## 6B.11B

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### 6B.11B

A student added solid Na2O to a volumetric flask of volume 200.0 mL, which was then filled with water, resulting in 200.0mL 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.

So for part a I have that [OH-] = .18M for the diluted and original solution. For part b, the equation would be Na2O(s) + H2O(l) --> 2NaOH(aq). How would I know how to calculate the mass of Na2O from the concentration of OH-? Does it matter if I use the diluted or original concentration of OH-?

Jacob Puchalski 1G
Posts: 97
Joined: Fri Aug 09, 2019 12:16 am

### Re: 6B.11B

For part a, since M is moles per liter, the [OH-] of the original solution would be 100 times stronger, since the 5.0 mL solution was diluted to 500 mL. So the original [OH-] would be 18M.

For part b, you use the equation you have to convert the 18M OH- to grams of Na2O.
18 mol/L x .2 L x 1 mol Na2O/2 mol NaOH x 61.98 g Na2O/mol = approximately 110 g Na2O.

ASetlur_1G
Posts: 101
Joined: Fri Aug 09, 2019 12:17 am

### Re: 6B.11B

You have to use the original concentration of OH- because they're asking for the initial amount of Na2O. Thus, you want the concentration of OH- before the dilution, which would be 18 M.