A 20 ml sample of a 0.2 M solution of the weak diprotic acid H2A is titrated with 0.25 M NaOH .Molarity of solution at the second equivalence point is:

1. 0.1 M NaHA
2. 0.153 M Na2A
3. 0.1 M Na2A
4. 0.0769 Na2A

H2A + NaOH = NaHA + H2ONaHA + NaOH = Na2A + H2OH2A + 2NaOH = Na2A + H2OIn 0.2 M of 20ml H2A we have 0.2*20/1000 = 4 mmoles Therefore moles of Na2A will be 4mmoles from the overall reaction.Now for reaching first end point M1V1 = M2V2 . Therefore volume of NaOH i.e. V2 = 16ml. Thus we have 20+16 = 36ml of volume at first end point.To reach second end point more 16ml of NaOH will be required since it is clear from the equation. Thus at end of second end point we have 36+16= 52 ml of volume which is of the solution of Na2A and H2OThus molarity of Na2A = no of moles of Na2A/ volume of solution= 4/52 = 0.0769 moles/L