Dilution Problems

Last updated 10-11-09

Dilution Problems

Reason: There are many reasons that we might want to go from a concentrated solution to a more diluted solution. It's usually because the concentrated solution is too strong. This image is from a government site that talks about diluting nitric acid. It says to first fill this 4 liter jug half full with "Milli-Q" water. That is water that is purified using Millipore brand water purifiers. The instructions say to add 200mL of the (concentrated) HNO3 to the jug that is half full of water. Like the sign says, "Add acid to water". That's because when acid dissolves in water, a lot of heat is generated that can change water into steam, which can propel the solution into one's face. If you are adding acid to water, the solution has much more water than acid, so that initial spray that might hit you is mostly water. If adding water to acid, that initial spray is mostly acid.
Now back to the dilution problem. The bottom image is a closeup of the concentrated nitric acid bottle. It says the concentration is 70.0-71.0%. They don't say weight to volume (w/v) or weight to weight (w/w), which is poor practice.

Their instructions say that after the 200mL of acid is added, bring the water up to the 4 liter mark. The label on the jug says the final concentration will be 5% nitric acid. They don't say weight to volume (w/v meaning grams per 100mL) but that is what they should have written. The bottom image is a closeup of the nitric acid bottle. It says the concentration is 70.0-71.0%. Again, they aren't saying w/v or w/w, but elsewhere on the bottle is 15.7M so that will help us determine which it is (w/v or w/w). The bottle also shows the specific gravity of 1.4 (That means a density of 1.4g/mL). Let's double-check their instructions. Let's assume they mean w/v which is grams per 100mL. So the question is, "To what volume in liters do you dilute 200mL of 70.0%-71.0% w/v nitric acid to get 5% w/v nitric acid?".
(See table below) Multiplying the starting concentration of 70.0g-71.0g/100mL times the 200mL gives us the total grams of the acid. In this case it is about 140 g. We want a 5% solution. So that means for every 5 grams, we need another 100mL. To find out how many 5 grams we have, we divide 5 into the 140. That goes 28 times. So that means we need 28 times 100mL. So by inverting the desired concentration, we get the division of the 5 grams and the multiplying of the 100mL. Unfortunately, the final answer shows 2.80-2.84 liters. So either their instructions are wrong or the 70% is not weight to volume but weight to weight. In other words nitric acid is 70 grams out of every 100 grams of the liquid.

	A	B	C	D	E	F	G	H	I	J
1	Concentration in g/100mL times its mL gives grams of HNO3				Ending 5% w/v inverted				Final volume
2	70.0-71.0	g HNO3	200	mL solution	100	mL	0.001	=	2.80-2.84	Liters
3	100	mL solution			5	g HNO3	milli

If nitric acid is 70 grams out of every 100 grams of the liquid, how many grams are we pouring when we add 200mL? That's where we can use the density which the bottle listed as 1.4g/mL. Density lets you convert volume to mass (or mass to volume). By column F we have calculated grams of HNO3, so the problem can continue the same as above. We get the range of 3.92 to 3.98 liters. Which is close to the 4 Liters they want. So it seems that the 70% on the bottle means 70% w/w not w/v. Let's see what using 15.7 M as the starting concentration gives us regarding how many liters we have to dilute it to.

Starting concentration

Volume x density finds grams of solution

End 5% w/v inverted

Final volume

70.0-71.0

g HNO3

200

mL solution

1.4

100

0.001

3.92-3.98

Liters

100

g Solution

g HNO3

milli

Again, on the bottle the concentration is also written in molarity (15.7 moles per liter). Starting with moles we will have to convert that using the molar mass of HNO3, which is 63.01 grams per mole. We also need to get rid of milli to cancel out the "m" in 200mL. Our answer here shows that we need to dilute it to 3.97 liters. So it seems that actual concentration of the bottle of nitric acid is 71% w/w (71 grams per 100 grams) or 15.7moles per liter. The confusing part is the final concentration is 5% w/v or 5 grams per 100mL when the starting concentration was 70% w/w (grams per 100grams). So this goes to show why we need to be diligent in showing w/v or w/w.

