If $[H_3O^{+}] = 3.5 \times 10^{-8} \ M$,calculate $[OH^{-}]$ and $pH$ of the solution.
(N/A) Given: $[H_3O^{+}] = 3.5 \times 10^{-8} \ M$.
Using the ion product of water at $298 \ K$,$K_w = [H_3O^{+}][OH^{-}] = 1.0 \times 10^{-14}$.
$[OH^{-}] = \frac{K_w}{[H_3O^{+}]} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-8}} \approx 2.86 \times 10^{-7} \ M$.
$pH = -\log[H_3O^{+}] = -\log(3.5 \times 10^{-8}) = 8 - \log(3.5) \approx 8 - 0.544 = 7.456$.
Using the ion product of water at $298 \ K$,$K_w = [H_3O^{+}][OH^{-}] = 1.0 \times 10^{-14}$.
$[OH^{-}] = \frac{K_w}{[H_3O^{+}]} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-8}} \approx 2.86 \times 10^{-7} \ M$.
$pH = -\log[H_3O^{+}] = -\log(3.5 \times 10^{-8}) = 8 - \log(3.5) \approx 8 - 0.544 = 7.456$.
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