Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Kimber on October 25, 2012, 10:58:55 PM
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Im really getting stuck on the math equations! I can figure alot of the questions out but sometimes I just get stuck on how to set up the equation from the word problem even if someone can just show me how to set up the problem and explain how/why it is set up that way it would help tons!
The question is
1. How many milliliters of 0.225 M NH4C2H3O2 are needed to make 750.0 mL or 0.1667 M NH4C2H3O2?
2. Describe how to make 750.0 mL of 0.1667 M NH4C2H3O2 from 0.225 M NH4C2H3O2?
3. If 25.9 mL of 3.00 M NaCl is diluted to 136 mL, what is the molarity of the dilute solution?
4. how many milliliters of 2.50 M KI must be diluted to make 365 mL of 0.432 M KI?
5. To how many mmilliliters must 29.6 mL of 4.25 M HCL be diluted to make 0.654 M HCl
6. What molarity of KNO3 should be used in order for dilution of 63.2 mL to yield 253 mL of 0.319 M KNO3
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Hello Kimber, there is a simple, yet very effective formula that you can apply to your stated questions,
M1V1 = M2V2, where M1 and V1 are the concentration of the original, more concentrated solution, and M2 and V2 are the concentration and volume of the diluted solution. Using this formula, you plug in the values and adjust accordingly. I'll use your first one, as an example, and afterwards, you give it a try, and post your answers.
1. How many milliliters of 0.225 M NH4C2H3O2 are needed to make 750.0 mL or 0.1667 M NH4C2H3O2?
In this problem, M1 will be 0.225 M, while M2 will be 0.1667 M, since it's more diluted. 750.0 mL will be V2. We are looking for the amount of volume that can be used to contain 0.225 M of NH4C2H3O2, V1.
Setup:
M1V1 = M2V2
Adjust according to the question asked:
V1 = (M2V2)/M1
Input the numerical values:
V1 = (0.1667M * 750.0 mL)/0.225 M
V1 = 556 mL is needed to produce 0.225 M of NH4C2H3O2.
Note: I applied the rules of significant figures in this problem. I don't know what methodology you use, but adjust accordingly, if my methods do not match with your approach.
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To add to MathisFun explanation - M1V1=M2V2 is just a way of saying "amount of substance before dilution and after dilution doesn't change".
This is another application of the mass conservation principle - matter is not produced nor destroyed during chemical processes. You use exactly the same mass conservation principle when balancing chemical reaction equations - number of atoms of each element must be identical before and after the reaction.
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To add to MathisFun explanation - M1V1=M2V2 is just a way of saying "amount of substance before dilution and after dilution doesn't change".
This is another application of the mass conservation principle - matter is not produced nor destroyed during chemical processes. You use exactly the same mass conservation principle when balancing chemical reaction equations - number of atoms of each element must be identical before and after the reaction.
Note though, that typically M here is a molar conc. of a species (say, 0.5 mol / L HCl) or mass conc. (gm / L ).
Neither of these quantities are conserved during chemical processes. So, you can strictly only use species conservation for dilutions etc.
Atom conservation is a stronger constraint; always valid. Unless you go into nuclear / radiochemistry that is. But then you can fall back on mass conservation which is an even stronger conservation law.
Scratched restating the obvious.
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To add to MathisFun explanation - M1V1=M2V2 is just a way of saying "amount of substance before dilution and after dilution doesn't change".
(...) you can strictly only use species conservation for dilutions
It was already stated, wasn't it?