Dilutions

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SMIYAZAKI_1B
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Joined: Sat Aug 24, 2019 12:15 am

Dilutions

Postby SMIYAZAKI_1B » Fri Oct 11, 2019 11:19 pm

So when it comes to dilutions, I understand how to calculate the molarity and concentration of the molecule or element; however, how can you solve for a volume of the water necessarily to dilute the solution to a given molarity with given volume? I feel like I have explained myself really badly so....
EX from the book: A student investigating the properties of solutions containing carbonate ions prepared a solution containing 8.124 g of Na2CO3 in a flask of volume 250.0 mL. Some of the solution was transferred to a buret. What volume of solution should be dispensed from the buret to provide 5.124 mmol Na2CO3?

Jasmine Vallarta 2L
Posts: 89
Joined: Sat Aug 17, 2019 12:18 am

Re: Dilutions

Postby Jasmine Vallarta 2L » Fri Oct 11, 2019 11:48 pm

I would use M(initial) x V(initial) = M(final) x V(final) and just manipulate this formula to get what you need

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

Re: Dilutions

Postby TYun_1C » Fri Oct 11, 2019 11:55 pm

You should use the M initial * v initial = M final * v final. However, that question seems kinda weird because we are solving for initial volume and have the initial molarity and final molarity but no final volume. They need to specify what the final volume of the solution is, so we can properly solve for the missing variable.

Tyler Angtuaco 1G
Posts: 111
Joined: Wed Sep 11, 2019 12:16 am

Re: Dilutions

Postby Tyler Angtuaco 1G » Sat Oct 12, 2019 12:14 am

This question should not be solved using the formula for dilutions. It states that part of the solution has been transferred, but this does not mean that more water has been added to the original solution to change the concentration. Assuming that the solution is homogeneous, the concentration should remain the same even after you transfer a portion of the original solution into another container. Thus, with moles and concentration given, use the formula V=mol/concentration to find the answer.

saigorijavolu2k
Posts: 91
Joined: Sat Sep 28, 2019 12:15 am

Re: Dilutions

Postby saigorijavolu2k » Sat Oct 12, 2019 11:09 am

I would use M1V1=M2V2.

I would start by making sure that all my numbers are in the correct units. Then I would plug in all the values that you know. From there I would solve for the unknown. Often times with dilutions you have to read the question carefully. In this case, just make sure that units are good

Siddiq 1E
Posts: 68
Joined: Fri Aug 09, 2019 12:15 am

Re: Dilutions

Postby Siddiq 1E » Sat Oct 12, 2019 11:28 am

For this problem, the equation M1V1=M2V2 is not appropriate. Instead use Molarity = Moles/Volume, manipulate so you get Volume = Moles/Molarity. Then convert grams to moles, use the molarity given in the equation, and solve for volume.

Rebekah Alfred 1J
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Joined: Thu Jul 11, 2019 12:15 am
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Re: Dilutions

Postby Rebekah Alfred 1J » Sat Oct 12, 2019 12:44 pm

When you are solving these type of problems, make sure you pay attention to the units! For this problem, some of the unit conversions you would need to do are converting the grams into moles and milliliters into liters.

emma brinton_3B
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Joined: Fri Aug 09, 2019 12:17 am

Re: Dilutions

Postby emma brinton_3B » Sat Oct 12, 2019 1:05 pm

use the equation M1V1=M2V2 in order to solve the amount of volume needed to get the same molarity when diluting.

AChoudhry_1L
Posts: 55
Joined: Sat Sep 07, 2019 12:17 am

Re: Dilutions

Postby AChoudhry_1L » Sat Oct 12, 2019 6:35 pm

I would also use the M1*V1=M2*V2 formula in this situation. Plug in the known values and solve for the unknown. The only thing I would say is to be careful of units. Sometimes the units provided and the units asked for are different and you must convert them.

Harry Zhang 1B
Posts: 92
Joined: Sat Sep 14, 2019 12:16 am

Re: Dilutions

Postby Harry Zhang 1B » Sun Oct 13, 2019 2:22 pm

When you see this kind of problem, the first thing to do is to just calculate the molarity of the original solution. Since the question is asking how much volume is needed to obtain a certain amount of moles of sodium carbonate, the logic here would be what volume of the original solution would contain the amount of moles of sodium carbonate that the question asks for. Therefore we should divide 5.124mmol by the original solution's molarity and you will have the volume needed.(millimole should be converted to moles first to cancel out the units and the unit left in the equation would be the volume unit for your answer: 5.124*10^-3 mol * L/mols of sodium carbonate in the original solution).

SimranSangha4I
Posts: 85
Joined: Sat Sep 14, 2019 12:17 am

Re: Dilutions

Postby SimranSangha4I » Sun Oct 13, 2019 2:28 pm

I would use M (initial) X Volume (initial) = M (final) X Volume (final)


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