Textbook 6B.11

[Elsie_Lin_2K]

I was wondering how to calculate the mass of Na2O when it was first added to the flask? The question is: A student added solid
to a volumetric flask of volume 200.0 mL, which was then filled with water, resulting in 200.0 mL of NaOH solution. Then 5.00 mL of the solution was transferred to another volumetric flask and diluted to 500.0 mL. The pH of the diluted solution is 13.25. What mass of Na2O was added to the first flask?

[Selene_Lam_1G]

hi! you can refer to the answer key in the macmillian textbook under the resources page. It should be titled "atkins....", and they have a step by step solution!

[Tony Chen 1F]

I got that the molarity of the initial solution = 17.78 mol/L. Since the volume of initial solution is 200 mL = .2 L, (17.78 mol/L)(.2L) = 3.556 mol Na2O. The molar mass of Na2O = 61.9789 g/mol, and since there are 3.556 mol you get (3.556 mol)(61.9789 g/mol) = 220.40 g Na2O, but the textbook answer says it's 110 g. Did I do something wrong?

[Tony Chen 1F]

Oops I think I calculated the mol of hydroxide ion instead in the original solution

[Tony Chen 1F]

Ok I got it now.

From 3.556 mol OH-

Na2O + H2O = 2Na + 2OH-

From stoichiometric coefficient, 1 mole Na2O : 2 mole OH-, so you divide 3.556 by 2 to find the moles of Na2O.
You get 3.556/2 = 1.778 moles. The molar mass of of Na2O is 61.9789 g/mol, so then (1.778 mol)(61.9789 g/mol) = 110 g Na2O