Calculate the volume of concentrated reagent required to prepare the diluted solution
17 M H3PO4 to prepare 939 mL of 6.7 M H3PO4
mL1 x M1 = mL2 x M2
Dr. Bob,
Can you explain this one please? Feel free to use made up numbers. I'm not looking for an answer, I'm just not understanding this question and looking for further help. Thank you
All you need to do is to substitute the numbers into the formula.
mL1 x M1 = mL2 x M2
mL x 17M = 939 x 6.7
solve for mL. And it doesn't make any difference which one you call mL1 and mL2 as long as you're consistent;i.e. you could have
939 x 6.7 = mL x 17 and solve for mL. You get the same answer.
Some profs like
c1v1 = c2v2 where c is concentration and v is volume. If you like that better it is
17 x v = 6.7 x 939 and solve for v. Same answer.
And there is another way that is good for the fundamentalist but I don't like it because it takes longer to do. Here it is. You know mols acid before dilution must equal mols acid after dilution. So how much do you want (of mols) in the diluted solution? That's mols H3PO4 in dil solution = M x L = ?
So you must have that amount (whatever that amount is) in the concentrated solution before dilution. mols = M x L. You know mols from the first part and know M from the problem (17 M in this case), calculate L and convert to mL and you will get the same answer as the two you obtained above.
To calculate the volume of concentrated reagent required to prepare the diluted solution, you can use the formula:
C1 * V1 = C2 * V2
Where:
C1 = Initial concentration of the reagent (in this case, the concentrated solution's concentration)
V1 = Volume of the initial concentrated reagent
C2 = Final concentration of the reagent (in this case, the diluted solution's concentration)
V2 = Volume of the final diluted solution
Let's plug in the values you've provided and solve for V1:
C1 = 17 M
V1 = unknown
C2 = 6.7 M
V2 = 939 mL (which is the same as 0.939 L)
17 M * V1 = 6.7 M * 0.939 L
Now we can solve for V1:
V1 = (6.7 M * 0.939 L) / 17 M
V1 ≈ 0.370 L
Therefore, approximately 0.370 liters (or 370 mL) of concentrated reagent would be required to prepare 939 mL of a 6.7 M H3PO4 solution.