What is the pH of a solution prepared by dissolving 3.1g of NaOH to make 305 mL of solution?

First convert grams to moles by dividing by the molar mass.

3.1g NaOH / (22.9+16+1 g/mol) = 0.0777 mol

Then divide the moles by the volume in liters.

305ml = 0.305 L

0.0777mol /0.305L = 0.2547M

You know that: pH = -log [H3O+] = -log (1 x 10^-12) = 12 or pOH = -log [OH-]

So using the second formula: -log (0.2547M) = 0.594

pH + pOH = 14
pH + 0.594 = 14
pH = 13.4 ??


If you just heard something that is my brain exploding; it's been a LONG time since I've done any of this, so you might want to double check my work.