The normality of H₂ SO₄ in the solution obtained on mixing 100 mL of 0.1 M H₂ SO₄ with 50 mL of 0.1 M NaOH is _______________ × 10⁻¹ N. (Nearest Integer)

Solution

Chemistry ma’am · 4 steps

Count milliequivalents, let the NaOH eat some of the acid, then share what is left over the total volume.

1. Milliequivalents of acid. H₂SO₄ gives 2 H⁺ per molecule, so its n-factor is 2. Its milliequivalents are 100 × 0.1 × 2 = 20 meq.
2. Milliequivalents of base. NaOH gives one OH⁻, so its n-factor is 1. Its milliequivalents are 50 × 0.1 × 1 = 5 meq.
3. Neutralise. Acid and base always react in equal milliequivalents. So 5 meq of acid is used up and 20 - 5 = 15 meq of H₂SO₄ is left.

   N = (meq left)/(Vₜₒₜₐₗ (mL)) = 15/(100 + 50) = 0.1 N

4. So. The normality of H₂SO₄ is 0.1 N = 1 × 10⁻¹ N, so the answer is 1.

The common slip: 0. Taking the n-factor of H₂SO₄ as 1 gives 10 - 5 = 5 meq left, so N = 5/150 = 0.33 × 10⁻¹, which rounds to 0.

The idea to keep. For acid-base mixing, work in milliequivalents (M × n × V in mL). Subtract the smaller from the larger, then divide by the total volume of the mixture.

Answer from the NTA final answer key. Solution by PrepWiser.