1. The problem statement, all variables and given/known data How would you prepare 5.00 L if an aqueous solution having a pH = 10.70, if you had a supply of 0.250 M NaOH? 2. Relevant equations I don't even know anymore.... 3. The attempt at a solution I've tried a million different ways but I never come up with the answer provided in the book, which is 10 mL. Here are two ways I've tried: pH = 10.7; so pOH = 14.0-10.7 = 3.3 antilog (-3.3) = 5.0x10^-4 M m1v1 = m2v2 v2 = m1v1 / m2 = (0.250 M x 5000 mL) / (5.0x10^-4) = 2500000 mL That's DEFINITELY not right.... Second attempt: 5000 mL has x moles NaOH 5.0x10^-4 M / 5 L = 1x10^-4 moles NaOH 1x10^-4 mol NaOH x (40.0 g/mol) = 0.004 g NaOH density of NaOH = 2.13 g/mL; so 0.004 g / 2.13 g/mL = 0.0018 mL And that's not right either... I'm missing SOMETHING and I've been working on this problem forever. Please help?