Chapter 24, Problem 101AP

Interpretation Introduction

Interpretation:

The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of ${25}^{°}\text{C}$ is to becalculated.

Concept introduction:

Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows:

$K_{\text{a}}=\frac{[\text{products}]}{[\text{reactants]}}$.

Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Liter of solution}}$.

Answer to Problem 101AP

Solution: pH is $5.72$.

Explanation of Solution

Given information: The white phosphorous burnt is $10.0\text{ g}$.

The volume of solution is $0.500\text{ L}$.

If the white phosphorus burns in excess of oxygen, the reaction is as follows:

${\text{P}}_4\left(s\right)+5{\text{O}}_2\left(g\right)\to {\text{P}}_4{\text{O}}_{10}\left(s\right)$

The product formed $\left({\text{P}}_4{\text{O}}_{10}\right)$ is dissolved in water and the reaction that takes place is as follows:

${\text{P}}_4{\text{O}}_{10}\left(s\right)+6{\text{H}}_2\text{O}\left(l\right)\to 4{\text{H}}_3{\text{PO}}_4\left(aq\right)$.

The moles formed by ${\text{H}}_3{\text{PO}}_4$ are to be calculated as follows:

$\text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}}$.

Substitute the values in the above equation:

$\begin{array}{l}{M}_{{\text{H}}_3{\text{PO}}_4}=10.0\cancel{{\text{g P}}_4}\times \left(\frac{\cancel{1{\text{ mol P}}_4}}{123.9\cancel{{\text{g P}}_4}}\right)\left(\frac{\cancel{1{\text{ mol P}}_4{\text{O}}_{10}}}{\cancel{1{\text{ mol P}}_4}}\right)\left(\frac{4{\text{ mol H}}_3{\text{Po}}_4}{\cancel{1{\text{ mol P}}_4{\text{O}}_{10}}}\right)\\ M_{{\text{H}}_3{\text{PO}}_4}=0.323{\text{ mol H}}_3{\text{Po}}_4\text{.}\end{array}$

Thus, the moles of ${\text{H}}_3{\text{PO}}_4$ are $0.323\text{ mol }$.

Molarity of the phosphoric acid solution is calculated as follows:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of Solution (L)}}$.

Substitute $0.323\text{ mol }$ for the moles of solute and $0.500\text{ L}$ for the volume of solution in the above equation as follows:

$\begin{array}{l}\text{Molarity}=\left(\frac{0.323\text{ mol}}{0.500\text{ L}}\right);\\ \text{Molarity}=0.646\text{ M}\text{.}\end{array}$

The equilibrium expression for a phosphoric acid solution is as follows:

$\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{ H}}_3{\text{PO}}_4\left(aq\right){\text{ + H}}_2\text{O}\left(aq\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(aq\right){\text{ + H}}_2{\text{PO}}^{-}\left(aq\right)\\ \text{initial}\left(\text{M}\right):\text{ 0}\text{.646 0 0}\\ \text{change}\left(\text{M}\right):-x+x+x\end{array}}\\ \text{Equilibrium}\left(\text{M}\right):\text{ 0}\text{.646 }-xxx\end{array}$

The value of $K_a$ for phosphoric acid is $7.5\times {10}^{-3}$, taken from the reference table $16.8$.

The $[{\text{H}}_3{\text{O}}^{+}]$ is calculated with the help of the equilibrium constant as follows:

$K_{\text{a}}=\frac{[{\text{H}}_3{\text{O}}^{+}][{\text{H}}_2{\text{PO}}_4^{-}]}{[{\text{H}}_3{\text{PO}}_4]}$.

Substitute $\left(0.646-x\right)$ for the concentration of phosphoric acid, $\left(x\right)$ for the concentration of hydronium ions, $\left(x\right)$ for the concentration of dihydrogen phosphate ions, and $7.5\times {10}^{-3}$ for $K_a$ in the above equation as follows:

$\begin{array}{c}7.5\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{\left(0.646-x\right)}\\ 7.5\times {10}^{-3}=\frac{\left(x^2\right)}{\left(0.646-x\right)}\\ x^2=\left(7.5\times {10}^{-3}\right)\left(0.646-x\right)\end{array}$.

Rearrange the above equation:

$\left(x^2+7.5\times {10}^{-3}x-4.85\times {10}^{-3}\right)=0$.

This is a quadratic equation, by solving:

$\begin{array}{c}x=0.066\text{ M}\text{.}\\ {\text{Therefore, [H}}_3{\text{O}}^{+}]=0.066\text{ M}\text{.}\end{array}$

The pH of the solution is to be calculated as follows:

$\text{pH}=-{\text{log[H}}^{+}]$.

Here, the concentration of hydronium ions $\left([{\text{H}}_3{\text{O}}^{+}]\right)$ is equal to the concentration of ${\text{[H}}^{+}]$.

$\begin{array}{l}\text{pH }=-\text{log}\left(0.066\right)\\ \text{pH}=\mathrm{1.18.}\end{array}$

Conclusion

The pH of the solution at ${25}^{°}\text{C}$ is $1.18$.