PH at the equivalence point?

Find the pH and volume (mL) of 0.0372 M NaOH needed to reach the equivalence point(s) in titrations of the fol

• 15.8 mL of 0.0868 M H2SO3 (two equivalence points) I already know that the first volume is 36.9 mL and the first pH is 7.13, however I can't figure out the second volume and pH. Thanks,

• Answer:

  0.0158 Litres @ 0.0868 Molar = 0.001371 moles acid @ 2 H+ ions / moole = 0.002743 moles H+ ion --------------------- 0.002743 moles H+ ion are destroyed by 0.002743 mol NaOH 0.002743 mol NaOH @ 0.0372 mol/Litre = 73.73 mls to reach the second equiv point & 1/2 it ...36.87 mls to reach the first (you are right) ------------------------- second pH comes from the second K... Ka2 Ka2 = 6.4 e-8 when the reaction is over, you have destroyed 0.001371 moles acid, and have produced 0.001371 moles of its conjugate base. this 0.001371 moles is diluted into 89.53 ml... giving a molarity of 0.0153 Molar (SO3)-2 ion the cojugate base (SO3)-2 reacts with water: (SO3)-2 & water --> HSO3- & OH- with a Kb = Kwater / Ka2 = 1e-14 / 6.4 e-8 = 1.56 e-7 1.56 e-7 = [HSO3-] [OH-] / [ (SO3)-2] 1.56 e-7 = [X][X] / [ 0.0153 Molar] X^2 = 2.39 e-9 X = OH = 4.89 e-5 pOH =4.31 pH = 14 -pOH = 9.69