Question:
If a 25.0mL sample of sulphuric acid is titrated with 50.0mL of 0.025M sodium hydroxide solution to a phenolphthalein end point, what is the molarity of the acid?
If a 25.0mL sample of sulphuric acid is titrated with 50.0mL of 0.025M sodium hydroxide solution to a phenolphthalein end point, what is the molarity of the acid?
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Always account for stoichiometric coefficients in acid–base titrations.
Updated On: Mar 19, 2026
- 0.020M
- 0.100M
- 0.025M
- 0.050M
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The Correct Option is C
Solution and Explanation
Step 1: Moles of NaOH used:
nNaOH = 0.025 × 0.050 = 0.00125mol
Step 2: Reaction:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Step 3: Moles of H₂SO₄:
nₐcid = (0.00125)/(2) = 0.000625mol
Step 4: Molarity of acid:
M = (0.000625)/(0.025) = 0.025M
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