The pH of a solution obtained by mixing 50mL of 1MHCl and 30mL of...

Milli equivalents of $HCl\left({N}_{a}{V}_{a}\right) = 50 \times 1 = 50$
Milli equivalents of $NaOH\left({N}_{b}{V}_{b}\right) = 30 \times 1 = 30$
Since ${N}_{a}{V}_{a} > {N}_{b}{V}_{b}$
\left[{H}^{+}\right] = \dfrac{{N}_{a}{V}_{a} - {N}_{b}{V}_{b}}{{V}_{a} + {V}_{b}} = \dfrac{50 - 30}{80} = \dfrac{20}{80} = 0 . 25 = 2 . 5 \times {10}^{- 1}$<br />$pH = - \log \left[{H}^{+}\right] = - \log \left(2 . 5 \times {10}^{- 1}\right) = 1 - 0 . 3979 = 0 . 6021$<br />$pH \times {10}^{4} = 0 . 6021 \times {10}^{4} = 6021