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