chemistry class

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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? or is the answer given wrong?

  • chemistry class -

    You didn't do anything wrong. The answer is wrong.
    For a pH of 8.4, the (H^+) [not OH^-)] must be 3.98E-9.
    You can prove that by
    pH = -log(H^+)
    -8.4 = log(H^+)
    Take antilog of both sides.
    3.98E-9 = (H^+)

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