What will be the resultant pH when 200 ml of an aqueous solution of HCl (pH = 2) is mixed with 300 ml of an aqueous solution of NaOH (pH = 12).

The HCl solution is 0.01 M, giving $$200 \times 0.01 = 2$$ mmol of $$\text{H}^+$$, and the NaOH solution is 0.01 M, giving $$300 \times 0.01 = 3$$ mmol of $$\text{OH}^-$$. After neutralisation, 1 mmol of $$\text{OH}^-$$ remains in 500 mL, i.e. $$[\text{OH}^-] = 0.002\text{ M}$$. Then $$\text{pOH} = -\log(2 \times 10^{-3}) \approx 2.7$$ and $$\text{pH} = 14 - 2.7 = 11.3$$.