Q.1 What will be the pH at the equivalence point during the titration of a 100 ml 0.2M solution of CH3COONa with 0.2M of solution of HCl?(Ka = 2*10^-5)

Question

Q.2 Aniline behaves as a weak base.When 0.1M,50ml solution of aniline was mixed with 0.1M,25ml solution of HCl the pH of resulting solution was 8.Then calculate the pH of 0.001M solution of anilinum chloride.

Answer 1

You are starting with M x L = 0.020 moles acetate. The equivalence point with 0.20 M HCl will be at 100 mL; therefore, you will have 0.020 moles acetic acid formed and it will be in a volume of 200 mL; therefore, the concn acetic acid will be 0.020/0.200. Set up an ICE chart and solve for H^+, then convert to pH.

2. I assume the information in the first part is to calculate pKa for aniline but I am not positive of that. phNH2 is aniline.
phNH2 + HCl ==> phNH3^+ + Cl^-

You started with 50 mL x 0.1 M aniline (5 mmoles) and added 25 mL of 0.1 M HCl (2.5 mmols). That will form 2.5 mmoles of the aniline hydrochloride salt and you will have 5.0-2.5 = 2.5 mmoles aniline left over and the pH of the buffered solution is 8. Use the Henderson-Hasselbalch equation to calculate pKa.
pH = pKa + log(base/acid)
8 = pKa + log (2.5/2.5)
8 = pKa.

phNH3^+ + HOH ==> phNH2 + H3O^+

Set up an ICE chart for hydrolyzing the salt, then substitute into Keq.
Ka = (H3O^+)(phNH2)/(phNH3^+)
I looked in a table of Kb values for aniline and found 3.97 x 10^-10 which is not pKa of 8 so I don't know which value you are supposed to use. If you don't use 8 I don't know why the firt part of q 2 was involved.