CR_Molarity Lesson

Molarity

image of beakerRecall that because the particles are free to move around and come into contact with each other, many reactions are carried out in solution. You have already been writing reactions with such chemicals. So far, our representation of such reactions has been only qualitative. We are now ready to add in the quantitative part. The most important information you need to know in order to be able to do this is something called molarity.  

We talk about a chemical being aqueous, or dissolved in solution. It should make sense to you that the amount of product that will be formed in a reaction that involves a solution will depend upon the amount of that chemical and the amount of water it is dissolved in. In any solution, the substance that does the dissolving is known as the  solvent. The  solute  is the substance that is dissolved. The ratio of solute to solvent or solution is known as the concentration  of the solution. Knowing the concentration of a solution will allow us to do stoichiometry problems that involve reactions that are in solution.

There are a number of ways to calculate concentration, the most common being molarity. The molarity of a substance is defined as the moles of solute divided by the liters of solution:

Molarity (M) = mol of solute / L of solution M = mol / L

Example: A solution of MgCl2 is made by dissolving 12.5 g of MgCl2 in enough water to make 67.0 mL of solution. What is the molarity of the solution? Often, the hardest part of doing a problem is knowing where to start. You should always start by writing down your data, including your unknown. m = 12.5 g (This lower case m is mass.) V = 67.0 mL M = ? Any time you see the word molarity (or the symbol M) you should use the molarity formula. So, write down the molarity formula. M = mole / 0.067 L Now, plug in what you know and circle what you ultimately want to solve for. M mol 0.067 L Notice that I plugged in L even though I was given mL. I just converted this in my head by dividing by 1000. This is something you will do often. I circled M because that is what I want to solve for. So, now I can easily see that I need moles to finish. Look back at your data. You don't have moles. But, you do have mass. Do you remember where to get the conversion between moles and mass? Hover for Answer 1 mol ?mol MgCl2 = 12.5g x = 0.131 mol 95.2 g Now you can plug this number into the formula above. M = 0.131 mol / 0.067 L = 1.96 M

Preparing Solutions

Analysis in any laboratory requires the use of what are known as standard solutions. Standard solutions are solutions of a precise, known concentration. The accuracy in the preparation of standard reflects accuracy of the results. There are two common methods for preparing standard solutions: 1) from a solid 2) from a dilution.

The second way is to take a concentrated solution and dilute it with solvent until the final volume is reached. Often this concentrated solution is called a standard stock solution.

Use a volumetric pipet to measure a small amount of the stock solution (in the beaker). A pipet is used because it allows you to measure volume very precisely. How to Make a Dilution This precise amount of stock solution is transferred to a volumetric flask. Then, distilled water is added until the meniscus reaches the line on the neck of the volumetric flask.

Can you see that the number of moles of solute in the pipet and then in the volumetric flask are the same?   Only the amount of water (solvent) has changed, resulting in a lower molarity.

For any dilution problem, the volume of the initial solution x its molarity is equal to the volume of the final diluted solution x its molarity.  

DILUTION EQUATION V₁ x Mi = Vfx Mf

Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution? Vi Mi = Vf Mt Vi (16 M) = (1.5 L) (0.10 M) V₁ = 0.0094 L = 9.4 mL

Remember to work on the module practice problems as you complete each section of content.  

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