# What is pH of a 0.002*mol*L^-1 solution of HCl(aq)?

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$
$\left[{H}_{3} {O}^{+}\right]$ $=$ $\frac{{10}^{-} 2 \cdot L \times 0.1 \cdot m o l \cdot {L}^{-} 1}{500 \cdot m L \times {10}^{-} 3 L \cdot m {L}^{-} 1}$
$=$ $2.0 \times {10}^{-} 3 m o l \cdot {L}^{-} 1$
And $- {\log}_{10} \left(2.0 \times {10}^{-} 3\right)$ $=$ $- \left(- 2.70\right) = 2.70$