600 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M H₂ SO₄. The pH of the mixture is ___________ × 10⁻². (Nearest integer)
[Given log 2=0.30
log 3=0.48
log 5=0.69
log 7=0.84
log 11=1.04]

Solution

Chemistry ma’am · 4 steps

Count the moles of H⁺ from each acid, remembering that every H₂SO₄ gives two, then divide by the total volume.

1. H⁺ from HCl. HCl is a strong monobasic acid. It gives 600 mL × 0.01 M = 6 mmol of H⁺.
2. H⁺ from H2SO4. H₂SO₄ is dibasic and is taken as fully ionised here, so each mole gives 2 moles of H⁺. That is 400 mL × 0.01 M × 2 = 8 mmol.
3. Total concentration. Total H⁺ is 6 + 8 = 14 mmol in 600 + 400 = 1000 mL, so [H⁺] = 0.014 M = 1.4 × 10⁻² M.

   pH = 2 - log 1.4 = 2 - (log 2 + log 7 - 1) = 2 - (0.30 + 0.84 - 1) = 1.86

4. So. The pH is 1.86 = 186 × 10⁻², so the answer is 186.

The common slip: 200. Treating H₂SO₄ as giving one H⁺ makes the total 10 mmol in 1 L, so pH = 2 and the answer comes out as 200.

The idea to keep. When strong acids are mixed, add their moles of H⁺, not their molarities, and divide by the total volume. A dibasic acid like H₂SO₄ counts twice.

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