So I've gotten close to the correct answer for this problem, but I think I'm accounting for the DI water incorrectly: In titrating a blank, 0.11 mL of sodium hydroxide solution was required to titrate 50.00 mL of DI water. It required 17.56 mL of the same sodium hydroxide solution to titrate 0.7711 g of potassium hydrogen phtalate (KHC8H4O4) dissolved in 20.00 mL of D water. Calculate the molarity of the NaOH solution. I accounted for the DI water by simply subtracting the NaOH required for the DI titration from the NaOH required for the KHP titration. I think I need to account for the change in volume? Not sure how to do that... My attempt: 17.56mL - .11 mL = .1745 mL .7711 g / 204.2234 g/mol = .003776 mol .003776 mol/ .01745 L = .2164 M Thanks!