Chapter 16, Problem 88QP

Calculate the pH at $\text{25°C}$ of a $0.25-M$ aqueous solution of phosphoric acid $\left({\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right)$. ( $K_{a_1}$, $K_{a2}$, and $K_{a3}$ for phosphoric acid are $7.5 \times {10}^{-3},6.25 \times {10}^{-5},\text{ and }4.8\times {10}^{-13}$, respectively.)

Interpretation Introduction

Interpretation:

The pH of the solution that contains phosphoric acid with given concentration is to be determined.

Concept introduction:

The first ionization of the polyprotic acid takes place as

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

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

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

The second ionization of the polyprotic acid takes place as

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

$K_{\text{a2}}$ is the measure of dissociation of the second proton of an acid and is known as the second acid-ionization constant, which is specific at a particular temperature.

$K_{\text{a2}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}$ …… (2)

The third ionization of the polyprotic acid takes place as

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

$K_{\text{a3}}$ is the measure of dissociation of the third proton of an acid and is known as the third acid-ionization constant, which is specific at a particular temperature.

$K_{\text{a3}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[{\text{HA}}^{-}\right]}$ …… (3)

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

$\%\text{ dissociation}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}}{{\left[{\text{H}}_2\text{A}\right]}_{\text{o}}}\times 100\%$ …… (4)

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

pH of the solution is calculated as

$\text{pH}=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$ …… (5)

Answer to Problem 88QP

Solution:

The pH of the solution is $1.40$.

Explanation of Solution

Given information:

The concentration of phosphoric acid $\left({\text{H}}_3{\text{PO}}_4\right)$ at $25{}^{\circ }\text{C}$ is $0.25M$. The ${\text{K}}_{\text{a1}}{\text{, K}}_{\text{a2}}{\text{, and K}}_{\text{a3}}$ values for phosphoric acid are ${\text{K}}_{\text{a1}}=7.5\times {10}^{-3},{\text{ K}}_{\text{a2}}=6.25\times {10}^{-8}{\text{, and K}}_{\text{a3}}=4.8\times {10}^{-13},\text{ respectively}$.

When phosphoric acid is dissolved in water, the dissociation takes place in three steps as it is a triprotic acid. First, one proton is partially dissociated because phosphoric 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 depicted as

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

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

$\begin{array}{ccccc}& {\text{H}}_3{\text{PO}}_4\left(aq\right)& {\text{H}}_2\text{O(}l\text{)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{H}}_2{\text{PO}}_4{}^{-}\left(aq\right)\\ \text{Initial concentration}\left(M\right)& 0.25& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.25-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (1) as

$K_{\text{a1}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}$

$K_{\text{a1}}=\frac{\left(x\right)\left(x\right)}{\left(0.25-x\right)}$

Because the value of $K_{\text{a1}}$ is small, the quantity of acid dissociated is less. Therefore, $\left(0.25-x\right)$ can be approximated as $0.25$. Now, substitute the value of ${\text{K}}_{\text{a1}}$ in the above equation:

$\begin{array}{c}7.5\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{0.25}\\ x^2=\left(7.5\times {10}^{-3}\right)0.25\\ x=\sqrt{1.875\times {10}^{-3}}\\ x=0.043\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_1\left(aq\right)=0.043M$ and ${\left[{\text{H}}_3{\text{PO}}_{\text{4}}\right]}_{\left(\text{O}\right)}=0.25M$.

Calculate the percent dissociation from equation (4) as

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

$\begin{array}{c}\%\text{ dissociation}=\frac{0.043}{\text{0}\text{.25}}\times 100\%\\ =17.32\%\end{array}$

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

$K_{\text{a1}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}$

$\begin{array}{c}7.5\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{\left(0.25-x\right)}\\ x^2+\left(7.5\times {10}^{-3}\right)x=\left(1.875\times {10}^{-3}\right)\\ x=-0.047,\text{ 0}\text{.0397}\end{array}$

Since concentration cannot be negative, so $x=0.0397$.

