If 0.50 moles of NaOH were added to reach the equivalence point in a titration how many miles of tartaric acid reacted?
Tartaric acid has two ionizable H^+ but k1 is very close to k2; I assume the 0.50 mols reduced both and that's the equivalence point to which you refer.
H2T + 2NaOH ==> Na2T + 2H2O
0.5 mol NaOH x (1 mol H2T/2 mols NaOH) = 0.5 x 1/2 = 0.25 mols H2T
To determine the number of moles of tartaric acid that reacted, we need to use the balanced chemical equation for the reaction between NaOH and tartaric acid. Let's assume that the balanced equation is:
NaOH + H2Tartaric -> NaTartaric + H2O
From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of tartaric acid. This means that the number of moles of tartaric acid that reacted is equal to the number of moles of NaOH added.
Given that 0.50 moles of NaOH were added, we can conclude that 0.50 moles of tartaric acid reacted.
Thus, 0.50 moles of tartaric acid reacted.