Calculate the pH of the solution that results from the addition of 0.020 moles of HNO3 to a buffer made by com

Question

Calculate the pH of the solution that results from the addition of 0.020 moles of HNO3 to a buffer made by combining
0.500 L of 0.290 M NH4Cl ( use Ka = 5.60 x 10 -10 )
and 0.500 L of 0.580 M

I get 9.39 and it's wrong.

Answer 1

Moles(HNO3) = 0.02

pH calculation without the addition of HNO3

[H+] = Ka x [0.58/0/5] / [0.29/0.5]
[H+] = 5.6 x 10^-10 x [0.29] / [ 0.145]
[H+] = 1.12 x10^-9
pH = - log(10)[1.12 x 10^-9]
pH = - -8.95 = 8.95

With the addition of 0.02 moles(HNO3), 0.02 moles of H+ is being added to the system. This removes chloride ions to HCl
so NH4Cl becomes less.
Hence
0.29 + 0.02 = 0.31 &
0.145 - 0.02 = 0.125
[H+] = 5.6 x10^-10 x 0.31 / 0.125
[H+] = 1.388 x 10^-9
pH = - log(10)[1.388 x 10^-9]
pH = - -8.86 = 8.86
pH = 8.86

NB the values of pH have reduced (slightly more acidic) from 8.95 (without addition of HNO3) to 8.86 (with addition of HNO3).
This is the whole point of 'buffers' they resist change in pH.
Hope this helps!!!!!