Chapter 16, Problem 148AP

Calculate the concentrations of all species in a $\text{0}\text{.100 }M{\text{ H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution.

Interpretation Introduction

Interpretation:

The concentrations of all the species present in the solution, which contains phosphoric acid, is to be determined.

Concept introduction:

The first ionization of the polyprotic acid takes place as

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

${\text{K}}_{\text{a1}}$ 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)

The second ionization of the polyprotic acid takes place as

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

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

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

The third ionization of the polyprotic acid takes place as

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

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

${\text{K}}_{\text{a3}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[{\text{HA}}^{-}\right]}$ …… (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.

Answer to Problem 148AP

Solution:

$\begin{array}{l}\left[{\text{H}}_3{\text{PO}}_4\right]=0.076\text{ M},\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]=0.0239\text{ M},\left[{\text{HPO}}_4{}^{2-}\right]\text{=}6.2\times {10}^{-8}\text{ M, }\\ \left[{\text{PO}}_4{}^{3-}\right]\text{=}1.1\times {10}^{-18}\text{M,}\left[{\text{H}}_3{\text{O}}^{+}\right]=0.0239\text{ M}\end{array}$

Explanation of Solution

Given information:

The concentration of phosphoric acid $\left({\text{H}}_3{\text{PO}}_4\right)$ at 25°Cis 0.100 M.

Refer to Table $16.8$ for ${\text{K}}_{\text{a1}}{\text{, K}}_{\text{a2}}{\text{, and K}}_{\text{a3}}$ values of phosphoric acid. The values are as given below:

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

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

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

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

$\begin{array}{ccccc}& {\text{H}}_3{\text{PO}}_4{}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{PO}}_4{{}^{-}}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 0.100& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.100-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (1) as

${\text{K}}_{\text{a1}}=\frac{\left(x\right)\left(x\right)}{\left(0.100-x\right)}$

Since the value of ${\text{K}}_{\text{a1}}$ is small, the amount of acid dissociated is less. Therefore, $\left(0.100-x\right)$ can be approximated as $0.100$. Now, substitute the value of ${\text{K}}_{\text{a1}}$ in above equation as

$\begin{array}{c}7.5\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{0.100}\\ x^2=\left(7.5\times {10}^{-3}\right)0.100\\ x=\sqrt{7.5\times {10}^{-4}}\\ x=2.74\times {10}^{-2}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{1\left(\text{eq}\right)}=2.74\times {10}^{-2}\text{M}$ and ${\left[{\text{H}}_3{\text{PO}}_{\text{4}}\right]}_{\left(\text{O}\right)}=0.100\text{ M}$

Calculate the percent dissociation from equation (4) as

$\begin{array}{c}\%\text{ dissociation}=\frac{2.74\times {10}^{-2}}{\text{0}\text{.100}}\times 100\%\\ =27.4\%\end{array}$

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

$\begin{array}{c}7.5\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{\left(0.100-x\right)}\\ x^2+\left(7.5\times {10}^{-3}\right)x=\left(7.5\times {10}^{-4}\right)\\ x=-0.031,0.0239\end{array}$

Since concentration cannot be negative, $x=0.0239$.

Thus,

$\begin{array}{c}{\left[{\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}\right]}_{\left(\text{O}\right)}=0.0239\text{ M}\\ {\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{\text{1}}=0.0239\text{ M}\end{array}$

Also,

$\begin{array}{c}\left[{\text{H}}_3{\text{PO}}_{\text{4}}\right]=\left(0.100-0.0239\right)\text{ M}\\ =0.076\text{ M}\end{array}$

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

${\text{H}}_2{\text{PO}}_4{{}^{-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{HPO}}_4{{}^{2-}}_{\left(\text{aq}\right)}$

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

$\begin{array}{ccccc}& {\text{H}}_2{\text{PO}}_4{{}^{-}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{HPO}}_4{{}^{2-}}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(\text{M}\right)& 0.0239& -& 0.0239& 0\\ \text{Change in concentration}\left(\text{M}\right)& -y& -& +y& +y\\ \text{Equilibrium concentration}\left(\text{M}\right)& 0.0239-y& -& 0.0239+y& y\end{array}$

Now, substitute these concentrations in equation (2) as

${\text{K}}_{\text{a2}}=\frac{\left(0.0239+y\right)\left(y\right)}{\left(0.0239-y\right)}$

Since the value of ${\text{K}}_{\text{a2}}$ is very small, the amount of acid dissociated is less. Therefore, $\left(0.0239-y\right)$ and $\left(0.0239+y\right)$ can be approximated as $0.0239$. Now, substitute the value of ${\text{K}}_{\text{a2}}$ in above equation as

$\begin{array}{c}6.2\times {10}^{-8}=\frac{\left(\cancel{0.0239}\right)\left(y\right)}{\cancel{0.0239}}\\ y=6.2\times {10}^{-8}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{2\left(\text{eq}\right)}=6.2\times {10}^{-8}\text{M}$ and ${\left[{\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}\right]}_{\left(\text{O}\right)}=0.0239\text{ M}$.

