Hydrogen peroxide, ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{2}}}$(${\mathbf{\text{pK}}}_{\mathbf{\text{a}}}\mathbf{\text{= 11.75}}$), is a bleaching agent and antiseptic. It is sold as 3%${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{2}}}$by mass and contains0.001%${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}$by mass to stabilize the solution. Which contributes more${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{+}}}$ to the solution, the${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{2}}}$or the ${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}$?

First, solve for the ${\text{K}}_{\text{a}}{\text{ofH}}_{\text{2}}{\text{O}}_{\text{2}}$ using the given ${\text{pK}}_{\text{a}}$.

$$\begin{gathered} {\text{pK}}_{\text{a}}\text{=}{\text{- logK}}_{\text{a}} \\ {\text{K}}_{\text{a}}{\text{= 10}}^{{\text{- pK}}_{\text{a}}} \\ {\text{K}}_{\text{a}}{\text{= 10}}^{\text{- 11.75}} \\ {\text{K}}_{\text{a}}{\text{=1.78×10}}^{-12} \end{gathered}$$

Next, solve for the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{inH}}_{\text{2}}{\text{O}}_{\text{2}}$.

In the solution, there exist ${\text{3\%H}}_{\text{2}}{\text{O}}_{\text{2}}$by mass. If one has a 100 g solution, then solve for the number of moles of${\text{H}}_2{\text{O}}_2.$

$\begin{array}{rcl}& & {\text{molesH}}_{\text{2}}{\text{O}}_{\text{2}}\text{=3.00g×}\frac{1\mathrm{mol}{\displaystyle {{\displaystyle \text{H}}}_{\text{2}}}{\displaystyle {{\displaystyle \text{O}}}_{\text{2}}}}{34\mathrm{g}{\displaystyle {{\displaystyle \text{H}}}_{\text{2}}}{\displaystyle {{\displaystyle \text{O}}}_{\text{2}}}}\\ & & {\text{=0.088molH}}_{\text{2}}{\text{O}}_{\text{2}}\end{array}$

Then, solve for the concentration. Note that the density of the water is 1g/mL

$\begin{array}{rcl}& & \text{M}\text{=}\frac{\text{mol}}{\text{L}}\\ & =& \frac{0.088}{100\mathrm{g}}\times \frac{\text{1g}}{\text{1mL}}\times \frac{\text{1000mL}}{\text{1L}}\\ & =& 0.88\mathrm{M}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\end{array}$

Write the dissociation reaction of${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{HO}}_{\text{2}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}$

Next, solve for the$\left[{\text{H}}_3{\text{O}}^{+}\right].$

The${\text{K}}_{\text{a}}$ is:

$\begin{array}{rcl}& & {\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{HO}}_{\text{2}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}\\ & & \text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.88 - x}}\end{array}$

We know that$\text{x}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{HO}}_{\text{2}}^{\text{-}}\right]$ since the reaction produced one mole of each. Since the${\text{K}}_{\text{a}}$ is very small, one can drop the x in the denominator. Then solve for x.

$\begin{array}{rcl}& & {\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.88}}\text{x}\\ & =& \sqrt{\frac{1.78\times {10}^{-12}}{0.88}}\\ \mathrm{x}& =& 1.42\times {10}^{-6}\mathrm{M}\end{array}$

Since$\text{x}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{HO}}_{\text{2}}^{\text{-}}\right]$, then the$\left[{\text{H}}_3{\text{O}}^{+}\right].$produced from${\text{H}}_2{\text{O}}_2$ is$1.42\times {10}^{-6}\mathrm{M}$

Next, solve for the concentration of${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{inH}}_{\text{3}}{\text{PO}}_{\text{4}}$.

In the solution, there exist${\text{0.001\%H}}_{\text{3}}{\text{PO}}_{\text{4}}$by mass. If one has a100 gsolution, then solve for the number of moles of${\text{H}}_3{\text{PO}}_4$.

Now, solve for the concentration. Note that the density of water is 1g/mL

$\begin{array}{rcl}& & \text{M}\text{=}\frac{\text{mol}}{\text{L}}\\ & =& \frac{1.02\times {10}^{-5}}{100\mathrm{g}}\times \frac{\text{1g}}{\text{1mL}}\times \frac{\text{1000mL}}{\text{1L}}\\ & =& 1.02\times {10}^{-1}{\text{MH}}_{\text{3}}{\text{PO}}_{\text{4}}.\end{array}$

Write the dissociation reaction of${\text{H}}_3{\text{PO}}_4$.

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{PO}}_{\text{1}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}$

Next, solve for the$\left[{\text{H}}_3{\text{O}}^{+}\right].$The${\text{K}}_{\text{a}}$is calculated as:

$\begin{array}{rcl}& & {\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}\\ & =& \frac{{\displaystyle {\text{x}}^{\text{2}}}}{1.02\times {10}^{-5}-\mathrm{x}}\\ & & {\text{TheK}}_{\text{a}}{\text{ofH}}_{\text{3}}{\text{PO}}_{\text{4}}\text{based on}\text{Appendix C:}\\ & & {\text{K}}_{\text{a}}{\text{=7.2×10}}^{-3}\end{array}$

We know that$\text{x}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]$since the reaction produced one mole of each. Now, solve for x.

$\begin{array}{rcl}\mathrm{K}& =& \frac{{\mathrm{x}}^2}{1.02\times {10}^{-4}-\mathrm{x}}\\ \left(1.02\times {10}^{-4}-\mathrm{x}\right)\left(7.2\times {10}^{-3}\right)& =& {\mathrm{x}}^2\\ 7.35\times {10}^{-7}-7.2\times {10}^{-3}\mathrm{x}& =& {\mathrm{x}}^2\\ {\mathrm{x}}^2+7.2\times {10}^{-3}\mathrm{x}-7.35\times {10}^{-7}& =& 0\\ \mathrm{x}& =& 1.01\times {10}^{-4}\end{array}$

Since, $\text{x}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]$the$\left[{\text{H}}_3{\text{O}}^{+}\right]$ produced from${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{is}1.01\times {10}^{-4}\mathrm{M}$ since.

Hence,${\text{H}}_3{\text{PO}}_4$contributes more$\left[{\text{H}}_3{\text{O}}^{+}\right]$to the solution despite its small percentage.