### Fundamentals G5

Posted:

**Thu Oct 03, 2019 6:02 pm**I was attempting to do problem G5 and I got a bit confused when it came to transferring volumes of solution to obtain a certain number of mmols, could someone explain?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=10&t=46161

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Posted: **Thu Oct 03, 2019 6:02 pm**

I was attempting to do problem G5 and I got a bit confused when it came to transferring volumes of solution to obtain a certain number of mmols, could someone explain?

Posted: **Thu Oct 03, 2019 6:30 pm**

First, you have to find the molarity of the sodium carbonate which is 0.08mol/L. For a), I converted mmol to moles (1 mmol = .001 mol). Since there are 2 Na’s in sodium carbonate, the molarity will be twice that of the solution which means 0.08 x 2 (the molarity will be 0.16mol/L). Then it will be (0.00215mol)/(0.16mol/L) which will give you 0.134L. For b), you would do the same thing but there is no need to multiply by 2 (there is only one carbonate ion in sodium carbonate). For c) convert mg to moles and divide it by 0.08M like before.

Posted: **Thu Oct 03, 2019 11:33 pm**

You kind of have to work backwards for this problem. You know you need 0.00215 mol Na at the end, which you need 0.001075 mol Na2CO3 since there is a ratio of 1:2. Since you are looking for volume and already have the final moles, you know you are looking for the molarity because of n=MV. The molarity is the same because nothing is being diluted, so if you convert 2.111g to moles and divide by the volume (0.25L), you get the molarity is 0.0796M. Then you plug this in to the missing piece of our original equation to find volume and get 0.001075/0.0796=0.0135L.

Posted: **Fri Oct 04, 2019 1:17 pm**

I was confused on that question too, but you have to make sure to do a unit conversion of mmol to mol and to use the M=n/v. This question also required the use of the ratio of Na:Na2CO3, which is 2:1 and CO3:Na2CO3, 1:1.