What will be the resultant pH when 200 ml of an aqueous solution of (pH = 2.0) is mixed with 300 ml of an aqueous solution of (pH = 12.0)?

Question

What will be the resultant pH when 200 ml of an aqueous solution of $HCl$ (pH = 2.0) is mixed with 300 ml of an aqueous solution of $NaOH$ (pH = 12.0)?

Prashant Kotian · Accepted Answer

Correct answer: B. 1. solution (pH = 2.0): [H+] = 10^-2 M = 0.01 M. Moles of H+ = 0.01 M × 0.2 L = 0.002 mol = 2.0 mmol.
2. solution (pH = 12.0 => pOH = 2.0): [OH-] = 10^-2 M = 0.01 M. Moles of OH- = 0.01 M × 0.3 L = 0.003 mol = 3.0 mmol.
When mixed: H+ + OH- -> . Excess OH- = 3.0 mmol - 2.0 mmol = 1.0 mmol = 0.001 mol.
Total volume V = 200 + 300 = 500 ml = 0.5 L.
Resulting [OH-] = 0.001 mol / 0.5 L = 0.002 M = 2.0 × 10^-3 M.
pOH = -log[OH-] = 3 - log 2 = 2.70.
pH = 14.0 - pOH = 11.30.

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