Question

The ionization constant of propanoic acid is $1.32 \times 10^{-5}$. Calculate the degree of ionization of the acid in its $0.05 \text{M}$ solution and also its \[\text{pH}\]. What will be its degree of ionization if the solution is $0.01 \text{M}$ in \[\text{HCl}\]?

Answer 1

Hint: We know that the dissociation constant for the complete reaction process is calculated by multiplying the dissociation constant of first and the second step. Thus we can say that the overall dissociation constant is used to calculate the hydrogen ion concentration.

Complete step by step answer:
We know that the dissociation constant can be expressed in terms of the degree of ionization for the weak acids and the total molar concentration of the acid. We can say that the value of degree of ionization for weak acids is very small.
We can take the value of \[\text{K}\] for propanoic acid from the question is $1.32 \times 10^{-5}$. Also it is given that the concentration of propanoic acid $0.05 \text{M}$, at this concentration value we have to calculate the degree of ionization of the acid and \[\text{pH}\] of the acid.
We have to suppose the degree of dissociation is $\alpha$. Therefore, we can calculate the degree of dissociation as follows.
$\begin{aligned} \alpha &=\sqrt{\frac{K}{c}} \\ &=\sqrt{\frac{1.32 \times 10^{-5}}{0.05}} \\ &=0.0162 \end{aligned}$
We know that the hydrogen ion concentration is calculated by the relation shown as follows. We will substitute the value of degree of dissociation and the concentration of acid to calculate the value of hydrogen ion concentration.
$[{H}^{+}]=c \times \alpha$
$={0.0162 \times 0.05}$
$=8.12 \times 10^{-4}$
Now, it is known that the pH is negative logarithm of hydrogen ion concentration, therefore, by putting the value of hydrogen ion concentration in the relation we can calculate the value of pH of the acid solution which is shown as follows.
$\begin{aligned} \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\ &=-\log \left(8.12 \times 10^{-4}\right) \\ &=3.09 \end{aligned}$
We can see that in the question the concentration of \[\text{HCl}\] at which the degree of dissociation is to be calculated is given as $0.01 \text{M}$. Therefore, we can calculate the required degree of ionization as follows.
$\begin{aligned} \alpha &=\sqrt{\frac{K}{c}} \\ &=\sqrt{\frac{1.32 \times 10^{-5}}{0.01}} \\ &=0.00132 \end{aligned}$

Thus, we can say that the degree of ionization of the propanoic acid in its $0.05M$ solution and its $pH$ is $0.0162$ and $3.09$ respectively. We have also calculated the degree of ionization if the solution is $0.01 M$ in $HCl$ is \[0.00132\].

Note:
We know that the degree of dissociation of an acid is a measure of its capacity to furnish hydrogen ions and hence a measure of its strength. It is known that the \[{pH}\] of the solution is the negative logarithm to the base ten of the concentration in moles per liter of the hydrogen ion which it contains.