A litre of buffer solution contains 0 . 1 mole of each of NH _ 3...

In resultant solution after addition of $HCl$ to the basic buffer:

${n}_{{NH}_{3}} = 0 . 1 - 0 . 02 = 0 . 08$

${n}_{{NH}_{4}Cl} = n{NH}_{4}^{+} = 0 . 1 + 0 . 02 = 0 . 12$

Using Henderson-Hasselbach equation:

$pOH = {pK}_{b} + log\dfrac{\left[{NH}_{4}^{+}\right]}{\left[{NH}_{3}\right]}$

$= 4 . 745 + log\dfrac{0 . 12}{0 . 08}$

$= 4 . 745 + log\dfrac{3}{2}$

$= 4 . 745 + 0 . 477 - 0 . 301$

$pOH = 4 . 921$
We know,

$pH = 14 - pH$

$= 9 . 079$