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?
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^+)
To find the pH of an aqueous solution, you need to use the equation:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions in the solution. However, in the question, you are given the concentration of hydroxide ions ([OH-]).
To find the concentration of hydrogen ions ([H+]), you can use the equation:
Kw = [H+][OH-]
where Kw is the ion product of water which is equal to 1.0 x 10^-14 at 25.0 °C.
Rearranging the equation, you get:
[H+] = Kw / [OH-]
[H+] = (1.0 x 10^-14) / (3.98 x 10^-9)
[H+] ≈ 2.51 x 10^-6 M
Now, you can use the pH formula to find the pH of the solution:
pH = -log[H+]
pH = -log(2.51 x 10^-6)
pH ≈ 5.60
Therefore, your initial calculation was correct with regards to finding the pH. The pH of the solution is approximately 5.60, not 8.400.
If the correct answer is given as 8.400, it may be an error, as the calculated value does not match that result. Double-check with your reference sources or consult your teacher or professor for clarification.