Calculate the pH of solution obtained by mixing 10 ml of 0.1 M HCl and 40 ml of 0.2 M ${\mathrm{H}}_2{\mathrm{SO}}_4$

Milliequivalent of ${\mathrm{H}}^{+}$ from HCl = 10 x 0.1 = 1

Milliequivalent of ${\mathrm{H}}^{+}$ from ${\mathrm{H}}_2{\mathrm{SO}}_4$ = 40 x 0.2 x 2 = 16

Total Meq of ${\mathrm{H}}^{+}$ in solution = 1 + 16  = 17

$\left[{\mathrm{H}}^{+}\right] = \frac{17}{50} = 3.4\times {10}^{-1}\left[\therefore \left[{\mathrm{H}}^{+}\right]=\frac{\mathrm{Meq}}{\mathrm{V} \mathrm{in} \mathrm{ml}}\right]$

$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=-\log \left[0.34\right]=0.4685$