Question

One mole of a weak acid HA was dissolved in 2.0 L of solution. After the system had come to equilibrium, the concentration of HA was found to be 0.45 M. Calculate K_a for HA.

Short answer

For the given acid,${\mathrm{K}}_{\mathrm{a}}=5.5\times {10}^{-3}$

Step-by-step solution

Calculating Initial Concentration

The concentration of the given weak acid:

$\begin{array}{c}\left[\mathrm{Acid}\right]=\frac{\text{Mole}}{\text{Volume(L)}}\\ =\frac{1}{2}\\ =0.5\text{\hspace{0.17em}mol/L}\end{array}$

Amount that is dissociated

The acid dissociates as:

$\begin{array}{l}\text{HA\hspace{1em}}\rightleftharpoons {\text{\hspace{1em}H}}^{\text{+}}{\text{+A}}^{\text{-}}\\ 0.500\\ 0.5-\mathrm{x}\mathrm{x}\mathrm{x}\end{array}$

The given concentration of HA at equilibrium =0.45

That means:

$\begin{array}{c}0.5-\mathrm{x}=0.45\\ \mathrm{x}=0.05\end{array}$

Ka Calculation

From the dissociation of the acid, we can say:

$\begin{array}{c}{\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{x}}^2}{0.5-\mathrm{x}}\\ =\frac{{(0.05)}^2}{0.45}\\ =5.5\times {10}^{-3}\end{array}$