[OH⁻] = 1.6 × 10⁻⁸ mol / dm³
By definition, [tex][\text{H}^{+}] = 10^{-\text{pH}}[/tex], where [tex][\text{H}^{+}][/tex] is the concentration of proton in the solution.
pH = 6.2 for this solution. As a result, [tex][\text{H}^{+} = 10^{-6.2} = 6.31 \times 10^{-7} \; \text{mol} \cdot \text{dm}^{-3}[/tex].
[tex][\text{H}^{+}] \cdot [\text{OH}^{-}] = \text{K}_w[/tex], where [tex][\text{OH}^{-}][/tex] the concentration of hydroxide ions and [tex]\text{K}_w[/tex] is the dissociation constant of water.
[tex]\text{K}_w = 10^{-14} \; \text{mol}\cdot \text{dm}^{-3}[/tex] at 0.10 MPa and 25 °C. As a result, [tex][\text{OH}^{-}] = \text{K}_w / [\text{H}^{+}] = 10^{14} / (6.31 \times 10^{-7}) = 1.6 \times 10^{-8} \; \text{mol}\cdot \text{dm}^{-3}[/tex].
