## G5 approach to question

Aishwarya Raich
Posts: 15
Joined: Fri Sep 25, 2015 3:00 am

### G5 approach to question

G5) A student prepared a solution of Na2CO3 by adding 2.111 g of solid to a 250 mL flask and adding water to the mark. Some was transferred to the buret. What volume of solution should the student transfer in a flask to obtain 2.15 mmol Na+.

So for this question would the approach of M1V1=M2V2 or Molarity= n/V work? and how does the factor of molar quantity in the ion work considering that there are 2 ions of Na+ in each compound. I was confused as to the set up of this problem since it shows up throughout section G. Can someone explain please? Thank you.

Chem_Mod
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### Re: G5 approach to question

The second equation you list is going to be very helpful here because we want to know the concentration of solution in the problem in order to find the answers to parts a, b, c, and d. The first equation is very applicable because we aren't diluting anything in the problem. Once you find the concentration of the starting solution (2.111 grams, converted to mols and divided by 0.250 L) you can take each started mols or grams given and convert to a volume based on the molarity.

Ritika Saranath 3I
Posts: 24
Joined: Fri Sep 25, 2015 3:00 am

### Re: G5 approach to question

Why is it that when in part a, when we are asked to obtain a 2.12 mmol Na+ solution, and in part b, when we are asked to obtain a 4.98 mmol CO3^2- solution, that the solutions manual sets up ratios between Na+ and Na2CO2 and CO3^2- and Na2CO2, rather than Na2CO3?

Hai Tran 1F
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Joined: Fri Jun 17, 2016 11:28 am

### Re: G5 approach to question

I still don't understand how the solutions manual did the problem. I understand that you have to use M1V1=M2V2 but I don't understand the steps the manual used including which variable it is solving for.

Ana Ordonez 1G
Posts: 13
Joined: Fri Jun 17, 2016 11:28 am

### Re: G5 approach to question

To the person who said they did not understand the way that the solution manual approached the question, they did it based on the formula Molarity(c)= mol(n)/ volume(V), not the other one.
I did it this way and what I did was first find the molarity of Na2CO3, which I did by converting 2.111 g to moles. Then I found the molarity of Na2CO3 with the formula c=n/v, since volume was given and n I just found all I had to do is plug them in. After getting those numbers I found part a by converting mmol to mol. Since part a is asking for the volume to obtain 2.15 mmol Na+ then I used the formula V=n/c, which n and c I have already found previously. Then I multiplied the answer that I get for V with the ratio 1 mol Na2CO3: 2 mol Na+ in order to get the appropriate volume necessary to obtain a 2.15 mmol Na+.
Hopefully this has helped you!