Calculate the pH of the solution that results?

Question

from the addition of 0.040 moles of HNO3 to a buffer made by combining 0.500 L of 0.260 M HC3H5O2 ( use Ka = 1.30 x 10 -5 ) and 0.500 L of 0.520 M

Answer 1

This is a 'Buffer' solution. A Buffer Sol'n is a sol'n that resists change in pH.
First calculate the pH without the addition of HNO3

[H+] = Ka x [HC3H5O2] / [HC3H4OO-]
[H+] = 1.3 x 10^-5 x [0.26 x 0.5] / [0.52 x 0.5]
[H+] = 1.3 x 10^-5 x 0.13 / 0.26
[H+] = 6.5 x 10^-6
pH = - log(10)[6.5 x 10^-6]
pH = - -5.187
pH = 5.19

On addition of 0.04 moles H+ (HNO3) this reduces the anion by 0.52 - 0.04 = 0.48
& increases the associated acid by 0.26 + 0.04 = 0.30

[H+] = 1.30 x 10^-5 x 0.30 / 0.48
[H+] = 8.125 x 10^-6
ph = - log (10)[8.125 x 10^-6]
pH = - -5.09
pH = 5.09

Comparing the two pH values 5.19 (without addition of HNO3) and 5.09 (with addition of HNO3). There is very little change (0.1), but the change is slightly acidic(caused by the addition of HNO3), but the 'buffer' has performed its function of resisting change.