Textbook Problem 6.B.11 Part b

[Jayden Arevalo 2B]

I am a bit confused on the choices they made when solving this problem. Part b of this problem asks us to find the mass of Na2O that was initially added to the original solution. The only information provided to us is this, "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." In part a, we calculate the hydroxide ion concentration within the diluted and original solution, in which they apply this in part b. But, I am unsure how the parts of the conversion equation relate to each other and why they used certain constants. I would greatly appreciate the help. :)

[Kaitlin Eblen 1I]

From the problem, we know that Na2O (s) is combined with water to form NaOH. We need to balance this reaction equation to solve part b. Using the info we found in part a (specifically that [OH-] = 17.8 M in the original solution), we can find the mass of Na2O. I attached a picture of my work but I first found the number of moles of OH- using the molarity equation (mL has to be converted to liters for this step). Then, since we know that NaOH dissociates into Na+ and OH- ions in water, we can assume that there are 3.65 moles of NaOH for every 3.56 moles of OH-. From there, we can use stoichiometry to find the mass of Na2O. Let me know if you have any other questions.