Chapter 16, Problem 156AP

Calculate the pH of a solution that is 1.00 M $\text{HCN}$ and 1.00 M HF. Compare the concentration (in molarity) of the CN^- ion in this solution with that in a 1.00 M $\text{HCN}$ solution. Comment on the difference.

Interpretation Introduction

Interpretation:

The pH of a solution that contains $\text{HF and HCN}$ is to be calculated. Also, the concentration of ${\text{CN}}^{-}$ ion in this solution is to be compared with a $1.00\text{ M HCN}$ solution.

Concept introduction:

pH is the measure of the acidity of a solution thatdepends on the concentration of hydronium ions and temperature of the solution.

The formula to calculate the pH of the solution from the concentration of hydronium ions is:

$\text{pH}=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$ …… (1)

The ionization of the weak acid takes place as follows:

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

$K_a$ is the measure of dissociation of an acid and is known as acid-ionization constant that is specific at a particular temperatures.

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

Answer to Problem 156AP

Solution: $\text{pH}$ is $1.57$

The concentration of ${\text{CN}}^{-}$ is higher in $1.00\text{ M HCN}$ as compared to the $1.00\text{ M HF and HCN}$ solution.

Explanation of Solution

Given information:

A solution of two weak acids that contains $1.00\text{ M HF and HCN}$.

Refer to table $16.6$ for the acid ionization constant of hydrocyanic acid as $K_a=4.9\times {10}^{-10}$ and for hydrofluoric acid as $K_a=7.1\times {10}^{-4}$.

As $K_a$ for $\text{HF}$ is much larger than the ${\text{K}}_{\text{a}}$ of $\text{HCN}$, the pH of the solution is largely dependent on the ionization of $\text{HF}$ and the ionization of $\text{HCN}$ can be assumed negligible.

When a weak acid is dissolved in water, only partial dissociation into ${\text{H}}_3{\text{O}}^{+}$ ions and its conjugate base takes place. Thus, the concentration of ${\text{H}}_3{\text{O}}^{+}$ is determined by $K_a$ of the acid.

The reaction of the weak acid $\text{HF}$ is depicted as follows:

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

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

$\begin{array}{ccccc}& \text{HF}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{F}}^{-}\left(aq\right)\\ \text{Initial concentration}\left(\text{M}\right)& 1.00& -& 0& 0\\ \text{Change in concentration}\left(\text{M}\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(\text{M}\right)& 1.00-x& -& x& x\end{array}$

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

$K_a=\frac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}$

Since the value of $K_a$ is very small, thus the amount of acid dissociated is less. Therefore, $\left(1.00-x\right)$ can be approximated as $1.00$. Now, substitute the value of $K_a$ in the above equation,

$\begin{array}{c}7.1\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{1.00}\\ x^2=\left(7.1\times {10}^{-4}\right)1.00\\ x=\sqrt{7.1\times {10}^{-4}}\\ x=0.027\end{array}$

Thus, the hydrogen ion concentration of the solution is $0.027\text{M}$.

Now, use equation (1) to calculate the pH of the solution as follows:

$\begin{array}{c}\text{pH}=-\log \left(0.027\right)\\ =1.57\end{array}$

The pH of the solution is $1.57$.

Since, $\text{HCN}$ is very weak, it will ionize to a very small extent. Thus, its concentration in the solution is approximately equal to $1.00\text{M}$.

Apply equation (2) to calculate the concentration of ${\text{CN}}^{-}$ as follows:

$\begin{array}{c}{K}_a=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}\\ 4.9\times {10}^{-10}=\frac{\left(0.027\right)\left[{\text{CN}}^{-}\right]}{1.00}\\ \left[{\text{CN}}^{-}\right]=\frac{\left(4.9\times {10}^{-10}\right)\left(1.00\right)}{\left(0.027\right)}\\ \left[{\text{CN}}^{-}\right]=1.8\times {10}^{-8}M\end{array}$

In only $1.00\text{ M HCN}$, the ionization of $\text{HCN}$ takes place as follows:

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

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

$\begin{array}{ccccc}& \text{HCN}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{CN}}^{-}\left(aq\right)\\ \text{Initial concentration}\left(\text{M}\right)& 1.00& -& 0& 0\\ \text{Change in concentration}\left(\text{M}\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(\text{M}\right)& 1.00-x& -& x& x\end{array}$

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

$K_a=\frac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}$

Since the value of $K_a$ is very small, thus the amount of acid dissociated is less. Therefore, $\left(1.00-x\right)$ can be approximated as $1.00$. Now, substitute the value of $K_a$ in above equation.

$\begin{array}{c}4.9\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{1.00}\\ x^2=\left(4.9\times {10}^{-10}\right)1.00\\ x=\sqrt{4.9\times {10}^{-10}}\\ x=2.2\times {10}^{-5}\end{array}$

Thus, $\left[{\text{CN}}^{-}\right]=2.2\times {10}^{-5}\text{M}$

The concentration of ${\text{CN}}^{-}$ is higher in $1.00\text{ M HCN}$ as compared to the $1.00\text{ M HF and HCN}$ solution.

According to the Le Chatelier’s principle, the higher concentration of ${\text{H}}_3{\text{O}}^{+}$ from $\text{HF}$ shifts the ionization of $\text{HCN}$ to the left, decreasing the concentration ${\text{CN}}^{-}$ in $1.00\text{ M HF and HCN}$.

Conclusion

The pH of the solution containing $1.00\text{ M HF and HCN}$ is $1.57$. The concentration of ${\text{CN}}^{-}$ in the solution containing $1.00\text{ M HF and HCN}$ is less in comparison to the $1.00\text{ M HCN}$ solution.