Calculate the pH of the solution that results from the addition of 0.020 moles of HNO3 to a buffer?

Question

made by combining 0.500 L of 0.240 M HC3H5O2 ( Ka = 1.30 x 10-5 ) and 0.500 L of 0.240 M NaC3H5O2.

Answer 1

To make life simple, call propionic acid PrOH and its anion PrO-. With both the acid and the anion in solution at a goodly conc, upsets in pH are resisted. In this case, the H+ from the nitric acid reacts with PrO- to form the undissociated PrOH.

Now to the math. For this weak acid,
[PrO-][H+]/[PrOH]= 1.3x10-5
When the buffer is first prepared, the amounts of PrOH and PrO- are relatively unchanged. Taking the volume increase into consideration, the buffer solution is 0.12 M in each. The effect of the HNO3 addition is to decrease PrO- to 0.10 and increase PrOH to 0.14.
Thus, 0.1 H+/0.14 = 1.3x10-5
H+ = 1.8 x10-5 (appx)
To convert, pH = 5 - log 1.8 = 4.74 appx.