Given the concentration of $\mathrm{OH}^{-}$ ions is $1.58 \times 10^{-6} \ \mathrm{M}$. First, calculate the concentration of $\mathrm{H}^{+}$ ions: $$ \begin{aligned} [\mathrm{H}^{+}] &= \frac{K_w}{[\mathrm{OH}^{-}]} \\ &= \frac{10^{-14}}{1.58 \times 10^{-6}} \\ &\approx 6.329 \times 10^{-9} \ \mathrm{M} \end{aligned} $$ Now, calculate the pH: $$ \begin{aligned} \mathrm{pH} &= -\log[\mathrm{H}^{+}] \\ &= -\log(6.329 \times 10^{-9}) \\ &\approx 8.20 \end{aligned} $$