$200 mL$ of $0.05 M$ $H X_{2} SO X_{4}$ (both protons fully ionised) is mixed with $300 mL$ of $0.1 M$ $NaOH$ and $100 mL$ of water. Calculate the pH of the final solution at 298 K to two decimal places.

Numerical answer type, tolerance 0.01.

Answer: 12.22

H X^{+}

= 200 \times 0.05 \times 2 = 20

mmol;

OH X^{-}

= 300 \times 0.1 = 30

mmol. Excess

OH X^{-}

= 10

mmol in

600 mL

[OH X^{-}] = 0.01667 M

pOH = 1.778

pH = 12.22