H2A is a diprotic acid with Ka1 =2.5x10–5,Ka2 =3.1x10–9. If 100.0mL of 1.00MNaOH is added to 200.0 mL of 1.00 M H2A, calculate the pH of the resultant solution.
So you ignore Ka2 because its past 10^-3
H2A+OH- - HA- + H2O Ka=2.5x10^-5 then you hit the half point and its just the pKa.

Am I interpreting properly?

Question

H2A is a diprotic acid with Ka1 =2.5x10–5,Ka2 =3.1x10–9. If 100.0mL of 1.00MNaOH is added to 200.0 mL of 1.00 M H2A, calculate the pH of the resultant solution.
So you ignore Ka2 because its past 10^-3 
H2A+OH- -> HA- + H2O Ka=2.5x10^-5 then you hit the half point and its just the pKa.

Am I interpreting properly?

vincent-lyon.fr · Answer

You are interpreting ok!
Congrats!

anonymous · Answer

Thanks again!