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

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 0.140 M HC3H5O2 ( Ka = 1.30 x 10-5 )
and 0.500 L of 0.140 M NaC3H5O2

Answer 1

Simple! 0!

Answer 2

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Answer 3

I'm assuming that the volume that is being added is negligible.

The nitric acid will react with the sodium propionate to produce propionic acid. You started with 0.070 moles of each, so you'll end up with 0.050 moles of NaC3H5O2 and 0.090 moles of HC3H5O2. Neglecting the volume increase from the nitric acid, the total volume is 1 L, so you have 0.050 M base and 0.090 M acid.

Plug those into the Henderson-Haselbach equation:

pH = pKa + log [base]/[acid]
pH = 4.89 + log (0.050/0.090) = 4.89 - 0.26 = 4.63

Answer 4

Ever heard of the Henderson-Hasselback equation? It makes this problem oh-so-very easy

pH = pKa + log ([base]/[acid])

pKa is found in the same way as the pH -> -log Ka.
As for the concentration divisions - just take you conjugate acid and your conjugate base, find the molarities of each and divide them. If acid or base is added to the solution, add and subtract moles of conjugate acid/base as necessary (and added acid eats up base and adds an equal amount of acid and vice versa), "letting" the acid/base reaction occur before calculating the pH.

Hope it helps!

Answer 5

ph=1.07