G5

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Malia Shitabata 1F
Posts: 127
Joined: Sat Aug 17, 2019 12:17 am

G5

Postby Malia Shitabata 1F » Sun Oct 06, 2019 2:56 pm

"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 the solution was transferred to a buret. What volume should the student transfer into a flask to obtain 2.15 mmol Na+?"

I understand that you have to convert the grams of Na2Co3 to moles and then divide by the given volume to obtain the molarity, and I got the right answer. But since you have to multiply by the molar ratio to get from moles of Na2Co3 to moles of Na+ how come you can only multiply in the last step when you're already finding the volume as opposed to when you're first converting the grams of Na2Co3? Please let me know if that needed clarification because it sounds kind of wordy.

Miriam Villarreal 1J
Posts: 105
Joined: Sat Aug 17, 2019 12:16 am

Re: G5

Postby Miriam Villarreal 1J » Sun Oct 06, 2019 5:59 pm

-1st step involves calculating the moles of Na2CO3

2.111gNa2CO3
_____________________ = .01992 mol NaCo3
105.991 g/mol Na2CO3

-2nd step
.01992 mol/.250L=.0769 mol/L or about .080mol/L

(a) Convert mmol to mol .002015 mol Na/0.16 mol/L = 0.013 L Na x 1000ml = 13 mL Na is needed to transfer

Hussain Chharawalla 1G
Posts: 100
Joined: Sat Jul 20, 2019 12:15 am

Re: G5

Postby Hussain Chharawalla 1G » Sun Oct 06, 2019 8:38 pm

I also don't understand this. Intuitively, for every one mole of Na2CO3, shouldn't there be 2 moles of Na? And why is the multiplication done in the last step?

2.11g Na2Co3= 0.0199 moles Na2Co3
(1 mole Na2Co3)*(2 mol Na/1 mole Na2Co3)= 0.0398M Na Solution
(0.0398)(x)=(2.15*10^3)(0.25)

x= 3.4 mL

Can someone explain please?

Ariel Fern 2B
Posts: 105
Joined: Fri Aug 30, 2019 12:17 am

Re: G5

Postby Ariel Fern 2B » Sun Oct 06, 2019 9:42 pm

Hi Hussain!
I was initially confused at first as well; it does make sense to have 2 moles of Na for every mole of Na2CO3.

I think when you multiplied this, it should be 0.0398 moles of Na rather than Molarity of the solution:
(1 mole Na2Co3)*(2 mol Na/1 mole Na2Co3) = 0.0398 moles of Na

Then, you would divide that by 0.250L to get a 0.1592M Na Solution

Instead of multiplying, I used the equation: Volume = Moles/Molarity:
Volume = (2.15*10^-3 moles) / 0.1592M = 13.5 mL

The Equation M1V1 = M2V2 is used for dilutions, so I don't think it would be the most convenient to use in this problem.

Hope this helps you!!

Jordan Ziegler 2J
Posts: 59
Joined: Sat Aug 17, 2019 12:15 am

Re: G5

Postby Jordan Ziegler 2J » Sun Oct 06, 2019 10:47 pm

To solve this, because the solutions are equimolar, I set up the proportion:



If you cross multiply to solve for volume, you'll get V=13.5ml

Naomi 3G
Posts: 9
Joined: Wed Sep 18, 2019 12:20 am

Re: G5

Postby Naomi 3G » Sun Oct 06, 2019 11:31 pm

Hi,

I am also confused why in the solutions book for a they multiply .07977 by 2 in the denominator of the equation

Matthew Chan 1B
Posts: 111
Joined: Sat Sep 07, 2019 12:16 am

Re: G5

Postby Matthew Chan 1B » Mon Oct 07, 2019 11:58 pm

Naomi 3G wrote:Hi,

I am also confused why in the solutions book for a they multiply .07977 by 2 in the denominator of the equation

They multiplied 0.07977 by 2 in the denominator because the 2 is from the 2 moles of Na+ that make up the 1 mole of Na2CO3. Hope this clarifies things!


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