Lesson 4: Molarity
Another way of expressing concentration, and the way that we will use most in this course, is called molarity.
Molarity is the number of moles of solute dissolved in one liter of solution. The units, therefore are moles per liter, specifically it's moles of solute per liter of solution. |
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Rather than writing out moles per liter, these units are abbreviated as M or M. We use a capital M with a line under it or a capital M written in italics. So when you see M or M it stands for molarity, and it means moles per liter (not just moles).
You must be very careful to distinguish between moles and molarity. "Moles" measures the amount or quantity of material you have; "molarity" measures the concentration of that material. So when you're given a problem or some information that says the concentration of the solution is 0.1 M that means that it has 0.1 mole for every liter of solution; it does not mean that it is 0.1 moles. Please be sure to make that distinction.
In this section of the lesson, we'll look at molarity calculations, the process of dilution (taking a more concentrated solution and adding more solvent to acheive a particular molarity), and species concentrations.
Molarity Calculations | Dilution | Species Concentrations
Molarity Calculations
There are several types of calculations that you need to be able to do with molarity.
- First, you should be able to calculate the molarity if you are given the components of the solution.
- Second, you should be able to calculate the amount of solute in (or needed to make) a certain volume of solution.
- Third, you might need to calculate the volume of a particular solution sample.
- Fourth, you might need to calculate the concentration of a solution made by the dilution of another solution.
In either of the first two cases, the amount of solute might be in moles or grams and the amount of solution might be in liters or milliliters. Please note that with molarity we are concerned with how much solute there is and with how much solution there is, but not with how much solvent there is.
Examples (Ex. 3)
The examples that follow are also shown in example 3 in your workbook. You may want to look through the workbook examples on your own and if they make perfect sense to you, test your understanding by doing exercise 4. Then check your answers below before continuing on to dilution calculations.
General Relationship (Ex. 3a)
Here is the general relationship that you will be using over and over again. The molarity is equal to the number of moles of solute divided by the volume of the solution measured in liters. If you like to think of numbers and units instead of quantities look at the second version of the equation. In this equation x, y and z represent numbers: 2, 6 and 3 as an example. |
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Calculating Molarity from Moles and Volume (Ex. 3b)
Here we are given something to figure out. To get the molarity we need to divide the number of moles of NaCl by the volume of the solution. In this case that is 0.32 moles NaCl divided by 3.4 L, and that gives 0.094 M NaCl. |
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Calculating Molarity from Mass and Volume (Ex. 3c)
This one is a bit more difficult. To get molarity we still need to divide moles of solute by volume of solution. But this time we're not given the moles of solute. We have to calculate it from the mass of NaCl. We multiply 2.5 g NaCl by the conversion factor of 1 mole NaCl over the formula weight of NaCl, 58.5 g. That tells us that we have 0.0427 mole of NaCl. I kept an extra digit here because we are not done with the calculations. When we are done I'll round off to two digits, the same as in the 2.5 g weight of NaCl. Now that we know the moles we can calculate the molarity. Moles of solute (0.0427) divided by the volume of the solution (0.125 L) gives us 0.34 M NaCl. |
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Calculating Mass of Solute from Molarity (Ex. 3d)
This question asks how you would prepare 400. ml of 1.20 M solution of sodium chloride. In this case what you need to find out is how much NaCl would have to be dissolved in 400 ml to give the concentration that is specified. This amount is going to have to be in grams because we don't have any balances that weigh in moles. So there is more than one step to this problem. |
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The approach shown here is a conversion factor approach. It involves remembering that molarity is a relationship between moles and liters. 1.20 M NaCl means there is 1.2 moles of NaCl per 1.00 liter of solution. We can use that as a conversion factor to set up the calculation that relates 400. ml (or .400 L) to the appropriate number of moles of NaCl. So we take .400 L and multiply by the conversion factor to get .480 moles NaCl. The next step is to find out how many grams that is. We change from moles of NaCl to grams by using the formula weight. It comes out to 28.1 g NaCl. The final answer is that you would make the solution by dissolving 28.1 g NaCl in enough water to make 400 ml of solution.
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There is more than one way to do this problem. If you like the algebraic approach, you would write down the general equation shown in part a, substitute in the known values, solve for moles of NaCl, and then change that into grams. |
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Calculating Moles of Solute from Molarity (Ex. 3e)
This question is a little easier. We do it the same way as the first step of the previous problem and then we stop. To find out how many moles of salt are contained in 300. ml of a 0.40 M NaCl solution, we start with the volume in liters (0.300 L) and multiply it by the number of moles per liter of solution, which is 0.40 moles over 1.00 L. The answer is 0.12 moles of NaCl. This could also have been done using algebra by writing down the general equation relating molarity, moles and liters, substituting the known values, and then solving the equation for moles. |
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Practice (Ex. 4)
Now you should take some time to review the calculations above (ex. 3). If you have any questions, check with your instructor. Once you are familiar with how those are done then you should try answering the following questions (exercise 4 in your workbook). Get help from the instructor if you need it. Check your answers below before you continue with the lesson.
