Introduction to Dilution Problems

In performing lab work, in science courses and in science-related jobs, it is sometimes necessary to prepare a solution of a specific, desired concentration by diluting another solution of higher concentration. These "dilution problems" are a source of repeated difficulty for many students. You must master how to solve these kinds of problems.

First let's get some basic terminology straight. A solution has two components: the solvent and the solute. The solute is the substance that is dissolved in the solvent. Most often the solvent will be water, but there are many others possible, as you will see especially in chemistry courses (various alcohols, ethers, e.g.). And most often the solute will be some type of solid (salts, carbohydrates, amino acids, e.g.), but the solute may be a gas or even another liquid. For example, carbon dioxide (CO2) gas and oxygen (O2) gas are important solutes in your blood; and if you dissolved 5 mL of ethyl alcohol in 100 mL of water, the alcohol (a liquid) would be the solute in that solution.

Concentration of a solution refers to how much of one component (solute usually) there is relative to the amount of the other component (solvent usually) or relative to the total amount of material present (solute plus solvent). Keep in mind, though, that one may view the solvent as well in terms of its fraction of the whole solution. The concept of concentration is a familiar one. For example, a cup containing one teaspoon of dissolved sugar has a lower sugar concentration that a cup containing two teaspoons of dissolved sugar. Though the volume of both solutions is one cup, the amount of solute (sugar) is different and the concentrations of the two are different.

There are many ways of expressing concentration, e.g. molarity, molality, normality, percentage, mass-per-volume. You'll deal with all of these in chemistry courses and, sooner or later, in biology courses too. In biology you will probably see molarity, percentage, and the mass-per-volume (mg/mL, g/L, etc.) type of expressions of concentration more often than others. Since percentage is already familiar to you, we'll use percentage expressions of concentration to introduce dilution problems. Percentage is always based on a fractional part of 100. A penny is 1% of a dollar, or 1 part in 100. A dime is 10% of a dollar, or 10 parts in 100. Percentage concentration expressions take various forms, as follows.

a. volume per volume (v/v), commonly used when both solute and solvent are liquids. Example: If you combine 5 mL of ethyl alcohol with enough water to make 100 mL (total volume) of solution, the concentration of alcohol is 5% (v/v).

b. weight per weight (w/w). Example: If you dissolve 5 g of table salt (NaCl) in enough water to make 100 g (total weight) of solution, then the salt concentration is 5% (w/w). Yes, you can weigh a liquid. Water weighs 1 g per mL. The w/w expression can also be used when solute and solvent are both liquids, but caution is needed because different liquids may have different densities. Water's density is 1g/mL, but alcohols and other types of liquids may have higher or lower densities. 5 mL (volume) of water and 5 mL (volume) of pure ethyl alcohol won't have exactly the same weight. So, a 5% (v/v) solution of ethyl alcohol and a 5% (w/v) solution of ethyl alcohol won't have exactly the same alcohol concentration.

c. weight per volume (w/v). Example: If you dissolve 5 g of table salt in enough water to make 100 mL (total volume) of solution, then the salt concentration is 5% (w/v). This is the most common of the three. **It will help to remember that water weighs one gram per mL, or 100 g/100 mL. Since pure water has no solute dissolved in it, the entire weight of 100 mL of pure water is accounted for by the water; the solute concentration is 0%, obviously. However, in 100 mL of a solution, some part of the weight and volume of that 100 mL total is due to the solute; the solute part of a solution does occupy space (volume).

Before going on, it's necessary to understand that in speaking of solutions, the concentration of the solute is the same everywhere within the solution; that is, the solute is uniformly distributed throughout the solvent; the solution is homogeneous. Consider one liter (about one quart) of 5% (w/v) sugar solution in a bottle. If you withdraw 1 mL samples from the bottle, every sample will have the same sugar concentration, and in every 1 mL sample there will be 0.05 g of sugar.

When you do the permeability exercise a few weeks hence, you will prepare a 50% (w/v) solution of sucrose (common table sugar). Then you'll dilute that stock solution to prepare sugar solutions of lower concentrations. Consider this. If you combined 20 mL of this sugar stock solution with 20 mL of water, the volume of the new solution would be 40 mL, and its concentration would be 25% (w/v). Perhaps it is obvious that by combining equal volumes of water and the stock solution, the sugar concentration is cut in half. Similarly, if you added 40 mL of water to 10 mL of that sugar stock solution, you would obtain 50 mL of 10% (w/v) sucrose solution. But maybe that answer is not so obvious, even though the numbers are "round" and easy to work with. What concentration do you get if you combine 17 mL of water with 26 mL of the 50% (w/v) sucrose solution? It's obvious that the volume of the new solution would be 43 mL, but what's the sugar concentration? That is not obvious at a glance.

