Titration?

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
A solution of 0.2522 M NaOH is used to neutralize 20.00 mL H2SO4 solution. If 35.36 mL NaOH solution is required to reach the endpoint, what is the molarity of the H2SO4 solution?

According to the equation, mole ratio H₂SO₄ : NaOH = 1 : 2
No. of milli-moles of NaOH reacted = (0.2522 mmol/mL) × (35.36 mL) = 8.918 mmol
No. of milli-moles of H₂SO₄ reacted = (8.918 mol) × (1/2) = 4.459 mmol
Molarity of H₂SO₄ solution = (4.459 mmol) / (20.00 mL) = 0.2230 M


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If 27.33 mL of 0.2184 M KOH is required to completely neutralize 20.00 mL of a CH3COOH solution, what is the molarity of the acetic acid solution?
CH₃COOH(aq) + KOH(aq) → CH₃COOK(aq) + H₂O(l)

According to the equation, mole ratio CH₃COOH : KOH = 1 : 1
No. of milli-moles of KOH = (0.2184 mmol/mL) × (27.33 mL) = 5.969 mmol
No. of milli-moles of CH₃COOH = 5.969 mmol
Molarity of CH₃COOH solution = (5.969 mmol) / (20 mL) = 0.2985 M