Thus,

$\begin{array}{l}\left[{\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}\right]=\text{0}\text{.0397 }M\\ {\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{\text{1}}=0.0397M\end{array}$

Now, the reaction of the second proton dissociation of phosphoric acid is depicted as

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

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

$\begin{array}{ccccc}& {\text{H}}_2{\text{PO}}_4{}^{-}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{HPO}}_4{}^{-}\left(aq\right)\\ \text{Initial concentration}\left(M\right)& 0.0397& -& 0.0397& 0\\ \text{Change in concentration}\left(M\right)& -y& -& +y& +y\\ \text{Equilibrium concentration}\left(M\right)& 0.0397-y& -& 0.0397+y& y\end{array}$

Now, substitute these concentrations in equation (2) as

$K_{\text{a2}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}$

$K_{\text{a2}}=\frac{\left(0.0397+y\right)\left(y\right)}{\left(0.0397-y\right)}$

Since the value of $K_{\text{a2}}$ is very small, the quantity of acid dissociated is less. Therefore, $\left(0.0397-y\right)$ and $\left(0.0397+y\right)$ can be approximated as $0.0397$. Now, substitute the value of ${\text{K}}_{\text{a2}}$ in the above equation as

$\begin{array}{c}6.25\times {10}^{-8}=\frac{\left(\cancel{0.0397}\right)\left(y\right)}{\cancel{0.0397}}\\ y=6.25\times {10}^{-8}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_2(aq)=6.25\times {10}^{-8}M$ and ${\left[{\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}\right]}_{\left(\text{O}\right)}=0.0397M$.

Calculate the percent dissociation from equation (4) as

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

$\begin{array}{c}\%\text{ dissociation}=\frac{6.25\times {10}^{-8}}{\text{0}\text{.0397}}\times 100\%\\ =0.00015\%\end{array}$

Since the percent dissociation is much less than $5\%$, the approximation taken is valid.

Now, the reaction of the third proton dissociation of phosphoric acid is depicted as

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

Prepare an equilibrium table and represent each of the species in terms of $z$ as follows:

$\begin{array}{ccccc}& {\text{HPO}}_4{}^{-}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{PO}}_4{}^{-}\left(aq\right)\\ \text{Initial concentration}\left(M\right)& 6.25\times {10}^{-8}& -& 6.25\times {10}^{-8}& 0\\ \text{Change in concentration}\left(M\right)& -z& -& +z& +z\\ \text{Equilibrium concentration}\left(M\right)& \left(6.25\times {10}^{-8}\right)-z& -& \left(6.25\times {10}^{-8}\right)+z& z\end{array}$

Now, substitute these concentrations in equation (3):

$K_{\text{a3}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[{\text{HA}}^{-}\right]}$

$K_{\text{a3}}=\frac{\left(\left(6.25\times {10}^{-8}\right)+z\right)\left(z\right)}{\left(\left(6.25\times {10}^{-8}\right)-z\right)}$

Since the value of ${\text{K}}_{\text{a3}}$ is very small, the quantity of acid dissociated is less. Therefore, $\left(\left(6.25\times {10}^{-8}\right)-z\right)$ and $\left(\left(6.25\times {10}^{-8}\right)+z\right)$ can be approximated as $6.25\times {10}^{-8}$. Now, substitute the value of $K_{\text{a3}}$ in the above equation as

$\begin{array}{c}4.8\times {10}^{-13}=\frac{\left(\cancel{6.25\times {10}^{-8}}\right)\left(z\right)}{\cancel{6.25\times {10}^{-8}}}\\ z=4.8\times {10}^{-13}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_3\left(aq\right)=4.8\times {10}^{-13}M$ and ${\left[{\text{PO}}_{\text{4}}{}^{-}\right]}_{\left(\text{O}\right)}=6.25\times {10}^{-8}M$.

Calculate the percent dissociation from equation (4) as

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

$\begin{array}{c}\%\text{ dissociation}=\frac{4.8\times {10}^{-13}}{6.25\times {10}^{-8}}\times 100\%\\ =0.00077\%\end{array}$

Since the percent dissociation is much less than $5\%$, the approximation taken is valid.

Also,

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_1+{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_2+{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_3\\ =0.0397+\left(6.25\times {10}^{-8}\right)+\left(4.8\times {10}^{-13}\right)\\ \simeq 0.0397M\end{array}$

Substitute this value in equation (5) to calculate pH of the solution:

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

$\begin{array}{c}\text{pH}=-\log \left[0.0397\right]\\ =1.40\end{array}$

Conclusion

The pH of the phosphoric acid solution is $1.40$ .