Homework problem G5

[Hannah Padilla 1H]

In the homework problem G5 says “a student prepared a solution on sodium carbonate by adding 2.11 g of a solid to a 250 mL volumetric flask and adding water to the mark. Some of the solution was transferred to a buret. What Volume of the solution should the student transfer into a flask to obtain (a) 2.15 mmol Na+ ; (b) 4.98 CO3 2- ; (c) 50.0 mg Na2CO3?”
I understand how to find the molaritu of Na2CO3, but I am getting confused on how the find the volume. Also, in part b, I am not sure what effect the 2- has on the formula.
Thank you!

[Ania Chavez Dis 4D]

To find the volume, you have to use the n/V= c equation, or in other words, the # moles divided by the volume gives you the molarity.

For Part a) you find the molarity of Na2CO3 and you are given the moles of Na+ which is 0.00215 mol Na.
To find the volume, you have to divide the number of moles (0.00215) by the molarity of Na2CO3 (0.07967), and this will give you a volume but you have to divide the whole thing by 2 because there are 2 moles of Na.

For part b) the 2- is the same as the + on the Na+ from part a. They are just ions but it shouldn't affect the problem.

[Hai-Lin Yeh 1J]

Ok, so the molarity of the sodium carbonate is 0.0797 M.

For part a, they have given you 2.15 mmol of Na+. That means, you have the number of moles and you have the molarity of the compound. So, you would just use the molarity formula: M (molarity) = n(number of moles) / V. You can plug in the numbers that you already know, so:

0.0797*2 (because there are 2 moles of Na in Na2CO3) = (0.00215 mol Na2CO3)/ V. Then, solve for V.

Part b, you do the same exact thing, except the 2 is the charge of the ion so you don't really have to worry about it. In essence, there is only one mole of CO3, not 2.