G.5 Concentration Calculation

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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!

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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).

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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
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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
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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.

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