Determine the volume of sodium hydroxide added when [HA] = [A-]?

Question

I have a graph which shows how the pH changes when 0.12M NaOH is added to 25cm³ of a solution of weak monoprotic acid HA. The graph shows pH/volume of NaOH.

I'm asked to calculate the intial concentration of HA. Does [H+] = [HA]? If so, I found it to be 0.02.

I'm also asked to determine the volume of NaOH when [HA] = [A-] but I don't know how to do this. Can anyone help me?

Answer 1

Since it is given that acid HA is a weak monoprotic acid, thus the graph given showing pH vs. added volume of 0.12M NaOH must show a tilted "S" shape. basically, pH increases relatively quickly at the beginning and the end of titration, and there is a single point along the curve where pH increases the slowest. That point is the middle of the buffering region, where [HA] = [A-]. The x-(volume of 0.12M NaOH) coordinate of this point indicates volume of 0.12M NaOH needed to reach the condition [HA] = [A-]. The y-(pH) coordinate of this point indicates kPa of this acid.
Since your graph is not visible to me, let this volume (x-coordinate) be represented by V ml. V ml is the volume of NaOH when [HA] = [A-].
I believe your answer to the first question is wrong. Note that 0.12*V (mMole) is the amount of NaOH used to reach this point [HA] = [A-], and this amount of NaOH is used to convert half of HA to A-. Hence 2*(0.12V) = 0.24V (mMol) is the intial amount of HA, and 0.24V (mMol)/ 25ml is the intial concentration of HA.