Situation: HA(aq) + OH-(aq) ? H2O + A-(aq)
Unknown acid (0.4g)+(50ml water) was titrated with NaOH to reach the equivalence point. Result : addition of 15ml NaOH was needed to reach the equivalence point.
Then this solution was added with 7.5ml of HCl which is 1/2 of the volume of NaOH. This will lead to the desired condition [HA] = [A-] to calculate Ka.
Ka = [H+][A-] / [HA] Since [HA] = [A-], Ka = [H+]
Question: If too much HCl is accidentally added after the titration with the correct amount of NaOH, what effect would this have on the Ka?
Explain your answer