Chapter 16, Problem 68QP

Compare the pH values for $0.10-M$ solutions of $\text{NaOH}$ and of ${\text{NH}}_{\text{3}}$ to illustrate the difference between a strong base and a weak base.

Interpretation Introduction

Interpretation:

The pH values of $0.10M{\text{ NaOH and NH}}_3$ solutions are to be compared.

Concept introduction:

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

${\text{BOH}}_{\left(\text{aq}\right)}\to {\text{B}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{OH}}^{-}{}_{\left(\text{aq}\right)}$

The ionization of a weak base takes place as:

${\text{B}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{OH}}^{-}{}_{\left(\text{aq}\right)}+{\text{HB}}^{+}{}_{\left(\text{aq}\right)}$

${\text{K}}_{\text{b}}$ is the measure of the dissociation of a base and is known as the base-ionization constant, which is specific at a particular temperature.

${\text{K}}_{\text{b}}=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{BH}}^{+}\right]}{\left[\text{B}\right]}$ …… (1)

The formula to calculate pOH is:

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$ …… (2)

pH is the measure of the acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution. The relationship between pH and pOH is:

$\text{pH}+\text{pOH}=14$ …… (3)

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

$\%\text{ dissociation}=\frac{{\left[{\text{OH}}^{-}\right]}_{\text{eq}}}{{\left[\text{B}\right]}_{\text{o}}}\times 100\%$ …… (4)

Here, ${\left[{\text{OH}}^{-}\right]}_{\text{eq}}$ is the hydroxide ions concentration at equilibrium and ${\left[\text{B}\right]}_{\text{o}}$ is the original base concentration.

Answer to Problem 68QP

Solution:

The pH of ${\text{NH}}_{\text{3}}$ is lower than that of $\text{NaOH}$ for equal concentrations as ${\text{NH}}_{\text{3}}$, being a weak base, produces less hydroxide ions than $\text{NaOH}$ does.

Explanation of Solution

Given information:

The concentration of the bases ${\text{NaOH and NH}}_3$ is given as $0.10M$.

$0.10M\text{ NaOH}$, being a strong base, ionizes completely into hydroxide ions and its conjugate acid, according to the reaction:

${\text{NaOH}}_{\left(\text{aq}\right)}\to {\text{Na}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{OH}}^{-}{}_{\left(\text{aq}\right)}$

Since there is complete dissociation of the base, the hydroxide ion concentration is equal to the amount of base dissolved. Thus,

$\left[{\text{OH}}^{-}\right]=0.10M$

Now, substitute this value in equation (2) to calculate the pOH of the solution, as follows:

$\begin{array}{c}\text{pOH}=-\log \left[\text{0}\text{.10}\right]\\ =1.00\end{array}$

Now, calculate the pH using equation (3), as follows:

$\begin{array}{c}\text{pH}+\text{1}\text{.00}=14\\ \text{pH}=13.00\end{array}$

Thus, pH of $0.10M\text{ NaOH}$ solution is $13.00$.

$0.10M{\text{ NH}}_3$, being a weak base, ionizes partially to give hydroxide ions, according to the reaction:

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

The concentration of ${\text{OH}}^{-}$ is determined by ${\text{K}}_{\text{b}}$ of the base. Now, prepare an equilibrium table and represent each of the species in terms of $x$, as follows:

$\begin{array}{ccccc}& {\text{NH}}_3{}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{OH}}^{-}{}_{\left(\text{aq}\right)}& {\text{NH}}_4{{}^{+}}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 0.10& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.10-x& -& x& x\end{array}$

Refer to table $16.7$ for the ${\text{K}}_{\text{b}}$ value of ${\text{NH}}_{\text{3}}$ as $1.8\times {10}^{-5}$.

Now, substitute these concentrations in equation (1), as follows:

${\text{K}}_{\text{b}}=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}$

Since the value of ${\text{K}}_{\text{b}}$ is very small, the amount of base dissociated is less. Therefore, $\left(0.10-x\right)$ can be approximated as $0.10$. Now, substitute the value of ${\text{K}}_{\text{b}}$ in the above equation:

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{0.10}\\ x^2=\left(1.8\times {10}^{-5}\right)0.10\\ x=\sqrt{1.8\times {10}^{-6}}\\ x=1.34\times {10}^{-3}\end{array}$

Thus, $\left[{\text{OH}}^{-}\right]=1.34\times {10}^{-3}M$

Calculate the percent dissociation from equation (4), as follows:

$\begin{array}{c}\%\text{ dissociation}=\frac{1.34\times {10}^{-3}}{\text{0}\text{.10}}\times 100\%\\ =1.34\%\end{array}$

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

Now, use equation (2) to calculate the pOH of the solution, as follows:

$\begin{array}{c}\text{pOH}=-\log \left(1.34\times {10}^{-3}\right)\\ =2.87\end{array}$

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

$\begin{array}{c}\text{pH}+\text{2}\text{.87}=14\\ \text{pH}=14-2.87\\ \text{pH}=11.13\end{array}$

Therefore, the pH of a $0.10M{\text{ NH}}_3$ solution is $11.13$.

The highpH of $\text{NaOH}$ shows that it ionizes more than ${\text{NH}}_3$. Thus, $\text{NaOH}$ is a strong base and ${\text{NH}}_3$ is weak base, which is justified by its lower pH value.

Conclusion

The pH values of a strong base $\text{NaOH}$ and a weak base ${\text{NH}}_3$, for equal concentrations, have been compared.