Answer:
At the second equivalent point 200 mL of NaOH is required.
Explanation:
at the first equivalent point:
H2A + OH- = HA- + H2O
initial mmoles y*100 y*100 - -
final mmoles 0 0 y*100 y*100
at the second equivalent point:
HA- + OH- = A2- + H2O
initial mmoles y*100 y*100 - -
final mmoles - - y*100 y*100
at the second equivalent point we have that y*100 mmoles of NaOH or 100 mL of NaOH ir required, thus:
at the second equivalent point 200 mL of NaOH is required.