$100 mL$ of $0.1 M$ $HCl$ and $200 mL$ of $0.05 M$ $H X_{2} SO X_{4}$ (treat both protons as completely ionised) are mixed with $100 mL$ of $0.28 M$ $NaOH$ . Find the pH to two decimals ( $lo g 5 = 0.699$ ).

Numerical answer type, tolerance 0.01.

Answer: 2.3

n (H X^{+}) = 10 + 20 = 30 mmol

;

n (OH X^{-}) = 28 mmol

. Excess

H X^{+} = 2 mmol

400 mL

gives

[H X^{+}] = 0.005 M

and

pH = 2.301 \approx 2.30