Most people cannot work such problems at first. But such problems do occur in many different settings and must, therefore, be dealt with. The good news is that everyone here can learn how to solve problems involving concentrations and dilutions.

It is important to realize that there is a logical way to solve dilution problems, to reason your way through them. Learning this is the sensible approach now. There's time later to learn a simpler method (shortcut) after you understand how to do these kinds of problems. {The danger in using shortcut methods without understanding why they work is that you'll use a shortcut in a situation where it does not apply.} Now let's reconsider the two sample problems introduced a moment ago.

Problem 1. If you combine 40 mL of water and 10 mL of 50% (w/v) sucrose solution, what is the sucrose concentration of the new solution? Clearly, all of the sucrose that there will be in the new solution must come from the 10 mL of the 50% (w/v) sucrose solution that you add to the 40 mL of water. (The 40 mL of water, itself, contains no sucrose.) Therefore, the total mass of sucrose in the new solution must be 10 mL X 0.5 g/mL = 5 g. Note that in the original sucrose solution 50% (w/v) means 50 g sucrose per 100 mL of solution, which is also 0.5 g/mL. Multiplying that gives you 5 g of sucrose. Again, that is the total amount (mass, weight) of sucrose that the new (diluted) solution will contain.

When you multiplied 10 mL by 0.5 g/mL, the mL units canceled out; so you know that the result in that first step has gram units. Think of it this way too: when you do that multiplication step, you're asking the question, "How much sucrose (mass) is in 10 mL of the 50% solution?" The answer to that cannot be "mL" or "g/mL." So, remember that you should never drop, lose track of, or neglect the units when you work problems. The units must come out right in the end. By faithfully keeping track of the units throughout a problem, you are actually checking the work for mistakes as you go. The units have to make sense.

At this point in working the problem, you have 5 g of sucrose. Obviously, that is not a concentration expression. Continuing, that 5 g of sucrose is now in a total volume (new solution) of 10 mL + 40 mL = 50 mL. Therefore, the concentration of the new solution is 5 g/50 mL = 0.1 g/mL, which is the same as 10 g/100 mL, which is 10% (w/v).

The shortcut is to multiply what you start with, i.e. the 50% (w/v) solution, by the dilution, which is 10 mL/50 mL, = 1/5. So, 50% (w/v) X 1/5 = 10% (w/v). Here the dilution is 1/5

Problem 2. If you combine 17 mL of water with 26 mL of 50% (w/v) sucrose solution, what is the concentration of the new solution?

Again, 50% (w/v) = 0.5 g sucrose per mL of solution (that is, 0.5 g/mL).

Then, 26 mL X 0.5 g/mL = 13 g of sucrose. Note that mL canceled. The result of this step is "grams of sucrose". In doing this step, you're asking: "If 1 mL of the 50% (w/v) sucrose solution contains 0.5 g sucrose, then how much sucrose is in 26 mL?" Obviously, the answer must be in grams, not mL or g/mL.

Now, this amount (mass) of sucrose is all of the sucrose that will be in the new solution. The new solution's volume is 26 mL + 17 mL. So, the new solution must have 13 g of sucrose in a total volume of 43 mL, that is, 13 g/43 mL = 0.30 g/mL. Note that "g/mL" is a concentration expression. Then, multiplying both numerator and denominator by 100, you get 30 g/100 mL of solution, which is 30% (w/v). Remember that your algebra long ago taught you that you may multiply numerator and denominator by the same thing without changing the value of the ratio.

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These sample problems are presented this way to show you how to reason through the working of such problems. Though these are short, simple problems, it's still possible to make mistakes. It is important to be organized and methodical in working problems.

It is not a good idea to skip steps in working problems, at least not now. It's too easy to lose track of the units of measurement when two or more mental steps are done without writing things down. The emphasis now is on understanding how to do these types of problems, how to reason through them. _____________________________________________________________________________

The first sample problem here introduced a type of expression you will see in dilution problems often; so, we must agree now on the meaning. If you read that a solution is supposed to be diluted "one to ten" or "by a factor of 10" or "1/10" or "1:10" (all 4 expressions are the same), the meaning is: "combine one part of the given solution with nine parts of the solvent, not ten parts of solvent." Likewise, the instruction to dilute a given solution 1:4 (or 1 to 4 or 1/4 or by a factor of 4) means to add 1 part of the given solution to 3 parts of solvent. Another example: if a 30% (w/v) salt solution were diluted 1/5 (1 to 5) with water, the new salt concentration would be 6%, which is 30% X 1/5.

