Concentration Calculations

[Tiffany Lin 2E]

Problem: A student prepared a solution of sodium carbonate by adding 2.111 g of the solid to a 250.0-mL volumetric flask and adding water to the mark. Some of this solution was transferred to a buret. What volume of solution should the student transfer into a flask to obtain (a) 2.15 mmol Na+ (b) 4.98 mmol CO3^2- (c) 50.0 mg Na2CO3?

I looked at the solution but I don't understand what it's doing. I found the molarity of sodium carbonate, but I do not know where to go from there. What's my next step?

[Eduardogonzalez1G]

Alright finding the molarity is the first step. For the Na2CO3 molarity is 0.08 M. You might want to convert the values from mmol to mols. but 1 mmol= .001 mol. To find a) you convert the mmol to mol to get 0.002015 mole Na. Since theres 2 Na's in Na2CO3 the molarity is gonna be twice so instead of 0.08, itll be 0.16mol/L. You could then divide the values after. so 0.002015/0.16 that is gonna be 0.013 L.
b) change 4.98 mmol to 0.00498 mol. Theres only one Na2CO3 molecule so no need to multiply by anything. You proceed to divide: 0.00498/0.08 = 0.062 L
c)convert mg to g and g to mol. So 50 mg= 4.717 x 10^-4 mol. You then divide: (4.717 x 10^-4)/ 0.08 = 0.0058 L

[Tinisha 1G]

How do we know when not to use the formula M(initial)V(initial) = M(final)V(Final)? That was my plan for this problem, but it does not work out.

[Adrian C 1D]

You should use M1V1=M2V2 when you have both the initial molarity and volume and you are trying to figure out the molarity or volume of the final solution, depending on whether they give you the M2 or V2. When the question gives you M2, you solve for V2 and vice versa. Although this question is asking for volume, similar to M1V1=M2V2 questions, it is not asking for the molarity of the final solution. It is asking for the volume needed from the 0.080 M Na2CO3 solution to reach a certain number of moles (part a&b) and mass (part c). Therefore, M1V1=M2V2 is not required. By dividing the given moles in parts a&b by 0.80 mol/L, the mole units cancel out and you're left with the Liter unit-giving you the volume. The same for part c, but you need to do an extra step and turn the mass into moles first.

[Madeline Louie 1I]

Does someone mind explaining why we can use the molarity of 0.08 M for part c? I don't quite understand why we can use it when the number of moles is very different from when we initially solved it for sodium carbonate using the 2.111 g.

[Kayko Lee 1C]

We use .08 moles in part c because since we are using the equation V = n/c to find the volume needed, the .08 is used as c because it represents the molarity from the original solution in the problem. Since we are solving for V, we just need to know what n is so the mg given in part c is used to find what the value of n is as the value for c is satisfied

[Alvin Lai 3J]

How did you find the molarity of Na2CO3?

[Mahnoor_Wani_1I]

To find the molarity of Na2CO3 you divide 2.111 grams of solid (the solid here is NA2CO3 since this is the solute in the aqueous solution) by its molar mass to find moles. Moles = Grams /molar mass
Then you do moles/volume to find the molarity

[haley f]

Alright finding the molarity is the first step. For the Na2CO3 molarity is 0.08 M. You might want to convert the values from mmol to mols. but 1 mmol= .001 mol. To find a) you convert the mmol to mol to get 0.002015 mole Na. Since theres 2 Na's in Na2CO3 the molarity is gonna be twice so instead of 0.08, itll be 0.16mol/L. You could then divide the values after. so 0.002015/0.16 that is gonna be 0.013 L.
b) change 4.98 mmol to 0.00498 mol. Theres only one Na2CO3 molecule so no need to multiply by anything. You proceed to divide: 0.00498/0.08 = 0.062 L
c)convert mg to g and g to mol. So 50 mg= 4.717 x 10^-4 mol. You then divide: (4.717 x 10^-4)/ 0.08 = 0.0058 L

How do they know there is 2 Na's and 1 CO3?

[Emily Ding 1J]

The subscripts of the molecular formula show you that there are 2 Na's and 1 CO3 in every 1 Na2CO3. I hope that makes sense. It's like the mole ratios of chemical equations, just inside of a molecule.

[Austin Aldujaili 2D]

We only know that the molecular formula of sodium carbonate is Na2CO3 by understanding charges in the molecules. CO3 has a -2 charge when by itself and Na has a +1 charge when isolated. Therefore, the formula would be Na2CO3 in order to result in a neutral charge.

[Giselle_zamora_1L]

for this question I am still confused on part b, are there not 3 CO molecules and therefore we should multiply 0.08 by 3 like in part a?

[Karen Harrison 1L]

for this question I am still confused on part b, are there not 3 CO molecules and therefore we should multiply 0.08 by 3 like in part a?

Hi!
For part b, we wouldn't multiple 0.08 mol/L by 3 because in the final answer there are already 3 oxygen moles and 1 carbon mole. This is compared to part a where we have 2 Na moles in our solution and the question started with only 1 Na mole (thus we have to multiply 0.08 by 2 for a).
Hope this helps future readers!

[Nathalia Garibay 1D]

Hi I have a question. Why do we multiply the molarity of sodium carbonate by two when we're finding Na+. I understand that there's two molecules of Na but if we multiplied the .08 by two, doesn't that also include the Co3 part of the sodium carbonate?

[Sevde Coban 2J]

You multiply the molarity of sodium carbonate by 2 for Na+ because in the Na2CO3 molecule, there are 2 moles of Na+ but only 1 mole of CO3, so you wouldn’t multiply the molarity of the molecule by 2 for CO3.

[Sneha Anantharaman2D]

It is based on the number of moles of each molecule or atom. Since in Na2CO3 you have 2 moles of Na+ and only 1 mole of carbonate, you would multiply by 2 moles for sodium and only by one for carbonate.