pH = pKa + log [base/acid] (H-H equation) Using the H-H equation, calculate the theoretical volume of 0.2 M NaOH required to raise the pH of 0.1 M Tris HCl (pk = 8.1) to raise the pH to its pK value. The way I see it is to get pH = pK so using H-H equation: 8.1 = 8.1 + log[0.2 M NaOH/0.1 M Tris HCl] 8.1-8.1 = log[0.2 M NaOH/0.1 M Tris HCl] 0 = log[0.2 M NaOH/0.1 M Tris HCl] Then taking anti log, 1 = [0.2 M NaOH/0.1 M Tris HCl] 1 = 0.5 means pH = 1, has base:acid ratio of 1/2 Thus if I have 0.1 mol/litre Tris HCL, I need half of 0.2 M NaOH. But how do I convert these to volumes? Please forgive me for asking this Q. I am really stuck. Am I on the right track? If so, how do I proceed to get the volumes. If not, what am I thinking that is wrong? Please give up a detailed understanding of what I did wrong? I would really appreciate it. Thanks. Then I am suppose to draw a curve but how can I? with only one point at the theoretical volume of 0.2 M NaOH. I just don't understand. Please help me understand what they want from me.