Also,

$\begin{array}{c}\left[{\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}\right]=\left(0.0239-6.2\times {10}^{-8}\right)\text{ M}\\ \approx 0.0239\text{ M}\end{array}$

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=0.0239\text{ M}$

Calculate the percent dissociation from equation (4) as

$\begin{array}{c}\%\text{ dissociation}=\frac{6.2\times {10}^{-8}}{0.0239}\times 100\%\\ =0.00026\%\end{array}$

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

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

${\text{HPO}}_4{{}^{2-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{PO}}_4{{}^{3-}}_{\left(\text{aq}\right)}$

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

$\begin{array}{ccccc}& {\text{HPO}}_4{{}^{2-}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{PO}}_4{{}^{3-}}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(\text{M}\right)& 6.2\times {10}^{-8}& -& 0.0239& 0\\ \text{Change in concentration}\left(\text{M}\right)& -z& -& +z& +z\\ \text{Equilibrium concentration}\left(\text{M}\right)& \left(6.2\times {10}^{-8}\right)-z& -& \left(0.0239\right)+z& z\end{array}$

Now, substitute these concentrations in equation (3) as

${\text{K}}_{\text{a3}}=\frac{\left(\left(6.2\times {10}^{-8}\right)+z\right)\left(z\right)}{\left(\left(6.2\times {10}^{-8}\right)-z\right)}$

Since the value of ${\text{K}}_{\text{a3}}$ is very small, the amount of acid dissociated is less. Therefore, $\left(\left(6.2\times {10}^{-8}\right)-z\right)$ can be approximated as $6.2\times {10}^{-8}$ and $\left(\left(0.239\right)+z\right)$ can be approximated as $0.0239$. Now, substitute the value of ${\text{K}}_{\text{a3}}$ in above equation as

$\begin{array}{c}4.8\times {10}^{-13}=\frac{\left(0.0239\right)\left(z\right)}{6.2\times {10}^{-8}}\\ z=\frac{\left(4.8\times {10}^{-13}\right)\left(6.2\times {10}^{-8}\right)}{0.0239}\\ z=1.1\times {10}^{-18}\text{M}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{3\left(\text{eq}\right)}=1.1\times {10}^{-18}\text{M}$ and ${\left[{\text{HPO}}_{\text{4}}{}^{2-}\right]}_{\left(\text{O}\right)}=6.2\times {10}^{-8}\text{ M}$

Calculate the percent dissociation from equation (4) as

$\begin{array}{c}\%\text{ dissociation}=\frac{1.1\times {10}^{-18}\text{M}}{6.2\times {10}^{-8}}\times 100\%\\ =0.0000\%\end{array}$

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

Also,

$\begin{array}{c}\left[{\text{HPO}}_{\text{4}}{}^{2-}\right]=\left(6.2\times {10}^{-8}-4.8\times {10}^{-13}\right)\text{ M}\\ \approx 6.2\times {10}^{-8}\text{ M}\end{array}$

And,

$\left[{\text{PO}}_4{}^{\text{3}-}\right]=1.1\times {10}^{-18}\text{M}$

Also,

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_1+{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_2+{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_3\\ =\left(0.0239\right)+\left(6.2\times {10}^{-8}\right)+\left(1.1\times {10}^{-18}\text{M}\right)\\ \simeq 0.0239\text{ M}\end{array}$

Therefore, the concentration of all the species in the phosphoric acid solution are as follows:

$\begin{array}{l}\left[{\text{H}}_3{\text{PO}}_4\right]=0.076\text{ M},\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]=0.0239\text{ M},\left[{\text{HPO}}_4{}^{2-}\right]\text{=}6.2\times {10}^{-8}\text{ M, }\\ \left[{\text{PO}}_4{}^{3-}\right]\text{=}1.1\times {10}^{-18}\text{M}\left[{\text{H}}_3{\text{O}}^{+}\right]=0.0239\text{ M}\end{array}$

Conclusion

Therefore, the concentration of all the species in the phosphoric acid solution are as follows:

$\begin{array}{l}\left[{\text{H}}_3{\text{PO}}_4\right]=0.076\text{ M},\left[{\text{H}}_2{\text{PO}}_4{}^{-}\right]=0.0239\text{ M},\left[{\text{HPO}}_4{}^{2-}\right]\text{=}6.2\times {10}^{-8}\text{ M, }\\ \left[{\text{PO}}_4{}^{3-}\right]\text{=}1.1\times {10}^{-18}\text{M}\left[{\text{H}}_3{\text{O}}^{+}\right]=0.0239\text{ M}\end{array}$