Questions about Homework Problem G5

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Junxi Feng 3B
Posts: 52
Joined: Sat Sep 14, 2019 12:17 am

Questions about Homework Problem G5

Postby Junxi Feng 3B » Fri Sep 27, 2019 6:20 pm

A student prepared a solution of sodium carbonate by
adding 2.111 g of the solid to a 250.0-mL volumetric fl ask 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 fl ask to obtain (a) 2.15 mmol Na+; (b) 4.98 mmol
CO3 2-; (c) 50.0 mg Na2CO3?

Can somebody explain me how to solve question (a)?
I remember something about two sodium ions, and that I need to multiply by two for its molarity, but I have not idea what's the reason.

Thanks!

Harry Zhang 1B
Posts: 101
Joined: Sat Sep 14, 2019 12:16 am

Re: Questions about Homework Problem G5

Postby Harry Zhang 1B » Fri Sep 27, 2019 6:38 pm

The reason that you need to multiply Na^+ by 2 is because in one molecule of Na2CO3, there are two Na^+ ions(which balances the negative 2 charge that CO3^2- ions carry), and so the number of sodium ions will always be 2X more than the number of Na2CO3, which means its molarity will also be 2X more than that of Na2CO3.

GFolk_1D
Posts: 101
Joined: Fri Aug 09, 2019 12:15 am

Re: Questions about Homework Problem G5

Postby GFolk_1D » Sat Sep 28, 2019 10:44 am

I had a question about this problem as well. I started out by converting all of the units to SI units and then dividing the amounts in grams by the solid's molar mass. However, I am confused as to where to go on from there. Thanks!

Rory Simpson 2F
Posts: 106
Joined: Fri Aug 09, 2019 12:17 am

Re: Questions about Homework Problem G5

Postby Rory Simpson 2F » Sat Sep 28, 2019 6:34 pm

For part A, first you'd want to find the molarity of the Na2CO3. The molar mass of Na2CO3 would be (22.99 g/mol * 2) + (12.01 g/mol ) + (16.00 g/mol * 3 ) = 105.99 g/mol. Since the amount of moles is equal to mass of solute divided by molar mass, the molarity would then be ( m / M ) / V, where m is mass of the solute, M is molar mass, and V is volume. Solving that would be (2.111g / 105.99 g/mol) / (0.250 L) = 0.07967 mol/L for the Na2CO3 solution.

Now, to get the volume needed 2.15 mmol of Na+ (we'll convert that to 2.15*10^-3 mol Na+ for SI units), you would rearrange the equation for molarity to find the volume.

So, we would use V = n/c, where V is volume, n is moles, and c is molarity. Since each molecule of Na2CO3 has 2 Na+ atoms, we would multiply the molarity of Na2CO3 by 2. Just make sure all of the units cancel out so that the only thing left is L!

0.07967 mol/L Na2CO3 * (2 mol Na+ / 1 mol Na2CO3) = 0.1593 mol/L Na+ (the mol Na2CO3 cancels)

V = n/c = (2.15*10^-3 mol Na+) / (0.1593 mol/L Na+) = 1.35 * 10^-2 L (the mol Na+ cancels, leaving only L)

Hope this helped a bit :)


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