Dilution Problems

Last updated 11-11-11

Dilution Problems Tutorial

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.

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). 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 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 time 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 27% solution	=	139	g of 27% 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 instead of measuring volume.

A better way to do the above problem is to start with a measurement that matches the answer. Since we want to know the grams of 27% solution needed to make the solution, the only data that includes "grams of 27% solution" is the concentration measurement of 27% w/w, which means 27 grams acetic acid per 100 grams of the 27% solution. We have to invert the concentration so that "grams of 27% solution" in on the top (the same as the answer in cell I3). The inverted concentration has 27 grams of pure acetic acid in the denominator. We need to cancel "grams of pure acetic acid" because the answer does not have "grams of pure acetic acid". To cancel "grams of pure acetic acid", we need the another measurement that has "grams of pure acetic acid". The only data that includes "grams of pure acetic acid" is the density of acetic acid, which is 5 grams of pure acetic acid per 100 mL of vinegar. We will use that to cancel out "grams of pure acetic acid" in cell B4 The concentration has 100 mL of vinegar in the denominator. We need that canceled also. So we write the 750 mL of vinegar in the numberator in cell F3, which will cancel out cell D4.

	A	B	C	D	E	F	G	H	I
1	Notice B3 matches I3		D3 cancels B4		F3 cancels D4			All units get canceled except for g of 27% solution
2	Concentration of 27% w/w inverted		5% w/v		volume of vinegar wanted			Final grams of the 27% w/w solution needed
3	100	g of 27% solution	5	g pure acetic acid	750	mL of vinegar	=	139	g of 27% solution
4	27	g pure acetic acid	100	mL of vinegar

The above calculation is a good example of dimensional analysis. Data is picked to match the unit of measurement in the answer. Then the problem progresses to get rid of any units of measurement that are not in the final answer. The color backgrounds show the pattern. The green is the start and end units. The yellow shows how one of the units at the start gets canceled. The purple shows how a new unit in the second fraction gets canceled. At the end only the unit of the answer should be present.

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. Our dilution problem is, "How many milliliters of the pure acetic acid is in this 473mL bottle of vinegar with a concentration of 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 (Note: this is a shortcut way of saying 1.049 grams of pure acetic acid per 1 mL of pure acetic acid).

Since our final answer is mL of pure acetic acid, let's use density of acetic acid, because it includes the units of mL of pure acetic acid. We have to invert the density of 1.049g/mL in order to get mL on top. Again: "mL" here means mL of pure acetic acid.
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 say "mL of vinegar " so they both cancel. The mL in F3 and I3 say "mL of pure acetic acid". That's why the answer is in mL of pure acetic acid. It doesn't cancel.

	A	B	C	D	E	F	G	H	I
1	B3 matches I3		D3 cancels B4		F3 cancels D4		All units cancel except mLof pure acetic acid
2	Density of pure acetic acid, inverted		Concentration of vinegar		Volume of vinegar		mL of pure acetic acid in 473 mL vinegar
3	1	mL of pure acetic acid	5	g of pure acetic acid	473	mL of vinegar	=	22.5	mL of pure acetic acid
4	1.049	g of pure acetic acid	100	mL of vinegar

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 (Na₃PO₄). How many mL of the stronger 0.1M solution do we need? Whenever you are given molarity "M", turn it to moles per liter. Life before, we start with the data that has the same units as the answer. The concentration of the strong solution is 0.1 moles per Liter of the strong solution. So we invert that concentration to get "Liter of strong solution" on top (to match the answer). Concentration of diluted solution is used next to cancel out moles of sodium phosphate (Na₃PO₄). Then the 150 mL of the diluted solution is next to cancel out "Liter of diluted solution" in D4. The "m", which is "milli" in F3 does not get canceled, so it is causes the liters of strong solution in the answer to actually be milliliters of strong solution..

	A	B	C	D	E	F	G	H	I
1	B3 matches I3		D3 cancels B4		F3 cancels D4		"milli" from F3 & "Liter of strong solution" from B3 don't cancel
2	0.1 M concentration inverted		diluted concentration		diluted volume
3	1	Liter of strong solution	0.02	moles Na₃PO₄	150	mL of diluted volume	=	30	mL of strong solution
4	0.1	moles Na₃PO₄	1	Liter of diluted 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	B3 matches I3		D3 cancels B4		F3 cancels D4		milli from F3 & Liter from B3
2	1 M concentration inverted		diluted concentration		diluted volume		mL strong solution needed
3	1	Liter of strong solution	0.08	moles K₂CrO₄	750	mL of diluted volume	=	60	mL of strong solution
4	1	moles K₂CrO₄	1	Liter of diluted solution

