Whats the pH of a solution created combining 11.10 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HC2H3O2(aq)?

Answer 1

You have 1.11 mmol of a base and 0.8 mmol of an acid, so once the two react, you'll end up with 0.31 mmol of unreacted base. Your total volume is now 19.10 mL, so the [OH-] = 0.31 mmol / 19.10 mL = 0.0162 M. pOH = 1.79, pH = 12.21