JEE Main 2023 · 13 April · morning · Numerical answer · +4 / −0
20 mL of calcium hydroxide was consumed when it was reacted with 10 mL of unknown solution of H₂ SO₄. Also 20 mL standard solution of 0.5 M HCl containing 2 drops of phenolphthalein was titrated with calcium hydroxide, the mixture showed pink colour when burette displayed the value of 35.5 mL whereas the burette showed 25.5 mL initially. The concentration of H₂ SO₄ is _____________ M. (Nearest integer)
Answer and solution
Answer 1
Solution
Chemistry ma’am · 4 steps
Standardise the lime water against HCl first, then let it tell you how strong the acid is.
- Read the burette. The base used is the final reading minus the initial reading: 35.5 - 25.5 = 10 mL of Ca(OH)₂. The pink colour of phenolphthalein marks the end point.
- Find base strength. Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. The HCl taken is 20 × 0.5 = 10 mmol, so Ca(OH)₂ is 5 mmol in 10 mL, which is 0.5 M.
Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2H₂O
- Titrate the acid. 20 mL of 0.5 M Ca(OH)₂ is 10 mmol. The ratio is 1:1, so H₂SO₄ is also 10 mmol, present in 10 mL.
- So. [H₂SO₄] = (10 mmol)/(10 mL) = 1 M, so the answer is 1.
The common slip: 2. Treating Ca(OH)₂ as reacting 1:1 with HCl makes the base 1 M, and then H₂SO₄ comes out as 2 M.
The idea to keep. In a two-titration problem, the first titration fixes the strength of the common reagent. Always write the balanced equation, because the mole ratio changes with each acid.
Answer from the NTA final answer key. Solution by PrepWiser.