Question

Drinking water is often disinfected with ${\mathbf{\text{CI}}}_{\mathbf{\text{2}}}$, which hydrolyzes to form HClO, a weak acid but powerful disinfectant:

${\mathbf{C}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{nHClO}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{Cl}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The fraction of HClO in solution is defined as
$\frac{\mathbf{\left[}\mathbf{HClO}\mathbf{\right]}}{\mathbf{\left[}\mathbf{HClO}\mathbf{\right]}\mathbf{+}\mathbf{\left[}{\mathbf{ClO}}^{\mathbf{-}}\mathbf{\right]}}$

(a) What is the fraction of HClO at pH 7.00(Ka of HClO=$\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{)}$?

(b) What is the fraction at pH 10.00?

Short answer

(a) The fraction of HCIO at $\mathrm{pH}=7.00$is $\mathrm{pH}=2.25\times {10}^{-2}$

(b) The fraction of HCIO at $\mathrm{pH}=10.00$ is $\mathrm{pH}=9.97\times {10}^{-1}$

Step-by-step solution

To find the fraction of HCIO at pH 7.00

(a)$7.00$

${\text{Cl}}^{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}\to {\text{HClO +H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ Cl}}^{\text{-}}$

HCIO will further dissociate in water.

$\text{HClO}+{\text{H}}_2\text{O}\rightleftharpoons {\text{ClO}}^{-}+{\text{H}}_3{\text{O}}^{+}$

Next, solve the$\left[{\text{H}}_3{\text{O}}^{+}\right]$in the${\text{K}}_{\text{a}}$solution from the given pH.

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}} \\ ={10}^{-7.00} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1\times {10}^{-7} \end{gathered}$$

Write and calculate theexpression from the${\text{K}}_{\text{b}}$ of_{${\text{NH}}_3$}.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\left[\mathrm{product}\right]}{\left[\mathrm{reactant}\right]} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{ClO}}^{-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[\mathrm{HClO}\right]} \end{gathered}$$

Since we know the value of localid="1663273953669" $\left[{\text{H}}_3{\text{O}}^{+}\right]$andlocalid="1663273969712" ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$let us try to derive the HClO+ ClO from thelocalid="1663273986935" ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$expression.

$$\begin{gathered} {\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{ClO}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\text{[HClO]}} \\ \frac{\text{[HClO]}}{\left[{\text{ClO}}^{\text{-}}\right]}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{{\text{K}}_{\text{a}}} \end{gathered}$$

To have an addition element, let us try to add a factor equivalent to 1 on both sides: denominator/denominator.

$\frac{\text{[HClO]}}{\left[{\text{ClO}}^{\text{-}}\right]}\text{+}\frac{\left[{\text{ClO}}^{\text{-}}\right]}{\left[{\text{ClO}}^{\text{-}}\right]}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{{\text{K}}_{\text{a}}}\text{+}\frac{{\text{K}}_{\text{a}}}{{\text{K}}_{\text{a}}}\frac{\text{[HClO] +}\left[{\text{ClO}}^{\text{-}}\right]}{\left[{\text{ClO}}^{\text{-}}\right]}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{+K}}_{\text{a}}}{{\text{K}}_{\text{a}}}$

Reverse both reaction to get the equation$\left[\frac{\text{HClO}]}{\left[\text{HClO}\right]+[\text{ClO}}\right]$

$\frac{\text{[HClO]}}{\text{[HClO] +}\left[{\text{ClO}}^{\text{-}}\right]}\text{=}\frac{{\text{K}}_{\text{a}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{+K}}_{\text{a}}}$

Sincewe obtained the formulae for

$\left.\frac{\text{HClO]}}{\text{[HClO] + ClO}}\right]\text{,we can use the values for}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{andK}}_{\text{a}}$

$$\begin{gathered} \frac{\left[\mathrm{HClO}\right]}{\left[\mathrm{HClO}\right]+\left[{\mathrm{ClO}}^{-}\right]}=\frac{{\mathrm{K}}_{\mathrm{a}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]+{\mathrm{K}}_{\mathrm{a}}} \\ =\frac{2.9\times {10}^{-8}}{1\times {10}^{-7}+2.9\times {10}^{-8}} \end{gathered}$$

$\frac{\left[\mathrm{HCIO}\right]}{\left[\mathrm{HClO}\right]+\left[{\mathrm{ClO}}^{-}\right]}=2.25\times {10}^{-2}$

Hence the fraction of HCIO at $\mathrm{pH}=7.00$is $\mathrm{pH}=2.25\times {10}^{-2}$

To find the fraction of HCIO at pH 10.00

(b)$10.00$

Since we already obtained the formula for$\frac{\left[\text{HCIO}\right]}{\left[\text{HClO}\right]+\left[{\text{ClO}}^{-}\right]}$

solve for the$\left[{\text{H}}_3{\text{O}}^{+}\right]$ of the solution first.

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=10^{-\mathrm{pH}} \\ ={10}^{-10} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1\times {10}^{-10} \end{gathered}$$

Lastly, solve for the$\frac{\left[\text{HCIO}\right]}{\left[\text{HClO}\right]+\left[{\text{ClO}}^{-}\right]}$

$$\begin{gathered} \frac{\left[\mathrm{HClO}\right]}{\left[\mathrm{HClO}\right]+\left[{\mathrm{ClO}}^{-}\right]}=\frac{{\mathrm{K}}_{\mathrm{a}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]+{\mathrm{K}}_{\mathrm{a}}} \\ =\frac{2.9\times {10}^{-8}}{1\times {10}^{-10}+2.9\times {10}^{-8}} \\ \frac{\left[{\mathrm{NH}}_3\right]}{\left[{\mathrm{NH}}_4^{+}\right]+\left[{\mathrm{NH}}_3\right]}=9.97\times {10}^{-1} \end{gathered}$$

Hence the fraction of HCIO at $\mathrm{pH}=10.00$ is $\mathrm{pH}=9.97\times {10}^{-1}$