Moles/Liter times Liters gives moles of HNO3

molar mass HNO3

Ending 5% w/v inverted

Final volume

15.7

moles

200

0.001

63.01

grams

100

0.001

3.97

Liters

milli

mole

g HNO3

milli

The manufacturer of the left bottle of acetic acid was smart enough to show "w/w" after the 27%. That means for every 100 grams of solution, 27 grams is acetic acid. So there's no confusion. It's hard to see but on the vinegar bottle, it says "5% acidity" meaning 5% w/v acetic acid. Some people think it means 5% w/w acetic acid, but my research points to 5% w/v. Since vinegar is close to 1gram per 1mL, they both are correct within 1 significant figure.
The vinegar bottle is 750mL. So a dilution problem might say, "How many milliliters of the 27% w/w concentration acetic acid do we need to make 750mL of 5% w/v acetic acid for the vinegar bottle?" Unfortunately, without knowing the density of the 27% w/w acetic acid, there's no way to find the milliliters needed. Remember density allows us to go from mass to volume and volume to mass. All we are given is the mass. I tried looking it up but only find density of pure acetic acid. That means the problem would have to ask "How many grams of the 27% w/w acetic acid solution is needed to make 750mL of the 5% w/v acetic acid for the vinegar bottle?" This is possible. One approach is to find out how many grams of acetic acid is in 750mL of 5% w/v acetic acid. This is easy because we can see that 5g/100mL times 750mL gives us grams. The harder part is to see that the 27% w/w (27g acetic acid per 100 g solution) needs to be inverted because our answer will then be grams of solution, which has to be on top.

	A	B	C	D	E	F	G	H	I
1	grams/mL times mL gives grams				Concentration of 27% w/w inverted to get grams of solution			Final grams of the 27% w/w solution needed
2	5% w/v		Vinegar volume		Concentration of 27% w/w inverted to get grams of solution			Final grams of the 27% w/w solution needed
3	5	g	750	mL	100	g of solution	=	139	g solution
4	100	mL			27	g pure acetic acid

Note that B3 is grams of pure acetic acid in the vinegar bottle which cancels with F4, which is the grams of pure acetic acid in the concentrated bottle. Our answer is grams of solution, which is what we would have to weigh out.

Here's another vinegar dilution problem. To the left is some vinegar we put on salads and food. If you look closely, it says "5% acidity". What they mean is that it contains 5% w/v acetic acid. The bottle at the bottom of the image is 100% acetic acid written in 7 languages. This is a typical 2.5 liter bottle that many chemicals come in. Our dilution problem is, "How many milliliters of the pure acetic acid is equal to the amount of acetic acid in this 473mL bottle of vinegar with 5% w/v acetic acid?" Normally, chemists don't want to weigh these corrosive liquids on their $2,000 analytical balances. They would prefer to just to measure out a certain volume with a $15 pipette. Like always to go from mass to volume, we need density. We can lookup the density of pure acetic acid (also called glacial acetic acid because it starts to freeze at 62°F and looks like ice). As a liquid, its density is 1.049 g/mL. Let's start by finding out the grams of acetic acid in the 473 mL. Then use density to turn that to volume. We put mL on top.

	A	B	C	D	E	F	G	H	I
1	5g/100mL times 473mL gives grams				Density of pure acetic acid, inverted		pure acetic acid
2	Vinegar conc.		Vol vinegar		Density of pure acetic acid, inverted		pure acetic acid
3	5	g	473	mL	1	mL	=	22.5	mL
4	100	mL			1.049	g

Note that the grams in B3 is grams of pure acetic acid, which cancels the grams in F4, which is also grams of pure acetic acid. The mL in B4 and D3 should say "mL of vinegar solution". They both cancel. The mL in F3 and I3 should say "mL of pure acetic acid". That's why the answer is in mL of pure acetic acid. It doesn't cancel.

To the left is a bottle of 0.1M sodium phosphate. Let's say we need to make up 150 mL of 0.02M sodium phosphate. How many mL of the 0.1M solution do we need? Whenever given molarity "M", turn it to moles per liter. The information of 150mL of 0.02 moles per liter is a good start because we can get find moles in that. What do we do with 0.1M (0.1 moles per liter)? We invert it so that we end up with liters on top because we want our answer in liters.

	A	B	C	D	E	F	G	H	I
1	Concentration times volume gives moles sodium phosphate (millimoles here)				Stronger solution conc. inverted cancels moles		Final mL of stronger solution
2	Diluted concentration		diluted volume		Stronger solution conc. inverted cancels moles		Final mL of stronger solution
3	0.02	moles	150	mL	1	Liter	=	30	mL
4	1	Liter			0.1	moles

Note: The liters in B4 and D4 are volumes of the diluted solution. The liters in F3 and I3 are referring to volumes of the stronger solution.

To the left are some chemical reagents of different molar concentrations. The right bottle is 1M potassium chromate. The middle bottle is 0.08M potassium chromate, which has about 250mL in it. We want to fill it. It's a liter bottle, so we need to make up 750mL of 0.08M solution. We will use the 1M solution. How many mL of the 1M solution do we need to make 750 of the 0.08M solution? This is the same setup as the previous problem.

