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?

You did an extra step of 14-8.4.

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

pH=8.400

To find the pH of an aqueous solution, you need to use the equation:

pOH = -log[OH-]

Then use the relation between pH and pOH:

pH + pOH = 14

In your calculation, you correctly calculated the pOH value:

pOH = -log(3.98 × 10^-9) = 8.4

However, you mistakenly subtracted the pOH from 14 instead of adding it to the pH. Let's correctly calculate the pH now:

pH + pOH = 14
pH = 14 - 8.4
pH = 5.6

So the pH of the aqueous solution at 25.0 °C containing 3.98 × 10^-9 M hydroxide ion is 5.6, not 8.4 as you mentioned. Please note that the correct answer is indeed 5.6.

To calculate the pH of the solution, you need to use the formula:

pOH = -log(OH- concentration)

Since pH + pOH = 14, you can rearrange to solve for pH:

pH = 14 - pOH

In your calculation, you correctly took the negative logarithm of the hydroxide ion concentration, but you made a mistake when subtracting it from 14.

Here's the correct calculation:

pOH = -log(3.98 × 10^-9) = 8.4

pH = 14 - 8.4 = 5.6

Therefore, the correct pH of the solution is 5.6.

If the correct answer is stated to be 8.400, it seems there might be some additional information or correction involved that may not be mentioned in the question. It's recommended to double-check your calculations and any additional instructions provided.