Posted by Delli on .
A hypothetical weak acid HA, was combined with NaOH in the following proportions: 0.20 mol HA, 0.08 mol NaOH. The mixture was then diluted to a total volume of 1L, and the pH measured.
(a) If pH=4.80, what is the pKa of the acid
(b) how many additional moles of NaOH should be added to the solution to increase the pH to 5.00
HA + NaOH ==> NaA + HOH
0.2 mol HA to begin.
0.08 mol NaOH used to neutralize.
0.2-0.08 = 0.12 mol HA reamins.
0.08 mol A^- formed.
M = mols/L.
Ka = (H^+)(A^-)/(HA)
You know (H^+), from pH, you know A^- and you know HA. Solve for Ka, then pKa. Post your work if you get stuck.