Show that for the titration of a weak acid, HA, with NaOH solution, at half-titer, pH = pKa.
Please explain your answer. Thank you.
HA + NaOH ==> NaA + HOH
Ka for HA = (H^+)(A^-)/(HA) and solve for H^+.
(H^+) = Ka*[(HA)/(A^-)
When HA is half neutralized, an equal amount of NaA has been formed, therefore, (HA) = (A^-) at the half-way point of the titration. (You can prove this over and over by starting with ANY volume of HA and adding half that volume of an equimolar NaOH solution.)Therefore, the term (HA)/(A^-) = 1 and
(H^+) = Ka
Take the negative log of both sides
-log(H^+) = -log Ka
pH = pKa.