A solution of NaOH(aq) contains 7.7 grams of NaOH in 93 mL of solution. Calculate the pH at 25◦C.

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7.7 grams x (1 mol/39.997 grams) = 0.19 mol
0.19 mol/0.093 L = 2.07 M
pOH = -log(2.07) = -0.316
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Am I doing anything wrong? Why am I getting a negative pOH/pH greater than 14?

MOST of the time the H^+ or OH^- is less than 1 so the log is negative but the - sign in the definition changes that to a + number. In fact, that's why the - sign was placed there in the definition so that the usually negative numbers would become positive. Here, however, the OH is >1 so the - sign changes that + log number to a negative number. Nothing wrong so far.

Some examples.
OH = 1; pOH = 0
OH = 0.1; pOH = -*-1 = +1
OH = 0.01; pOH = -*-2 = +2
BUT if OH = 10, pOH = -*1 = -1
etc.

It seems like you made a calculation error while converting the concentration of NaOH from molarity to pOH. Let's go through the correct steps together:

First, we need to calculate the molarity (M) of the NaOH solution:

Molarity (M) = moles of NaOH / volume of solution in liters

moles of NaOH = mass of NaOH / molar mass of NaOH
moles of NaOH = 7.7 grams / 39.997 g/mol = 0.19 mol

volume of solution in liters = 93 mL / 1000 mL/L = 0.093 L

Molarity (M) = 0.19 mol / 0.093 L = 2.04 M

Now, to find the pOH, we can use the formula:

pOH = -log10(Molarity)

pOH = -log10(2.04) = 0.69

Since pOH + pH = 14, we can find the pH:

pH = 14 - pOH
pH = 14 - 0.69 = 13.31

So, the pH of the NaOH solution at 25°C is approximately 13.31.

By following these steps correctly, you should obtain a pH value between 0 and 14, as expected for a basic (alkaline) solution.