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 "percent by volume" which is written % v/v. So 95% means 95 mL ethanol per 100mL of liquid. Notice the flammable warning. The top of the bottle has 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% v/v (49 mL pure ethanol per 100 mL of solution)? (Alcohol does not burn under 50% (100 proof). Give answer in milliliters.
Our approach has been to start with a value (data) that has the units that match the answer. Since we want to know the final milliliters of the weaker (diluted) alcohol, we need to use the concentration of the weaker alcohol (49% v/v) because that is 49 mL of pure ethanol per 100 mL of the weaker alcohol solution. We invert its concentration to get "mL of weaker alcohol" on top.

	A	B	C	D	E	F	G	H	I
1	B3 matches I3		D3 cancels B4		F3 cancels D4		mL of weaker alcohol
2	weaker concentration inverted		concentration strong alcohol		Volume strong alcohol		Volume weaker alcohol
3	100	mL weaker alcohol	95	mL pure ethanol	237	mL stronger alcohol	=	459	mL weaker alcohol
4	49	mL pure ethanol	100	mL strong alcohol

Note: It's not enough just to indicate "mL" in these kind of problems. It's important to indicate what specific solution the mL is referring to. This problem is different than the above problems, but it still has the same strategy. The green boxes show the starting and ending units match. After than you start cancelling out units you don't want.

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 as stated in the textbook:

"Typical blood serum is about 0.14M NaCl. What volume of blood contains 1.0mg NaCl?"
First of all, 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. Since the question is asking about "What volume" we know we start with the volume of liters that's we see in "0.14 moles per liter". However, because this is "per liter," we will have to invert it to make it "Liters per mole". So we put 1 liter in the numerator (A2&B2) and 0.14 moles in the denominator (A3&B3) . Note, since this blood serum, the "Liter" means "Liter of blood." Now B3 matches the units of our answer in I3 (green boxes below). Now we need to cancel out the moles of NaCl in B3 (yellow box). This is where the molar mass of NaCl is handy because it relates moles of NaCl to grams of NaCl. Molar mass is usually given as grams per mole, but we need moles in the numerator. So we invert the molar mass to get moles on top (D3). This puts the grams of NaCl on the bottom (purplish box). To cancel grams of NaCl, we can put the 1 mg of NaCl in the numerator of the next fraction (purplish box). 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	B3 matches answer in I3		D3 cancels B4		F3 cancels D4			"m" from F3 doesn't cancel, nor does "Liter of blood"
2	Concentration 0.14 moles per liter inverted		molar mass NaCl		mass NaCl			Volume of blood that has 1mg NaCl
3	1	Liter of blood	1	mole NaCl	1.0	mg NaCl	=	0.13	mL of blood
4	0.14	moles NaCl	55.44	g NaCl

The question didn't say if it wanted the answer in mL or Liters. If they wanted "Liters" we would need to cancel out the "milli" in the "mg" in F3. The below spreadsheet does that.

	A	B	C	D	E	F	G	H	I	J
1	B3 matches answer in I3		D3 cancels B4		F3 cancels D4		Cancels m in mg in F3		Answer is now in Liters
2	Concentration 0.14 moles per liter inverted		molar mass NaCl		mass NaCl		Cancels m in mg in F3		Volume of blood that has 1mg NaCl
3	1	Liter of blood	1	mole NaCl	1.0	mg NaCl	0.001	=	0.00013	Liters of blood
4	0.14	moles NaCl	55.44	g NaCl			milli

Summary:
These kind of problems are not very easy, but there are some standard approaches.
1. Start with the data that matches the units of the answer. These have been indicated with the green boxes.
2. The starting fraction usually has something in the denominator that needs to be canceled. So the next fraction is set up to cancel the unit in the denominator. These have been indicated with yellow boxes.
3. In these problems, it usually takes a third fraction that is setup to get rid of the denominator in the second fraction. These have been indicated with the purplish boxes.
4. If prefixes like "milli" need to added or canceled, use a fraction that uses the prefix and its equivalent decimal fraction.

Textbook Readings on Concentrations and Dilutions

Remember, the authors of these books may be doing the calculations differently than I do, but the readings will give you more information on concentrations and dilutions.

Zumdaul: 6th Edition: pages, 140-148: 7th Edition: pages 133-140

Tro: 1st Edition: pages 144-148

Tro: 2nd Edition: pages 140-145

<-CHM151

Since Oct., 2009