	A	B	C	D	E	F	G	H	I
1	Concentration times volume gives moles sodium phosphate (millimoles here)				Stronger solution conc. inverted cancels moles		Final mL of stronger solution
2	Diluted concentration		diluted volume		Stronger solution conc. inverted cancels moles		Final mL of stronger solution
3	0.08	moles	750	mL	1	Liter	=	60	mL
4	1	Liter			1	mole

So this means we take 60mL of the 1M sodium phosphate and add water until we reach 750mL. The final volume of 750mL is 12.5 times more than the 60mL, which is why the concentration of 1M reduces by 12.5 times down to 0.08M.
There are some easy formulas for doing dilution that your book might show. Those are fine when all units match. Like here both concentration were in moles per liter and both are measured in milliliters. However, if any one of these didn't match, the shortcut dilution formulas won't work. Dimensional analysis allows you to adjust for those curves that get thrown at you.

Here's a bottle of Everclear. This is in volume percent (v/v). So 95% means 95 mL ethanol per 100mL of liquid. Notice the flammable warning. The top of the bottle as a metal flame arrestor (probably a wire mesh). This prevents a flame from going down into the bottle. Your friends are smokers and you don't want them to catch their drink on fire. So if you started with 1 cup (237 mL) of 95% v/v (190 proof) what volume would it have to be to be 49% (98 proof)? (Alcohol does not burn under 50% (100 proof). Give answer in liters or milliliters.
Our approach has been to find the actual quantity of the solute in question using the given concentration and given volume. So we begin with that.

	A	B	C	D	E	F	G	H	I
1	Concentration times volume calculates mL pure ethanol				Weaker solution conc. inverted cancels mL pure ethanol		Final mL of weaker solution
2	strong concentration		volume		Weaker solution conc. inverted cancels mL pure ethanol		Final mL of weaker solution
3	95	mL pure ethanol	237	mL	100	mL of weaker solution	=	459	mL
4	100	mL strong solution			49	mL pure ethanol

Note: B4 and D3 are mL of the strong solution. F3 and I3 are mL of the weaker solution. This this means you would take the 237mL of the 95% liquor and raise the volume to 459mL to get it to 49%. In other words, you would add 222 mL (459mL-237mL) of water to dilute it to the point it won't catch fire.

I'd like to do a problem from the textbook but use my spreadsheet method rather than the algebra method used in the textbook. This is the problem:
"Typical blood serum is about 0.14M NaCl. What volume of blood contains 1.0mg NaCl?"
Whenever I see "M" as a concentration, I always turn it into "moles per liter" because that's the units you will need to work with. So the 3 units we have are moles, liters, and grams with a metric "milli" prefix. We know we can always convert between grams and moles using molar mass. We know we can get rid of milli if we need to. Since the question is asking about "What volume" we know we will need to use the liters that's we see in "0.14 moles per liter". However, because this is "per liter" we will have to invert it. So we put 0.14 moles in the denominator (A3/B3), and 1 liter in the numerator (A2,B2). We might not know whether to put the other data "1.0 mg NaCl in the numerator or denominator at this point. However, when we do grams-to-mole conversion (E/F) we see we have to put moles on top (F2) so that it cancels the moles in B3. That causes the grams NaCl (F3) to be on the bottom. To cancel grams, our 1.0mg NaCl will then have to be on top (C2/D2). If we keep the milli in "mg" our answer will be in milliliters, which is good because the answer in liters would be quite small. Even though this seems difficult, I think it is better than the way the book showed it by doing algebra followed by dimensional analysis. My approach does it all with dimensional analysis.

	A	B	C	D	E	F	G	H	I
1	Concentration 0.14 moles per liter inverted		mass NaCl		grams-to-moles			Volume of 0.14M blood that has 1mg
2	1	Liter	1.0	mg NaCl	1	mole NaCl	=	0.13	mL blood
3	0.14	moles NaCl			55.44	g NaCl

Summary:
These kind of problems are not very easy, but there are some standard approaches.
1. First realize that the solute (substance dissolved) is the same amount in both the stronger solution and after it is diluted.
2. A problem always gives the concentration and volume of either the diluted solution or the stronger solution.
3. Start your dimensional analysis with those two knowns (the given concentration and its volume)
4. Use the other concentration to find the unknown volume. This concentration is usually inverted to do the proper canceling of the units we want to cancel and to leave the volume units.

	A	B	C	D	E	F	G	H	I
1	Known concentration		Known volume or grams		Other concentration inverted			Final volume
2	Given	moles, grams, mL of solute	Given	grams or mL of solution1	1 or 100	grams or volume of solution2	=	???	mL of solution2
3	1 or 100	grams or mL of solution1			Given	moles, grams, or mL of solute

<-CHM151

Since Oct., 2009