Chapter 16, Problem 140AP

A 10.0-g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500.0 mL of solution. Calculate the $\text{pH}$ of the solution at $\text{25ºC}$.

Interpretation Introduction

Interpretation:

The pH of the solution formed by the combustion of phosphorus and consequential dissolution in water is to be determined.

Concept introduction:

The first ionization of a poly-protic acid takes place as

${\text{H}}_3{\text{A}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{A}}^{-}{}_{\left(\text{aq}\right)}$

${\text{K}}_{\text{a1}}$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific to a particular temperature.

${\text{K}}_{\text{a1}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}$ …… (1)

Percent ionization is the percentage of acid that gets dissociated upon addition to water. It depends on the hydronium ion concentration.

$\%\text{ dissociation}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}}{{\left[{\text{H}}_2\text{A}\right]}_{\text{o}}}\times 100\%$ …… (2)

Here, ${\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}$ is the concentration of hydronium ions at equilibrium, and ${\left[{\text{H}}_2\text{A}\right]}_{\text{o}}$ is the original acid concentration.

pH of the solution is calculated as

$\text{pH}=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$ …… (3)

Molarity $\left(M\right)$ is the concentration of any substance given by number of moles $\left(n\right)$ of the substance divided by the volume of solution $\left(V\right)$ in liters.

$M=\frac{n}{V}$ …… (4)

Answer to Problem 140AP

Solution:

The pH of the solution is $1.18$.

Explanation of Solution

Given information:

$10.0\text{ g}$ white phosphorus is burned in excess oxygen and the product is then dissolved in enough water to make a $500.0\text{ mL}$ solution.

White phosphorous is burned in oxygen according to the reaction:

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

According to the stoichiometry, $124\text{ g}$ of phosphorous reacts to produce $284\text{ g}$ of ${\text{P}}_4{\text{O}}_{10}$.

So, $10.0\text{ g}$ of phosphorous will produce.

$\begin{array}{c}m\left({\text{P}}_4{\text{O}}_{10}\right)=\frac{284}{124}\times 10.0\\ =22.9\text{ g}\end{array}$

Molar mass of ${\text{P}}_4{\text{O}}_{10}$ is $284\text{ g/mol}$.

Number of moles of ${\text{P}}_4{\text{O}}_{10}$ is calculated as

$\begin{array}{c}n\left({\text{P}}_4{\text{O}}_{10}\right)=\frac{m\left({\text{P}}_4{\text{O}}_{10}\right)}{\text{MM}\left({\text{P}}_4{\text{O}}_{10}\right)}\\ =\frac{22.9\cancel{\text{g}}}{284\cancel{\text{g}}\text{/mol}}\\ =0.081\text{ mol}\end{array}$

Now, ${\text{P}}_4{\text{O}}_{10}$ is dissolved in water to form $500.0\text{ mL}$ phosphoric acid solution, according to the reaction:

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

According to the stoichiometry, 1 mole of ${\text{P}}_4{\text{O}}_{10}$ produces 4 moles of ${\text{H}}_3{\text{PO}}_4$. So, $0.081{\text{ mol P}}_4{\text{O}}_{10}$ will give:

$\begin{array}{c}n\left({\text{H}}_3{\text{PO}}_4\right)=\frac{4}{1}\times 0.081\\ =0.323\text{ mol}\end{array}$

Use equation (4) to calculate molarity of the solution as

$\begin{array}{c}M=\frac{0.323\text{ mol}}{\text{0}\text{.5}\text{L}}\\ =0.646\text{ mol}\end{array}$

Refer to table $16.8$ for ${\text{K}}_{\text{a1}}{\text{, K}}_{\text{a2}}{\text{, and K}}_{\text{a3}}$ values of phosphoric acid as

$\begin{array}{l}{\text{K}}_{\text{a1}}=7.5\times {10}^{-3}\\ {\text{K}}_{\text{a2}}=6.2\times {10}^{-8}\\ {\text{K}}_{\text{a3}}=4.8\times {10}^{-13}\end{array}$

When phosphoric acid is dissolved in water, dissociation takes place in three steps as it is a triprotic acid. First, one proton is partially dissociated, becausephosphoric acid is a weak acid. Thus, the pH of the solution is determined by the contribution made by all the three proton dissociation steps.

The reaction of the first proton dissociation of phosphoric acid is represented as

${\text{H}}_3{\text{PO}}_4{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{PO}}_4{{}^{-}}_{\left(\text{aq}\right)}$

Prepare an equilibrium table and represent each of the species in terms of $x$ as

$\begin{array}{ccccc}& {\text{H}}_3{\text{PO}}_4{}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{PO}}_4{{}^{-}}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 0.646& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.646-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (1) as

${\text{K}}_{\text{a1}}=\frac{\left(x\right)\left(x\right)}{\left(0.646-x\right)}$

Since the value of ${\text{K}}_{\text{a1}}$ is small, the amount of acid dissociated is less. Therefore, $\left(0.646-x\right)$ can be approximated as $0.646$. Now, substitute the value of ${\text{K}}_{\text{a1}}$ in above equation as

$\begin{array}{c}7.5\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{0.646}\\ x^2=\left(7.5\times {10}^{-3}\right)0.646\\ x=\sqrt{4.84\times {10}^{-3}}\\ x=0.067\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{1\left(\text{eq}\right)}=0.067M$, ${\left[{\text{H}}_3{\text{PO}}_{\text{4}}\right]}_{\left(\text{O}\right)}=0.646M$

Calculate the percent dissociation from equation (2) as

$\begin{array}{c}\%\text{ ionization}=\frac{0.067}{\text{0}\text{.646}}\times 100\%\\ =10.77\%\end{array}$

Since the percent dissociation is more than $5\%$, the approximation is not valid. Thus, again solve for the value of $x$ as

$\begin{array}{c}7.5\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{\left(0.646-x\right)}\\ x^2+\left(7.5\times {10}^{-3}\right)x=\left(4.84\times {10}^{-3}\right)\\ x=-0.059,\text{ 0}\text{.066}\end{array}$

Since concentration cannot be negative, so $x=0.066$.

Thus,

$\begin{array}{c}\left[{\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}\right]=\text{0}\text{.066 }M\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=0.066M\end{array}$

The second and third ionization constants for phosphoric acid have a low value. Thus, it will have negligible contribution in the hydronium ion concentration.

Now, use equation (3) to calculate pH of the solution as

$\begin{array}{c}\text{pH}=-\log \left[0.066\right]\\ =1.18\end{array}$

Conclusion

The pH of the solution is calculated to be $1.18$.