Chapter 16, Problem 147AP

Calculate the pH of a $\text{2}\text{.00 }M{\text{ NH}}_{\text{4}}\text{CN}$ solution.

Interpretation Introduction

Interpretation:

The pH of the solution with dissolved salt is to be calculated, with given concentration.

Concept introduction:

The relationship between $K_{\text{a}}\text{ and }{K}_{\text{b}}$ of a weak acid and its conjugate base or vice versa is:

$K_{\text{w}}=K_{\text{a}}\times K_{\text{b}}$…… (1)

$K_a$ is the measure of dissociation of an acid and is known as acid-ionization constant, which is specific at a particular temperature:

$K_a=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[\text{B}\right]}{\left[{\text{BH}}^{+}\right]}$…… (2)

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

$K_b=\frac{\left[{\text{OH}}^{-}\right]\left[\text{HA}\right]}{\left[{\text{A}}^{-}\right]}$…… (3)

The formula to calculate the pOH of the solution from the concentration of hydroxide ions is given as follows:

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$ …… (4)

The relation between pH and pOH is as follows:

$\text{pH}=14-\text{pOH}$…… (5)

The formula to calculate the pH of a solution is:

$\text{pH}=-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=-\text{log}\left[{\text{H}}^{+}\right]$ as $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=\left[{\text{H}}^{+}\right]$

Answer to Problem 147AP

Solution: $11.80$.

Explanation of Solution

Given information:

The concentration of ${\text{NH}}_4\text{CN}$ is $2.00\text{ M}$.

In the salt ${\text{NH}}_4\text{CN}$, the ions present are ${\text{NH}}_4{}^{+}{\text{ and CN}}^{-}$.

The cation ${\text{NH}}_4{}^{+}$ comes from a weak base ${\text{NH}}_3$. Thus, it will recombine with the water molecule to produce a weak base and hydronium ions.

Refer to table $16.7$ for $K_b$ value of base ammonia as:

$K_b=1.8\times {10}^{-5}$

The reaction is as follows:

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

$K_a$ of ${\text{NH}}_4{}^{+}$ that is the conjugate acid of base ammonia is calculated by equation (1) as follows:Substitute the values of $K_b$ and $K_{\text{w}}$ as:

$K_{\text{w}}=K_{\text{a}}\times K_{\text{b}}$

$\begin{array}{c}{K}_a\times \left(1.8\times {10}^{-5}\right)=1.0\times {10}^{-14}\\ K_a=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ K_a=5.6\times {10}^{-10}\end{array}$

Now, prepare an equilibrium table and represent each of the species in terms of $x$ as shown below.

$\begin{array}{ccccc}& {\text{NH}}_4{}^{+}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{NH}}_3\left(aq\right)\\ \text{Initial concentration}\left(\text{M}\right)& 2.00& -& 0& 0\\ \text{Change in concentration}\left(\text{M}\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(\text{M}\right)& 2.00-x& -& x& x\end{array}$

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

$K_a=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[\text{B}\right]}{\left[{\text{BH}}^{+}\right]}$

$5.6\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(2.00-x\right)}$

Assume $\left(2.00-x\right)$ as $2.00$ due to low ionization of the weak acid.

$\begin{array}{c}5.6\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(2.00\right)}\\ x^2=1.12\times {10}^{-9}\\ x=3.35\times {10}^{-5}\end{array}$

Thus, $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]=3.35\times {10}^{-5}\text{ M}$.

Similarly, the anion ${\text{CN}}^{-}$ comes from a weak acid $\text{HCN}$. Thus, it will recombine with the water molecule to produce a weak acid and hydroxide ions.

Refer to table $16.6$ for $K_a$ value of hydrocyanic acid as:

$K_a=4.9\times {10}^{-10}$

The reaction is given as follows:

${\text{CN}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

$K_b$ of ${\text{CN}}^{-}$ thatis the conjugate base of hydrocyanic acid is calculated as follows by equation (1) and the values of $K_a$ and $K_{\text{w}}$ are substituted.

$K_{\text{w}}=K_{\text{a}}\times K_{\text{b}}$

$\begin{array}{c}{K}_b\times \left(4.9\times {10}^{-10}\right)=1.0\times {10}^{-14}\\ K_b=\frac{1.0\times {10}^{-14}}{4.9\times {10}^{-10}}\\ K_b=2.0\times {10}^{-5}\end{array}$

Now, prepare an equilibrium table and represent each of the species in terms of $y$ as shown below.

$\begin{array}{ccccc}& {\text{CN}}^{-}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{OH}}^{-}\left(aq\right)& \text{HCN}\left(aq\right)\\ \text{Initial concentration}\left(\text{M}\right)& 2.00& -& 0& 0\\ \text{Change in concentration}\left(\text{M}\right)& -y& -& +y& +y\\ \text{Equilibrium concentration}\left(\text{M}\right)& 2.00-y& -& y& y\end{array}$

Now, substitute these concentrations in equation (3) as follows:

$K_b=\frac{\left[{\text{OH}}^{-}\right]\left[\text{HA}\right]}{\left[{\text{A}}^{-}\right]}$

$2.04\times {10}^{-5}=\frac{\left(y\right)\left(y\right)}{\left(2.00-y\right)}$

Assume $\left(2.00-y\right)$ as $2.00$ due to low ionization of the weak base.

$\begin{array}{c}2.04\times {10}^{-5}=\frac{\left(y\right)\left(y\right)}{\left(2.00\right)}\\ y^2=4.0\times {10}^{-5}\\ y=6.32\times {10}^{-3}\end{array}$

Thus, $\left[{\text{OH}}^{-}\right]=6.32\times {10}^{-3}\text{ M}$.

${\text{CN}}^{-}$ is a stronger base and ${\text{NH}}_4{}^{+}$ is an acid. Thus, some of the ${\text{OH}}^{-}$ will be used to neutralize the ${\text{H}}_3{\text{O}}^{+}$ in the solution according to the reaction given as follows:

${\text{OH}}^{-}\left(aq\right)+{\text{H}}_3{\text{O}}^{+}\left(aq\right)\to {\text{H}}_2\text{O}\left(aq\right)$

The equilibrium table for this reaction is shown below.

$\begin{array}{cccc}& {\text{OH}}^{-}\left(aq\right)& {\text{H}}_3{\text{O}}^{+}\left(aq\right)& {\text{H}}_2\text{O}\left(aq\right)\\ \text{Initial concentration}\left(\text{M}\right)& 6.32\times {10}^{-3}& 3.35\times {10}^{-5}& 0\\ \text{Change in concentration}\left(\text{M}\right)& -3.35\times {10}^{-5}& -3.35\times {10}^{-5}& +3.35\times {10}^{-5}\\ \text{Equilibrium concentration}\left(\text{M}\right)& 6.29\times {10}^{-3}& 0& 3.35\times {10}^{-5}\end{array}$

Thus, $\left[{\text{OH}}^{-}\right]=6.29\times {10}^{-3}\text{ M}$.

Use equation (4) to calculate the pOH of the solution as follows:

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

$\begin{array}{c}\text{pOH}=-\log \left(6.29\times {10}^{-3}\right)\\ =2.20\end{array}$

Now, use equation (5) to calculate the pH of the solution as follows:

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

$\begin{array}{c}\text{pH}=14-2.20\\ =11.80\end{array}$

Conclusion

The pH of the salt solution containing ${\text{NH}}_4\text{CN}$ is $11.80$.