Question

A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was diluted to 1.500 L. The measured pH of the solution was 3.85. What is the pKa of the weak acid?

Answer 1

Answer : The value of $pK_a$ of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

$\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}$

$\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol$

Now we have to calculate the value of $pK_a$ of the weak acid.

The equilibrium chemical reaction is:

                          $HA+KOH\rightleftharpoons HK+H_2O$

Initial moles     0.25     0.03        0

At eqm.    (0.25-0.03)   0.03      0.03

                     = 0.22

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[HK]}{[HA]}$

Now put all the given values in this expression, we get:

$3.85=pK_a+\log (\frac{0.03}{0.22})$

$pK_a=4.72$

Therefore, the value of $pK_a$ of the weak acid is, 4.72