Fundamental G Exercise 5

[Barbara Soliman 1G]

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) 2.15mmol Na⁺
(b) 4.98mmol CO₃^2-
(c) 50mg Na₂CO₃?

Could someone explain this to me? I found the initial molarity, but I'm a bit stuck after that. Could you use M_initalV_initial=M_finalV_final for this?

Thanks!

[amara ajon 1d]

Hi,
I won't do the entire problem, only part a, but yes it is possible to use MiVi=MfVf to solve all of these - you just need to take into account the appropriate ratios. For example, there are 2 moles of Na per Na2CO3 molecule. After finding the molarity (Mi) of the initial Na2CO3 solution to be 0.0796M, we can solve:
MiVi=MfVf
(0.0796M)(Vi)=(0.00215M Na)(2)(0.250L) - in this step, we multiply the side with Na by 2 since there are 2 Na in a Na2CO3 molecule.
Vi=(0.00215)(2)(0.250)/(0.0796)
Vi=0.0135L or 13.5mL of the initial solution.
Something similar can then be done for parts b and c.

Full Disclosure: I did not do it this way personally, but I think this might still work..?

[Shivani Sakthi 1l]

Hi! This response is going to focalize on part B. To answer your question, MinitialVinitial= MfinalVfinal can be used to find the volume of solution needed to to transfer into a flask to obtain 4.98 mol of Co3^2-. One must pay attention to the molar ratios in order to complete this problem successfully, however. Since only 1 mole of CO3 is required for every 1 mole Na2CO3, the ratio does not greatly affect the solution.

Step 1: Convert the given grams of NaCO3 into moles. 2.111g x (1 mol Na2Co3/ 105.99 g Na2Co3) = .01992 mol Na2CO3.

Step 2: Since the MinitialVinitial= MfinalVfinal requires molarity, we will use the calculated mols of Na2CO3 and the given liters of solution to obtain molarity. So, .01992 moles/ .250 L = .0797 M Na2Co3.

Step 3: Since there is 1 mole of CO3 is required for every 1 mole Na2CO3, the molarity of Co3 would be .0797 M. Now we have M initial.

Step 4: Now lets analyze our knowns. We know M initial. We know M final. M final is 4.98 mmol Co3. To get this to moles/L, we divide by 1000, giving us .00498 M Co3. We are also given our final volume of .250 L within the problem. Since the problem is asking the volume of solution that should be transferred into the flask, we can delineate that our unknown is V initial.

Step 5: Plug in values.

MinitialVinitial= MfinalVfinal
.07968(V initial)= .0498 (.250 L)
V inital= .156 L

[Vy Le 1G]

Hi,

This post will explain Part C of the problem. While you can use M1V1=M2V2, I found it easier to work with the molarity formula.

Step 1: Convert Na2CO3 to moles by taking the 2.111g and dividing by the molar mass.

Step 2: Find the molarity of Na2CO3 with your newly found moles of Na2CO3 and the given, 250 mL (convert this to liters first). You should get around 0.08M.

Step 3: Take the given 50mg (from Part C) and convert that to moles. You would do it by first converting it to grams and then dividing that by the molar mass. You should get an answer around 0.000472 mol.

Step 4: Plug the known values into the Molarity formula. 0.08M = (0.000472 mol)/x. With variable x being the value you are looking for, once you solve the simple algebraic equation, that is your answer!

Step 5 (optional): The answer is a decimal in liters and changing it to mL might be a better choice!

Note: This is just how I did the question without using M1V1 = M2V2. With estimating I was about 0.03 off from the textbook answer but it shouldn't be too much of a difference.

Hope it helped!

[SarahOMalley1D]

Hi! My response will focus on part c. I initially thought of using M1V1=M2V2. However, I realized that I needed to use M=n/V instead, where n is the moles of sodium carbonate needed, V is the volume of solution being transferred to the flask, and M is the molarity of the solution in the buret.

With this in mind, I first solved for the moles of sodium carbonate needed by converting 50.0 mg to moles.
mol=(50.0mg)*(1 g/1000mg)*(1 mol/105.99g)=4.71x10^-4 mol sodium carbonate.

Next, I manipulated M=n/V to solve for V and got V=n/M.

Lastly, I plugged in my solved value of n and the value of M I used in the previous question parts.
V=(4.71x10^-4 mol)/(0.07967 M)=0.00592 L

In mL, the answer would be 5.92 mL of solution.

Update: Sorry, I finished writing my post right when the post before was posted, so I apologize if we have any repeating remarks. It looks like we did describe our process a bit differently, though.