chem class

What is the pH of an aqueous solution at 25.0 °C that contains 3.98 × 10-9 M hydroxide ion?

i posted this before and i tried solving it but i get it wrong

14 = ph + - log (3.98 × 10-9)
14- 8.4 = ph
5.6 = ph
but the correct answer is 8.400
what did i do wrong?

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  1. You did an extra step of 14-8.4.

    pH= -log[H+]
    -log[3.98x10^-9] = 8.400

    pH=8.400

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