Chapter 16, Problem 85QP

Compare the pH of a $\text{0}\text{.040 }M\text{ HCl}$ solution with that of a $0.040M{\text{ H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution. (Hint: ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is a strong acid: $K_{\text{a}}$ for ${\text{HSO}}_{\text{4}}^{\text{-}}\text{=1}{\text{.3×10}}^{\text{-2}}\text{)}\text{.}$

Interpretation Introduction

Interpretation:

The pH values of $0.040M{\text{ HCl and H}}_2{\text{SO}}_4$ solutions are to be compared.

Concept introduction:

A strong acid is an electrolyte that gets completely dissociated when dissolved in water to produce hydronium ions and its conjugate base.

${\text{AH}}_{\left(\text{aq}\right)}+{\text{H}}_3{\text{O}}^{+}{}_{\left(\text{aq}\right)}\to {\text{A}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_3{\text{O}}^{+}{}_{\left(\text{aq}\right)}$

The first ionization of the diprotic acid takes place as:

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

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

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

The second ionization of the diprotic acid takes place as:

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

${\text{K}}_{\text{a2}}$ 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.

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

The formula to calculate the pH is:

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

Percent ionization is the percentage of base that gets dissociated upon addition in water. It depends on the hydroxide ion concentration.

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

Here, ${\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}$ is the hydroxide ions concentration at equilibrium and ${\left[\text{AH}\right]}_{\text{o}}$ is original base concentration

Answer to Problem 85QP

Solution:

The pH of ${\text{H}}_2{\text{SO}}_4$ is lower than that of $\text{HCl}$ for equal concentrations.

Explanation of Solution

Given information:

The concentration of both the acids ${\text{HCl and H}}_2{\text{SO}}_4$ is given as $0.040M$.

For $0.040M\text{ HCl}$, being a strong acid, it ionizes completely into hydronium ions and its conjugate base according to the reaction:

${\text{HCl}}_{\left(\text{aq}\right)}+{\text{H}}_3{\text{O}}^{+}{}_{\left(\text{aq}\right)}\to {\text{Cl}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_3{\text{O}}^{+}{}_{\left(\text{aq}\right)}$

Since, complete dissociation of the acid takes place, the hydronium ion concentration is equal to the amount of acid dissolved. Thus,

$\left[{\text{H}}_3{\text{O}}^{+}\right]=0.040M$

Now, substitute this value in equation (3) to calculate the pH of the solution as:

$\begin{array}{c}\text{pH}=-\log \left[\text{0}\text{.040}\right]\\ =1.40\end{array}$

For $0.040M{\text{ H}}_2{\text{SO}}_4$,

$\begin{array}{l}{\text{K}}_{\text{a1}}=\text{ very large}\\ {\text{K}}_{\text{a2}}=1.3\times {10}^{-2}\end{array}$

Since, ${\text{K}}_{\text{a1}}$ of sulfuric acid is very large, its first ionization is complete and all the sulfuric acid is dissociated into ${\text{HSO}}_{\text{4}}{}^{-}{\text{ and H}}_3{\text{O}}^{+}$ according to the reaction,

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

Thus,

$\begin{array}{c}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_1=\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right]\\ =0.040M\end{array}$

Also,

$\begin{array}{c}{\left[{\text{HSO}}_4{}^{-}\right]}_1=\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right]\\ =0.040M\end{array}$

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

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

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

$\begin{array}{ccccc}& {\text{HSO}}_4{{}^{-}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{SO}}_4{{}^{-}}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 0.040& -& 0.040& 0\\ \text{Change in concentration}\left(M\right)& -y& -& +y& +y\\ \text{Equilibrium concentration}\left(M\right)& 0.040-y& -& 0.040+y& y\end{array}$

Now, substitute these concentrations in equation (2) as:

${\text{K}}_{\text{a2}}=\frac{\left(0.040+y\right)\left(y\right)}{\left(0.040-y\right)}$

Since the value of ${\text{K}}_{\text{a2}}$ is small, the amount of acid dissociated is less. Therefore, $\left(0.040-y\right)$ and $\left(0.040+y\right)$ can be approximated as $0.040$. Now, substitute the value of ${\text{K}}_{\text{a2}}$ in the above equation as:

$\begin{array}{c}1.3\times {10}^{-2}=\frac{\left(\cancel{0.040}\right)\left(y\right)}{\cancel{0.040}}\\ y=1.3\times {10}^{-2}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{2\left(\text{eq}\right)}=1.3\times {10}^{-2}M$, ${\left[{\text{HSO}}_{\text{4}}{}^{-}\right]}_{\left(\text{O}\right)}=0.040M$

Calculate the percent dissociation from equation (4) as:

$\begin{array}{c}\%\text{ dissociation}=\frac{1.3\times {10}^{-2}}{\text{0}\text{.040}}\times 100\%\\ =32.5\%\end{array}$

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

$\begin{array}{c}\text{1}\text{.3}\times {\text{10}}^{-2}=\frac{\left(0.040+y\right)\left(y\right)}{\left(0.040-y\right)}\\ \left(5.2\times {10}^{-4}\right)-\left(\text{1}\text{.3}\times {\text{10}}^{-2}\right)y=0.040y+y^2\\ 0=y^2+0.053y-\left(5.2\times {10}^{-4}\right)\\ y=-0.061,\text{ 8}\text{.46}\times {\text{10}}^{-3}\end{array}$

Since concentration cannot be negative so, $y=\text{8}\text{.46}\times {\text{10}}^{-3}$.

Thus,

$\begin{array}{l}\left[{\text{HSO}}_{\text{4}}{}^{-}\right]=\text{8}\text{.46}\times {\text{10}}^{-3}M\\ {\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_2=\text{8}\text{.46}\times {\text{10}}^{-3}M\end{array}$

Also,

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_1+{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_2\\ =0.040+\left(\text{8}\text{.46}\times {\text{10}}^{-3}\right)\\ \simeq 4.846\times {\text{10}}^{-2}M\end{array}$

Use equation (3) to calculate the pH of the solution as:

$\begin{array}{c}\text{pH}=-\log \left[4.846\times {\text{10}}^{-2}\right]\\ =1.31\end{array}$

Therefore, the pH of $0.040M\text{ HCl}$ is higher than the pH of $0.040M{\text{ H}}_2{\text{SO}}_4$.

Conclusion

The pH values of the monoprotic and diprotic strong acids for equal concentration have been compared. The pH of $0.040M\text{ HCl}$ is higher than the pH of $0.040M{\text{ H}}_2{\text{SO}}_4$.