posted by gaurav on .
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)
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.
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.