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 - DrBpb222, Monday, November 22, 2010 at 10:36pm
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^+)