Molarity Calculations: Practice
a. |
How would you prepare 100. mL of 0.25 M KNO3 solution? |
b. |
A chemist dissolves 98.4 g of FeSO4 in enough water to make 2.000 L of solution. What is the molarity of the solution? |
c. |
How many moles of KBr are in 25.0 mL of a 1.23 M KBr solution? |
d. |
Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are in 400. mL of this solution? |
Answers (Ex. 4)
Here are the answers to exercise 4.
a |
How would you prepare 100. mL of 0.25 M KNO3 solution? Dissolve 2.53 g of KNO3 in enough water to make 100 ml of solution. |
b |
A chemist dissolves 98.4 g of FeSO4 in enough water to make 2.000 L of solution. What is the molarity of the solution? 0.324 M |
c |
How many moles of KBr are in 25.0 mL of a 1.23 M KBr solution? |
d |
Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are in 400. mL of this solution? 120 g |
Dilution
Up until now, the concentration calculations that we have done essentially involved preparing a solution from scratch. We started with separate solvent and solute and figured out how much of each you would need to use.
Quite often, however, solutions are prepared by diluting a more concentrated solution. For example, if you needed a one molar solution you could start with a six molar solution and dilute it. Consequently, you also need to be familiar with the calculations that are associated with dilutions.
There is an element of simplicity in calculations of these types and the element of simplicity is that the number of moles of solute stays the same, as shown here. The number of moles of solute in the concentrated solution (indicated by the subscripted molescon) is equal to the number of moles in the dilute solution. You have simply increased the amount of solvent in the solution. |
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Of course you know that the number of moles of solute in the concentrated solution is equal to the molarity of the concentrated solution times the volume of the concentrated solution. Also, the number of moles of solute in the dilute solution is equal to the molarity of the dilute solution times the volume of the dilute solution. Since we are really interested in the molarities and volumes we can substitute and use the equation shown in the second line (or third line, depending on your preference for how to show multiplication). Let's use this equation in a few examples. |
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Examples (Ex. 6)
Calculating New Concentration (Ex. 6a)
In this example you are asked what the concentration of a solution would be if it were made by diluting 50.0 ml of 0.40 M NaCl solution to 1000. ml. As a general procedure, I recommend that you first write down the equation that is given in the first line: the molarity of the concentrated solution times the volume of the concentrated solution is equal to the molarity of the dilute solution times the volume of the dilute solution. Next, substitute the known values into the equation as shown in the next line. Then rearrange the equation to solve for the unknown value. In this case that is done by dividing both sides of the equation by 1000. ml. Last, carry out the necessary calculations to get, in this case, .020 M. |
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Calculating Initial Volume (Ex. 6b)
The question in this example is different only in that you are asked to determine a volume instead of a concentration. The process used to answer the question is the same as in the previous example. Write down the algebraic equation that represents the relationship. Rearrange the equation to solve for the unknown quantity. Substitute numbers (and units) for the known values. (The second and third steps can be reversed if you wish.) Calculate the unknown value (which now becomes known). |
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If you have any questions about how to do these kinds of problems, please ask the instructor.
Practice (Ex. 7)
Now you should practice working on dilution problems by answering the following questions (from exercise 7 in your workbook). Do those now and check your answers before you continue.
Dilution Calculations: Practice
a |
How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M NaCl solution? |
b |
What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M KOH to 250 mL? |
c |
What would be the concentration of a solution made by adding 250 mL of water to 45.0 mL of 4.2 M KOH? |
d |
How much 0.20 M glucose solution can be made from 50. mL of 0.50 M glucose solution? |
Answers (Ex.7)
Here are the answers to the questions in exercise 7.
a |
How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M NaCl solution? 19 mL (2 s.d. from 18.75 mL) |
b |
What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M KOH to 250 mL? 0.76 M (2 s.d.) |
c |
What would be the concentration of a solution made by adding 250 mL of water to 45.0 mL of 4.2 M KOH? 0.64 M |
d |
How much 0.20 M glucose solution can be made from 50. mL of 0.50 M glucose solution? 130 mL (2 s.d. from 125 mL) |
Species Concentrations
There are times when you need to deal with the concentration of chemical species in solution rather than the concentration of a chemical compound. Perhaps these examples will show why this is important.
When electrolytes dissolve in water, they dissociate into ions. When sodium chloride dissolves in water, sodium and chloride ions are formed. When calcium chloride dissolves in water, calcium and chloride ions are formed. Note that different amounts of chloride ions are formed. Equal concentrations of NaCl and CaCl2 generate different concentrations of Cl- ion. |
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Na+ + Cl-
With weak electrolytes, this issue is complicated by the fact that some of the chemical remains undissociated. For example when HF dissolves in water HF(aq), H+(aq) and F-(aq) are all formed and present in the solution. Depending on the conditions under which the solution is formed, the concentration of each of those chemical species might be different.
When dealing with the concentrations of chemical species, it is customary to use brackets around the formula of the chemical specie as a symbol for the concentration of that specie. For example, [Cl-] = 2 M means the concentration of chloride ion is 2 M.
Later in this lesson (e.g. equilibrium constants) and in later lessons (e.g. reaction rates) we will need to focus on the concentration of the particular chemical species that are reacting with one another or of particular interest to us for some other reason. Molarity can be used for this. It is simply a matter of specifying that the concentration refers to a particular chemical specie rather than a chemical compound.