G.5 Concentration Calculation

Moderators: Chem_Mod, Chem_Admin

ayushibanerjee06
Posts: 177
Joined: Thu Jul 11, 2019 12:16 am
Been upvoted: 1 time

G.5 Concentration Calculation

Postby ayushibanerjee06 » Wed Oct 02, 2019 10:51 am

Hi! I am having a problem with this question: 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? After calculating the molarity of Na+ I don't know what to do, and the solutions manual did not make much sense to me. Can someone please explain how to solve the problem after finding the molarity? Thanks!

Chem_Mod
Posts: 18879
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 714 times

Re: G.5 Concentration Calculation

Postby Chem_Mod » Wed Oct 02, 2019 11:23 am

Once you find the molarity, you have a way to convert between moles and liters. So for a), convert mmol of Na+ to mol, then convert from moles of Na+ to moles of Na2CO3, then divide by the molarity to get the volume needed. The same procedure applies for b) and c).

TYun_1C
Posts: 52
Joined: Wed Sep 11, 2019 12:16 am

Re: G.5 Concentration Calculation

Postby TYun_1C » Wed Oct 02, 2019 11:43 am

Remember that Na2Co3 will dissociate in a ratio of two Na+ ions to one CO3 polyatomic ion. That way, you know that in a certain molarity sodium carbonate solution, the concentration of pure Na+ ions is doubled. Use the M initial x V initial = M final x V final equation of find the volume needed of obtain the certain molarities. Also remember that mmol= millimole, so do the appropriate conversions.

Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

Re: G.5 Concentration Calculation

Postby Chris Tai 1B » Wed Oct 02, 2019 11:57 am

When you're trying to find the volume needed to get moles of Na+, it's best to use the equation Molarity = moles/Volume.
You know that the number of moles of Na2CO3 in the given amount of grams is 2.111g/106gmol^1 = 0.01992. You can then calculate the molarity of Na2CO3 within the given 250 mL solution by dividing here: 0.01992mol/0.25L = 0.07968 M or mol/L.
Since the question is asking about the necessary volume for 2.15 mmol of Na+, I find it easiest to first convert 2.15 mmol to 0.00215 mol of Na+. Because there are two of these Na+ ions within each Na2CO3 molecule, you have to divide 0.00215/2 to get the necessary around of Na2CO3 molecules that you need to GET 2.15 mmol of Na+. Hence, you need 0.00215/2 = 0.001075 mols of Na2CO3 to obtain 0.00215 mols of Na+.
Once you have all this, you can plug values into the equation like this -
M = m/V
0.07968 mol/L = 0.001075mol/V
Solving for V: V = 0.01349 mL or 13.49 mL -> sig figs -> 13.5 mL

Brandon Valafar
Posts: 112
Joined: Sat Aug 17, 2019 12:16 am
Been upvoted: 1 time

Re: G.5 Concentration Calculation

Postby Brandon Valafar » Wed Oct 02, 2019 12:12 pm

Once you find the molarity, I would first go about changing each of the units from milli-moles to moles. For example, this would make 2.15 mmol Na+ into 0.00215 mol Na+. Once you do that, for a, you have to find the volume needed for Na+. Since there are two Na+ atoms, you would have to divide the amount of moles given by two, to get the value for one mole of Na+. You would then fill in the formula of M=n/v and solve for v. You do this same process for part b, except the part of dividing the given moles by two since there is only one CO3. For part c there is an extra step in which you have to convert the grams to moles first. After you do that then you just solve for the volume again.


Return to “Molarity, Solutions, Dilutions”

Who is online

Users browsing this forum: No registered users and 1 guest