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What volume of 0.123 M NaOH(aq) must be added to 125 mL of 0.197 M H2SO3(aq) to reach the following points?

(a) the first stoichiometric point
(b) the second stoichiometric point

2 answers

2nd.
H2SO3 + 2NaOH ==> Na2SO3 + 2H2O
mols H2SO3 = M x L = ?
mols NaOH = 2 times mols H2SO3
It takes that many mols NaOH to neutralize BOTH H ions of H2SO3.
Then mols NaOH = M x L; you know mols NasOH and M, solve for L NaOH.

Take half that volume to find L for the first stoichiometric point.
moles of H2SO3 = 0.0246
moles of 2NaOH = 0.04925

Second stiochiometric point volume: 400. mL
First stiochiometric point volume: 200. mL

Thanks!