Question

0.1M weak acid HA (pH = 3) is titrated with .05M NaOH soln. Calculate the pH when 25% of acid solution has been neutralised.

Answer 1

first of all, we have to find $K_a$ of given acid.

$HA+H_2O\rightleftharpoons H_3O^++A^-$

so, $K_a=\frac{[H_3O^+][A^-]}{[HA]}$

here, pH = 3

from formula , $pH=-log[H_3O+]$

so, 3 = - log[H3O^+]

or, [H3O^+] = 10^-3 mol/L

at equilibrium,

[H3O^+] = [A^-] = 10^-3 mol/L

and [HA] = initial [HA] - [H3O+]

= 0.1 - 10^-3

= 0.1 - 0.001

= 0.099 mol/L

so, $K_a=\frac{10^-3\times10^-3}{0.099}$

Ka = 1.01 × 10^-5

and pKa = -log(1.01 ×10^-5) ≈ 5

Suppose we have taken 10 ml of 0.1M weak acid, HA . For complete neutralisation 20 ml of 0.05 NaOH will be required.

so for 25% neutralisation 5ml of NaOH will be required.

at this point,

number of millimole of HA present in the titration mixture = 0.75 millimole

The number of m.moles of salt present in the mixture = 0.25 millimole.

now, $pH=pK_a+log\frac{[salt]}{[acid]}$

= 5 + log(0.25/0.75)

= 5 + log(1/3)

= 5 - log3

= 4.522