01-\mathrm{x}}\Rightarrow \mathrm{x}=1.37\times {10}^{-3} \end{gathered}$$

Calculating the pH and fraction of dissociation

Substituting the obtained values of pH , x and F in the formula

$$\begin{gathered} \mathrm{pH}=-\mathrm{logx}=-\log (1.37\times {10}^3)=2.86 \\ \mathfrak{a}=\mathrm{x}/\mathrm{F}=(1.37\times {10}^3)/0.01=0.14=14\% \end{gathered}$$