Question

D Question 1 2 pts 0.635g of NaOH was placed in a 250 mL volumetric flask and distilled water was added to the 250.00 mL mark. What volume of 0.231M sulfuric acid, in mL, is required to titrate a 25.00 mL aliquot of the NaOH solution? Round your answer to two decimal places. Molar Mass: Na 23.0 g/mol, S 32.1 g/mol, O 16.0 g/mol, H 1.0g/mol

Answer 1

moles of NaOH = 0.635 g / 40.0 g/ mol

                           = 0.0159

volume = 250 mL = 0.250 L

molarity = moles / volume

            = 0.0159 mole / 0.250 L

           = 0.0635 M

H2SO4 + 2 NaOH ---------------------------> Na2SO4 + 2H2O

C1V1 / n1 = C2 V2 / n2

0.231 x V1 / 1 = 0.0635 x 25 / 2

V1 = 3.44 mL

volume of H2SO4 needed = 3.44 mL