10 ml of 0.5 M NaOH is added to 100 ml of 0.2 M formic acid , containing no formate anion.. (the pka of formic acid is 3.75) the resulting mixture has a ph close to:

Question

The answer is 3.3 but how do you get this answer?

Answer 1

millimoles NaOH = mL x M = 10 x 0.5 = 5
millimoles formic acid (HCOOH) = 100 x 0.2 = 20
HCOOH forms HCOONa which is sodium formate and that's the formate anion.
..............HCOOH + NaOH ==> HCOONa + H2O
I...............20.............0..................0................0
add...........................5...................................
C...........-5...............-5...................5...............5
E............15...............0..................5.................5

So here you have a buffer containing 15 mmols formic acid (the acid) and 5 mmols sodium formate (the salt or base), so plug into the Henderson-Hasselbalch equation and solve for pH. I get 3.27 which is close to 3.3.
You have to recognize that when you have a weak acid and you add a base, then a salt of the weak acid is formed. IF the acid is in excess then you see you always have a salt and some acid and that's a buffer.

Answer 2

30 mL of 1 M HCl were added to 200 mL of 0.1 M lysine solution having pH = pKa of the side-chain amino group. What is the new pH? pKa values for lysine are 2.18, 8.95 and (R-group) 10.53.

Answer is suppose to be 5.57