200 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M H₂ SO₄. The pH of the mixture is _________.
Given: log 2=0.30, log 3=0.48, log 5=0.70, log 7=0.84, log 11=1.04

Solution

Chemistry ma’am · 4 steps

Count moles of H⁺ from each acid, add them, then divide by the total volume of 600 mL.

1. H+ from HCl. HCl is a strong monobasic acid, so each mole gives one mole of H⁺. Moles of H⁺ = 200 mL × 0.01 M = 2 mmol.
2. H+ from H2SO4. H₂SO₄ is treated as a strong dibasic acid here, so each mole gives two moles of H⁺. Moles of H⁺ = 2 × 400 mL × 0.01 M = 8 mmol.

   [H⁺] = (2 + 8 mmol)/(200 + 400 mL) = 10/600 M = 1/60 M

3. Take the log. pH = -log 1/60 = log 60 = log 2 + log 3 + log 10 = 0.30 + 0.48 + 1 = 1.78.
4. So. The pH of the mixture is 1.78, option (b).

The idea to keep. For a mixture of strong acids, add the moles of H⁺ (count two per H₂SO₄), divide by the total volume, then take -log.

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