Assuming complete dissociation,calculate the $pH$ of the following solutions:
$(i) \ 0.003 \ M \ HCl$
$(ii) \ 0.005 \ M \ NaOH$
$(iii) \ 0.002 \ M \ HBr$
$(iv) \ 0.002 \ M \ KOH$

$(i) \ 0.003 \ M \ HCl$
$HCl \rightarrow H^{+} + Cl^{-}$
$[H^{+}] = 0.003 \ M$
$pH = -\log[H^{+}] = -\log(3 \times 10^{-3}) = 3 - \log(3) = 3 - 0.477 = 2.523 \approx 2.52$
$(ii) \ 0.005 \ M \ NaOH$
$NaOH \rightarrow Na^{+} + OH^{-}$
$[OH^{-}] = 0.005 \ M$
$pOH = -\log[OH^{-}] = -\log(5 \times 10^{-3}) = 3 - \log(5) = 3 - 0.699 = 2.301$
$pH = 14 - pOH = 14 - 2.301 = 11.699 \approx 11.70$
$(iii) \ 0.002 \ M \ HBr$
$HBr \rightarrow H^{+} + Br^{-}$
$[H^{+}] = 0.002 \ M$
$pH = -\log[H^{+}] = -\log(2 \times 10^{-3}) = 3 - \log(2) = 3 - 0.301 = 2.699 \approx 2.70$
$(iv) \ 0.002 \ M \ KOH$
$KOH \rightarrow K^{+} + OH^{-}$
$[OH^{-}] = 0.002 \ M$
$pOH = -\log[OH^{-}] = -\log(2 \times 10^{-3}) = 3 - 0.301 = 2.699$
$pH = 14 - pOH = 14 - 2.699 = 11.301 \approx 11.30$