Question

0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was found to be 0.013 M. Determine the value of Ka for the weak acid.

Answer 1

Answer: The value of $k_{a}$ for the weak acid is $3.07 \times 10^{-4}$.

Explanation:

First, we will calculate the molarity of HA as follows.

     [HA] = $\frac{\text{no. of moles}}{\text{volume}}$

             = $\frac{0.03 mol}{2 L}$

             = 0.015

At equilibrium,

              $HA + H_{2}O \rightleftharpoons H_{3}O^{+} + A^{-}$

Initial:   0.015                  0          0

Change:  -x                     +x         +x

Equilibm: 0.015 - x          x           x

It is given that, at equilibrium

          [HA] = 0.015 - x = 0.013

             - x = 0.013 - 0.015

                  = 0.002

Now, expression for $k_{a}$ of this reaction is as follows.

          $k_{a} = \frac{[A^{-}][H_{3}O^{+}]}{[HA]}$

                      = $\frac{x^{2}}{[HA]}$

                      = $\frac{(0.002)^{2}}{0.013}$

                      = $3.07 \times 10^{-4}$

Thus, we can conclude that the value of $k_{a}$ for the weak acid is $3.07 \times 10^{-4}$.