Problem 3. Did you notice that the previous paragraph's examples didn't say what "part" means and didn't say how much (what volume) of new solution to make. That's because "part" is relative. So, here's the rest of it. Suppose you're given a 30% (w/v) solution of sucrose and asked to prepare 125 mL of a 1 to 5 (1/5) dilution. The 1 to 5 means that we split the 125 mL into 5 parts, so that each "part" is 25 mL. Then 1 part (25 mL) of the new solution will be the given 30% solution and the other 4 parts (100 mL) will be the added solvent (water). Now the new solution is 30% X 1/5 = 6% (w/v) and we have 125 mL of it. [If the problem had asked for 300 mL of a 1 to 10 (1/10) dilution, each "part" would be 30 mL (i.e. 300 mL ÷ 10), and we would add 30 mL of the 30% given solution to 270 mL (i.e. 9 X 30 mL) of solvent. The result would be 300 mL of 3% (w/v) sucrose solution.]

Problem 4. Let's use the first method we learned here to check the answer to that last problem about preparing 125 mL of the 1 to 5 dilution of the 30% solution. The concentration of that 125 mL solution was 6% (w/v). That means 6 g sucrose per 100 mL solution or 0.06 g/mL solution. Suppose the problem had not said anything about 1 to 5 dilution, but instead had asked us to prepare 125 mL of a 6% (w/v) solution from the 30% (w/v) solution. How do we do that? We reason as follows.

a. First, we know that 6% (w/v) = 0.06 g sucrose/mL of solution.

b. We need to make 125 mL (total volume) of that concentration. Therefore, we know that the new solution will contain 125 mL X 0.06 g/mL = 7.5 g sucrose (total mass).

c. We will have to get that amount (mass) of sucrose from the given 30% (w/v) solution, of course; there's no place else! So, how much (how many mL?) of that 30% solution do we need to give us the 7.5 g sucrose.

d. 7.5 g sucrose X 1 mL solution/0.3 g sucrose = 25 mL solution. (Note that "g sucrose" cancels.) What is this telling us? It says that 25 mL of the 30% solution contains 7.5 g of sucrose.

e. So, now we need to add enough solvent to that 25 mL to get a final volume of 125 mL of solution. That's 100 mL, of course. And that's exactly what we did in the short problem a moment ago. We should have gotten the same answer, of course, and we did.

Serial dilution problems are common in biology; one solution is diluted to make another, which in turn is diluted to give a third, and so on. The rules just explained above about dilution problems apply in these cases too. Example: if a 20 mg/mL sugar solution were diluted 1:4 and then the new sugar solution were diluted 1:5, the concentration of the third solution would be 1 mg/mL. Try working this as a sequence of two separate dilution problems, and then you'll see why the shortcut works: 20 mg/mL X 1/4 X 1/5 = 1 mg/mL. The total dilution here is 1 to 20.

A common sort of procedure used in working with viruses and bacteria is to begin with a solution that contains bacterial cells or virus particles and to prepare a set of serial dilutions of 1/10, 1/100, 1/1000, and so on. A typical way to do this is to add 1 mL of the original solution to 9 mL of solvent in a test tube; that's a 1/10 dilution. After thoroughly mixing that new solution, add 1 mL of it to 9 mL of solvent in another test tube; that's a 1/100 dilution of the original solution. Then adding 1 mL of this solution to 9 mL of solvent in another test tube produces a 1/1000 dilution of the original solution. Here's that rule again for doing serial dilutions, you multiply the dilution factors: 1/10 X 1/10 X 1/10 = 1/1000.

Practice problems.....

1. Solution A has 1.20 mg of protein per mL. Solution B has 3.10 mg of protein per mL. If you combine 34 mL of solution A with 19 mL of solution B, what is the protein concentration of the new solution? Round the answer to 2 decimal places.

Ans. 1.88 mg/mL

2. The concentration of a salt solution is 3.8 mg/mL. 124 mL of this is placed in a beaker and allowed to evaporate until 92 mL remains. What is the salt concentration then? (We assume, of course, that water evaporates but that the salt does not.) Round the answer to 2 decimal places.

Ans. 5.12 mg/mL

3. If you're asked to dilute 67 mL of a 21% (w/v) sugar solution 1 to 7, how much solvent must you add? Round the answer to the nearest whole number (integer).

