Calculate the H3O + concentration at the halfway point when 38.8 mL of 0.15 M HBr is titrated with 0.1 M KOH. Assume additive volumes. Answer in units of M

Answer 1

HBr + KOH ==> KBr + H2O .. balanced equation

moles HBr = 38.8 ml x 1 L / 1000 ml x 0.15 mol / L = 0.00582 mols

At halfway point, moles KOH = 1/2 x 0.00582 = 0.00291 mols KOH

Volume of KOH needed to provide 0.00291 mols = 0.00291 mols x 1 L / 0.1 mols = 0.0291 L = 29.1 mls

TOTAL VOLUME = 38.8 mls x 29.1 mls = 67.9 mls = 0.0679 L

HBr + KOH ==> KBr + H2O

0.00582...0.00291...0........0..........Initial

-0.00291...-0.00291......................Change

0.00291......0.................................Equilibrum

[HBr] = 0.00291 mols / 0.0679 L

[HBr] = 0.0429 M

Since HBr is a strong acid and ionizes 100%, the [H₃O⁺] will be 0.0429 M @ the halfway point.