Volume of Solution

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Duby3L
Posts: 72
Joined: Fri Sep 28, 2018 12:26 am

Volume of Solution

Postby Duby3L » Thu Oct 04, 2018 11:18 pm

A student prepared a solution of sodium carbonate by adding 2.111g of the solid to a 250.0mL 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+
So I converted to units of moles but how would I calculate the volume for Na+?

Henri_de_Guzman_3L
Posts: 88
Joined: Fri Sep 28, 2018 12:25 am

Re: Volume of Solution

Postby Henri_de_Guzman_3L » Thu Oct 04, 2018 11:29 pm

So I'm assuming you converted grams of Na2CO3 to 39.84 mmoles of Na+. So now you have mmoles of Na+ in a volume of 250.0 ml and you want 2.15 mmol Na+. The mmoles of Na+ that you converted from the grams of Na2CO3 and the 250.0 ml are your "initial" mols and volume, respectively.

Think of it as a ratio:
39.84 mmoles Na+ / 250.0 ml = 2.15 mmoles Na+ / x ml

and find x

Nicholas Carpo 1L
Posts: 36
Joined: Fri Sep 28, 2018 12:19 am

Re: Volume of Solution

Postby Nicholas Carpo 1L » Thu Oct 04, 2018 11:30 pm

If you are talking about question G5 in the textbook, these are the steps you would need to take.

First solve for the molar mass of Na2CO3.
Na2CO3= (22.99x2)+12.01+(16.00x3) = 105.99g/mol

Solve for the molarity of Na2CO3.
M = 2.111g / [(105.99g/mol)(0.2500 L) = 0.07967 M

You would then need to incorporate the molar ratios between Na and Na2CO3 along with the given to solve for the needed volume of Na. Since there are two moles of Na in Na2CO3, the ratio is 2:1. So the solution for volume would be as follows:

V = ( 0.00215mol Na) / 0.07967 M Na2CO3) x (1mol Na2CO3 / 2mol Na) = 1.35 x 10^-2 L


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