the balanced equation of for the above reaction is as follows;
2KOH + H₂SO₄ ---> K₂SO₄ + 2H₂O
stoichiometry of KOH to H₂SO₄ is 2:1
the number of KOH moles reacted - 0.124 mol/L x 0.046 L = 0.0057 mol
for complete neutralisation the base should react with acid in 2:1 molar ratio
if 2 mol of KOH reacts with 1 mol of H₂SO₄
then 0.0057 mol of KOH reacts with - 1/2 x 0.0057 mol = 0.0029 mol
the number of H₂SO₄ moles in 28.1 mL - 0.0029 mol
Molarity is number of H₂SO₄ moles in 1 L
therefore number of H₂SO₄ moles in 1 L - 0.0029 mol / 0.0281 L = 0.103 M
molarity of H₂SO₄ is 0.103 mol/L
