Q155CP - Chemistry: Molecular Nature Of Matter And Change

Question

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 contains  0.001% ${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}$ 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  ${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}$?

Step-by-Step Solution

Answer

 ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$contributes more  ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}$to the solution

Step 1: Solve for the K a  of H 2 O 2

First, solve for the ${\text{K}}_{\text{a}}{\text{ of H}}_{\text{2}}{\text{O}}_{\text{2}}$  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 ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{ in H}}_{\text{2}}{\text{O}}_{\text{2}}$  .

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 ${\text{H}}_2{\text{O}}_2.$

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

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 ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ .

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

Step 2: To solve for [ H 3 O  +  ]

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 $1.42\times {10}^{-6}\mathrm{M}$

Step 3: Solve for the concentration of H 3 O  +   in H 3 PO 4

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

In the solution, there exist${\text{0.001\% H}}_{\text{3}}{\text{PO}}_{\text{4}}$  by mass. If one has a  100 g solution, 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:

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

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.