2007-07-29 15:26:55 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
If you use the henderson-hasselbach equation, which says,
pH = pKa + log ( [ base]/[acid])
at the half-equivalence point, the base and acid are in equal proportions, so [base] = [acid].
log [base]/[acid] = log 1 = 0
so
pH = pKa
2007-07-29 15:43:15 · answer #1 · answered by wholikeshomework 2 · 0⤊ 0⤋
Well the other poster is correct if it is a weak acid/strong base titration which i assume is what you meant. In a strong acid/strong base titration the pH at the half eq point is found by finding the new concentration of acid by dividing the concentration in half and then using this as M1 for the M1V1=M2 V2 to find the new concentration of acid paying attention to both the neutralization and dilution effects of the base.
2007-07-29 23:04:38 · answer #2 · answered by Anonymous · 0⤊ 0⤋
You mean a weak acid, strong base titration. Strong acids have Ka >1, which implies a pH of 0.
2007-07-29 22:53:25 · answer #3 · answered by cattbarf 7 · 0⤊ 1⤋