For the second stoichiometric point of H
2S
2O
3, basically I did some calculations and got that for S
2O
32- + H
2O
HS
2O
32- + OH
-, the [OH
-]=4.30*10
-8, which is what the answer key had. I didn't have any problems with the calculations up to that point, it was after this where I have questions.
Basically, the answer key says that since 4.30*10
-8 < [OH
-]
water, so the pH=7.0 at the second stoichiometric point. This seems a bit odd to me because if this were the case, is the pH of the second stoichiometric point of all diprotic acids be pH=7.0 or higher? And also, If you calculated [H
+] from [OH
-]=4.30*10
-8, shouldn't it give a [H
+]>10
-7, so pH<7.0?