Chapter 16, Problem 68QP

Interpretation Introduction

Interpretation:

The pH of the solutions prepared by the dissolution of the given weak bases are to be determined at $25{}^{\circ }\text{C}$.

Concept introduction:

The ionization of the weak base takes place as

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

The formula to calculate the pOH of the solution from the concentration of hydroxide ions is

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$ …… (1)

K_b is the measure of dissociation of a base and is known as the base-ionization constant, which is specific at a particular temperature.

$K_{\text{b}}=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{BH}}^{+}\right]}{\left[\text{B}\right]}$ …… (2)

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

$\%\text{ dissociation}=\frac{{\left[{\text{OH}}^{-}\right]}_{\text{eq}}}{{\left[\text{B}\right]}_{\text{o}}}\times 100\%$ …… (3)

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

pH is the measure of 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

$\text{pH}+\text{pOH}=14$ …… (4)

Answer to Problem 68QP

Solution:

(a)

$11.11$

(b)

$8.96$

Explanation of Solution

a) $0.10M$.

Refer to Table $16.7$ for the $K_{\text{b}}$ value of ammonia as $1.8\times {10}^{-5}$.

When a weak base is dissolved in water, only partial dissociation takes place into ${\text{OH}}^{-}$ ions and its conjugate acid. Thus, the concentration of ${\text{OH}}^{-}$ is determined by the K_b of the base.

The reaction of ammonia is depicted as

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

First, prepare an equilibrium table and represent each of the species in terms of $x$ as follows:

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

Now, substitute these concentrations in equation (2) as

$K_{\text{b}}=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{BH}}^{+}\right]}{\left[\text{B}\right]}$

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

Since the value of K_b is very small, the quantity of base dissociated is less. Therefore, $\left(0.10-x\right)$ can be approximated as $0.10$. Now, substitute the value of K_b in the above equation:

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{0.10}\\ x^2=\left(1.8\times {10}^{-5}\right)0.10\\ x=\sqrt{1.8\times {10}^{-6}}\\ x=1.3\times {10}^{-3}\end{array}$

Thus, $\left[{\text{OH}}^{-}\right]=1.3\times {10}^{-3}M$

Calculate the percent dissociation from equation (3) as follows:

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

$\begin{array}{c}\%\text{ dissociation}=\frac{1.3\times {10}^{-3}}{\text{0}\text{.10}}\times 100\%\\ =1.34\%\end{array}$

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

Now, use equation (1) to calculate the pOH of the solution:

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

$\begin{array}{c}\text{pOH}=-\log \left(1.3\times {10}^{-3}\right)\\ =2.89\end{array}$

Now, use equation (4) to calculate the pH of the solution:

$\text{pH}+\text{pOH}=14$

$\begin{array}{c}\text{pH}+2.89=14\\ \text{pH}=14-2.89\\ \text{pH}=11.11\end{array}$

Therefore, the pH of the solution is $11.11$.

b) $0.050M$, $K_{\text{b}}$ value of pyridine is $1.7\times {10}^{-9}$.

When a weak base is dissolved in water, only partial dissociation takes place into ${\text{OH}}^{-}$ ions and its conjugate acid. Thus, the concentration of ${\text{OH}}^{-}$ is determined by the K_b of the base.

The reaction of pyridine is depicted as follows:

${\text{C}}_5{\text{H}}_5\text{N(}aq\text{)}+{\text{H}}_2\text{O(}l\text{)}\rightleftharpoons {\text{OH}}^{-}(aq)+{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}(aq)$

First, prepare an equilibrium table and represent each of the species in terms of $x$ as follows:

$\begin{array}{ccccc}& {\text{C}}_5{\text{H}}_5{\text{N}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{OH}}^{-}{}_{\left(\text{aq}\right)}& {\text{C}}_5{\text{H}}_5{\text{NH}}^{+}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 0.050& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.050-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (2):

$K_{\text{b}}=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{BH}}^{+}\right]}{\left[\text{B}\right]}$

$K_{\text{b}}=\frac{\left(x\right)\left(x\right)}{\left(0.050-x\right)}$

Since the value of K_b is very small, the quantity of base dissociated is less. Therefore, $\left(0.050-x\right)$ can be approximated as $0.050$. Now, substitute the value of K_b in the above equation:

$\begin{array}{c}1.7\times {10}^{-9}=\frac{\left(x\right)\left(x\right)}{0.050}\\ x^2=\left(1.7\times {10}^{-9}\right)0.050\\ x=\sqrt{8.5\times {10}^{-11}}\\ x=9.2\times {10}^{-6}\end{array}$

Thus, $\left[{\text{OH}}^{-}\right]=9.2\times {10}^{-6}M$

Calculate the percent dissociation from equation (3):

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

$\begin{array}{c}\%\text{ dissociation}=\frac{9.2\times {10}^{-6}}{\text{0}\text{.050}}\times 100\%\\ =0.018\%\end{array}$

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

Now, use equation (1) to calculate the pOH of the solution:

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

$\begin{array}{c}\text{pOH}=-\log \left(9.2\times {10}^{-6}\right)\\ =5.04\end{array}$

Now, use equation (4) to calculate the pH of the solution:

$\text{pH}+\text{pOH}=14$

$\begin{array}{c}\text{pH}+5.04=14\\ \text{pH}=14-5.04\\ \text{pH}=8.96\end{array}$

Therefore, the pH of the solution is $8.96$.