Mixing equal volumes of pH 10 (basic) and pH 4 (acidic) solutions results in complete neutralization, giving a final pH of 7.00 in the 200 ml mixture. This occurs because H⁺ and OH⁻ ions react stoichiometrically at equal concentrations. The calculation assumes strong acid/base behavior at 25°C where Kw = 10-14.

Step-by-Step Calculation

pH 4 solution has [H⁺] = 10-4 M, so 100 ml (0.1 L) contains 10-4 × 0.1 = 10-5 mol H⁺.

pH 10 solution has pOH = 4, so [OH⁻] = 10-4 M, yielding 10-5 mol OH⁻ in 100 ml.

H⁺ + OH⁻ → H₂O neutralizes exactly 10-5 mol each, leaving no excess ions in 200 ml; thus [H⁺] = 10-7 M from water.

Final pH = −log(10-7) = 7.00.

Common Misconceptions Explained

  • Averaging pH values (e.g., (10+4)/2 = 7) works coincidentally here but fails generally, as pH is logarithmic.
  • For two acids (e.g., pH 4 + pH 6), total H⁺ moles average yields pH ≈3.30, not 5.
  • Weak acids/bases require equilibrium constants; strong ones neutralize directly.

Verification Table

Parameter Acid (pH 4, 100 ml) Base (pH 10, 100 ml) Mixture (200 ml)
[H⁺] or [OH⁻] (M) 10-4 H⁺ 10-4 OH⁻ 10-7 H⁺
Moles 10-5 10-5 0 excess
pH 4 10 7.00

CSIR NET Exam Tips

  • Practice mole-based neutralization over pH averaging for ionic mixtures.
  • Use Python/Mathematica for log calculations in complex cases.
  • Similar questions test H⁺/OH⁻ balance in buffers or dilutions.

Tags: pH calculation, acid-base neutralization, CSIR NET chemistry, pH mixture, strong acid strong base