Ans. 402 mL

4. In the previous problem (#3), what is the sugar concentration of the new solution?

a. expressed as percent (w/v)? Round to the nearest integer. Ans. 3% (w/v)

b. expressed as g/mL? Round to 2 decimal places. Ans. 0.03 g/mL

Expressing concentrations as molarity

A mole is the amount of a chemical substance that contains the Avogadro number (approx. 6.02 X 1023) of particles. The particles may be atoms, ions, or molecules, depending on the substance. For instance, we may speak of a mole of sodium chloride molecules (NaCl) or a mole of sodium ions (Na+). Though different molecules may differ greatly in their molecular weights, a mole of each contains this same number of molecules; this is an important concept.

The gram molecular weight or gram formula weight of a chemical substance is the molecular (or formula) weight expressed as grams. The weights of sodium chloride (NaCl) and glucose (C6H12O6), for example, are 58.3 daltons and 180 daltons, respectively. Their gram molecular weights are 58.3 grams and 180 grams, respectively. That means that 58.3 g of NaCl contains 6.02 X 1023 molecules of NaCl, and that 180 g of glucose contains that same number of glucose molecules. Thus, a mole of a substance is equal to its molecular weight in grams.

Molarity, which is symbolized by the letter "M," expresses the concentration of a solute in solution in terms of the number of moles of that solute per liter of solution. As an example, a "one molar" solution (written as 1 M) of glucose contains glucose dissolved in water to the extent of one mole of glucose (= 180 g) in one liter of solution. To prepare such a solution, one would add water to the glucose, with stirring, until all of the glucose was dissolved, and then add enough additional water to bring the final volume of the solution to exactly one liter. [Customarily a volumetric flask is used to do this.] Since concentration is a ratio of solute mass to solution volume, it applies to any volume of solution. For example, you may picture a 1 M glucose solution filling a tiny test tube or filling a tea cup or a bath tub or a swimming pool. The total volume of the solution in these cases is very different, but the concentration is the same, 1 M.

This is an important point to understand. Among other things, it shows us how to relate molarity to the weight-per-volume concentration expression discussed above. Staying with the glucose example, if you dissolve 18 g of glucose in enough water to make exactly 100 mL of solution, then you have a 1 M solution (same as 180 g in 1 liter of solution). However, notice that this is also an 18% (w/v) glucose solution. Therefore, if you know the molecular weight of the solute of interest, you can interconvert molarity expressions of concentration and weight-per-volume expressions of concentration.

Problem 6. Here's an example using the common amino acid called glycine, whose molecular weight is 75.1 daltons. Given a 0.35 M glycine solution, what is its concentration expressed as weight per volume (w/v)? This problem turns into a dimensional analysis problem. Notice first that 0.35 M means 0.35 mole of glycine per liter of solution. (Glycine is abbreviated GL here.)

 0.35 mole GL L soln. X 75.1 g GL mole GL X 1 L soln. 103 mL soln. = 0.026 g GL mL soln. X 100 100 = 2.6% (w/v)

You see what cancels out: "mole GL" and "L soln." The g/mL form of the answer is correct, but if we want the answer in percentage (w/v) form, then we need to convert the answer to "parts of solute per 100 parts of solution." To do that we need only multiply both numerator and denominator by 100, as shown (there's algebra again). The 0.026 g/mL form of the answer could be written as 2.6 X 10-2 g/mL. [Note that the issue of significant digits is ignored here to be sure we all see exactly what's happening with the numbers.]

Problem 7. Now try this problem. Given an 8.5% (w/v) glycine solution, convert that to a molarity concentration expression. First, we know that 8.5% (w/v) = 8.5 g glycine/100 mL solution.

 8.5 g GL 100 mL soln. X 103 mL soln. L soln. X 1 mole GL 75.1 g GL = 1.13 mole GL L soln. = 1.13 M

Again, you see that this is a dimensional analysis problem.

Problem 8. Let's return to dilution problems, but work now with molarity expressions instead of percentage concentration expressions. An example: If you add 85 mL of water to 38 mL of 0.6 M glycine solution, what is the concentration of the new solution? Clearly the volume of the new solution is 123 mL.

 0.6 mole GL L soln. X 1L soln. 103 mL soln. X 75.1 g GL mole GL X 38 mL soln. = 1.71 g GL

This sequence of steps calculates the amount (mass) of glycine in 38 mL of the 0.6 M solution. Be sure you see how units canceled out to give "g of glycine." We know that the 1.71 g glycine is now in 123 mL of solution, so the rest of the problem is to convert that to molarity, as follows.

 1.71 g GL 123 mL soln. X 103 mL soln. L soln. X 1 mole GL 75.1 g GL = 0.19 mole GL L soln.. = 0.19 M

If you understand this procedure well, you may be able to see that a shortcut can be devised by combining the two parts of the problem's solution:

0.6 M GL X 38 mL / 123 mL = 0.19 M

**Note again that the units always work out correctly if you perform the dimensional analysis correctly.