chemistry question

Question

A dibase acid, H₂J has the concentration of 0.5 mol dm^-3. Letter J is not the actual symbol of the element.

What is the volume of potassium hydroxide, KOH, 1.0mol dm^-3 that can neutralize 25.0 cm³ of the H₂J acid solution?

Answer 1

H₂J + 2KOH ==> K₂J + 2H₂O ... balanced equation

mols H₂J present = 25.0 cm³ x 1 dm³ / 1000 cm³ x 0.5 mol / dm³ = 0.0125 mol H₂J

mols KOH needed = 0.0125 mols H₂J x 2 mols KOH / 1 mol H₂J = 0.025 mols KOH

volume of KOH needed: 0.025 mols KOH x 1 dm³ / 1.0 mol = 0.025 dm³ = 25 cm³

(If you are using correct sig. figs., I'd report the answer as 0.03 dm³ or 30 cm³ (1 sig. fig.)