Question

A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. After 25.0 mL of base is added, the pH of the solution is 3.42.Estimate the pKa of the weak acid.

Answer 1

Explanation:

The given reaction is as follows.

        $HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Hence, number of moles of NaOH are as follows.

        n = $0.05 L \times 0.1 M$

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = $0.025 L \times 0.1 M$

                = 0.0025 mol

According to ICE table,

         $HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = $\frac{0.0025 mol}{V}$

        [NaA] = $\frac{0.0025 mol}{V}$

       $[A^{-}] = [NaA] = \frac{0.0025 mol}{V}$

Now, we will calculate the $pK_{a}$ value as follows.

          pH = $pK_{a} + log \frac{A^{-}}{HA}$

       $pK_{a} = pH - log \frac{[A^{-}]}{[HA]}$

                  = $3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}$

                  = 3.42

Thus, we can conclude that $pK_{a}$ of the weak acid is 3.42.