Question

A 100 ml solution of pH 10 was well-mixed with a 100 ml solution of pH 4. The pH of the resultant 200 ml solution is _______ (rounded off to two decimal places).

Hint:

When mixing equal volumes of acid and base solutions with very different pH, the stronger contributor (lower pH) dominates. Use average moles and divide by final volume for precise pH.

Answer 1

Step 1: Calculate [OH$^-$] in pH 10 solution.
pH = 10 $⇒$ pOH = 14 - 10 = 4.
\([OH^-] = 10^{-4}\, \text{M}\). Moles of OH$^-$ in 100 ml: \[ n_{OH^-} = 10^{-4} \times 0.1 = 1.0 \times 10^{-5}\, \text{mol}. \] Step 2: Calculate [H$^+$] in pH 4 solution.
pH = 4 $⇒ [H^+] = 10^{-4}\, \text{M}. \] Moles of H$^+$ in 100 ml: \[ n_{H^+} = 10^{-4} \times 0.1 = 1.0 \times 10^{-5}\, \text{mol}. \] Step 3: Net neutralization.
\(n_{H^+} = n_{OH^-}\). So they neutralize exactly. Step 4: Residual [H$^+$].
Effective [H$^+$] after mixing = \(\dfrac{10^{-4} + 10^{-10}}{2}\, \text{M} \approx 5 \times 10^{-5}\, \text{M}\). \[ pH = -\log(5 \times 10^{-5}) \approx 4.30. \]