Question

The pH of a 0.78M solution of 4-pyridinecarboxylic acid HC6H4NO2is measured to be 2.53.Calculate the acid dissociation constant Ka of 4-pyridinecarboxylic acid. Round your answer to 2 significant digits.

Answer 1

Answer : The value of acid dissociation constant is, $1.1\times 10^{-5}$

Solution :  Given,

Concentration pyridinecarboxylic acid = 0.78 M

pH = 2.53

First we have to calculate the hydrogen ion concentration.

$pH=-\log [H^+]$

$2.53=-\log [H^+]$

$[H^+]=2.95\times 106{-3}M$

Now we have to calculate the acid dissociation constant.

The equilibrium reaction for dissociation of (weak acid) is,

                        $HC_6H_4NO_2\rightleftharpoons C_6H_4NO_2^-+H^+$

initially conc.         0.78                0             0

At eqm.               (0.78-x)              x             x  

The expression of acid dissociation constant for acid is:

$k_a=\frac{[C_6H_4NO_2^-][H^+]}{[C_6H_4NO_2]}$

As, $[H^+]=[C_6H_4NO_2^-]=x$

So,  $x=2.95\times 106{-3}M$

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

$k_a=\frac{(x)\times (x)}{(0.78-x)}$

$k_a=\frac{(2.95\times 106{-3})\times (2.95\times 106{-3})}{(0.78-2.95\times 106{-3})}$

$K_a=1.1\times 10^{-5}$

Therefore, the value of acid dissociation constant is, $1.1\times 10^{-5}$