Solutions and Molarity

 

The amount of material in a solution has to expressed in terms of concentration. You can see how confusing it would be if solutions were discussed in terms of moles.

If beaker A was labeled

"1 mole of sugar"

because 1 mole of sugar was placed in 2 L of water there would be no way of knowing how much sugar was left after some of the solution was removed.

So, it would be better if the jar was labeled "342 g of sugar in 2 L of solution". Then we would be able to tell how much sugar was left after some of the solution was removed.

Solutions are always described in terms of amount of solute dissolved in amount of solvent or in terms of amount of solute dissolved in amount of solution.


Molarity

There are a variety of ways to express concentration.

Commercial products are often labeled

% vol. (hydrogen peroxide, wine)

or

% mass (hydrochloric acid)

Polymer chemists use

g/100 mL

when performing viscosity experiments to determine the molar mass of their polymers.

Typically, chemists use molarity. Molarity is abbreviated (M) and it is defined as follows:

 

Believe it or not, molarity was invented for convenience.

Let's imagine, for a moment, a world in which concentration was measured in grams per liter of solution.

You are instructed to make 58 g NaCl by mixing together HCl and NaOH solutions.

In the lab you find two bottles. One bottle is labeled 36.5 g HCl per L of solution, and the other is labeled 20.0 g of NaOH per L of solution.

So, how much of each solution do you need to mix together?

The balanced equation is the only way to relate NaCl, HCl, and NaOH; therefore, the amounts of HCl, NaCl, and NaOH must be related using moles. That means all the concentrations must be converted to moles each time the solutions are used.

Now, let's imagine, for a moment, a world in which concentration was measured in grams per liter of solvent.

Now the problem is the volume... If you add a liter of water to 20.0 g of NaOH what is the volume of the solution? I do not know and neither would you because the NaOH will change the volume. So, there would be more than 1 L of solution

Molarity eliminates both of these problems.

Solution preparation is simplified because solvent is added until the proper volume of solution is obtained.

Gram to mole conversions are eliminated once the solution is prepared.


Molarity Problems

First kind: Determine the concentration of a solution.

What is the concentration (molarity) of a solution made by dissolving 0.20 mol ammonium nitrate (NH4NO3) in enough water to make 500.0 mL of an NH4NO3 solution.

Since molarity is just moles per liter if you know how many moles are present in a certain number of liters then you know the molarity.

 

Second kind: Determine the number of moles in a certain volume of a solution of known concentration.

How many moles of albumin are in 3.00 x 102 mL of a 0.015 M solution?

Essentially the problem is a conversion from mL to mol.
Molarity relates moles to L (or mL); so, you could use molarity as a conversion factor!

Use the frame advance button to step through the QuickTime movie.



Third kind: making a solution.

Make 250 mL of a 2.5 M aqueous CH3CH(OH)CH3 (isopropanol) solution.

You need to now how much isopropanol to mix with the water. So, you need to determine the number of moles of isopropanol that you need, which will allow you to calculate the number of grams of isopropanol are needed.

Start with 250 mL because that is how much you want to make, and convert to L because that is what molarity uses.

 

Use the frame advance button to step through the QuickTime movie.

Since isopropanol is a liquid you could convert that number to mL and measure 48 mL isopropanol. Does that mean you need to add 202 mL H2O? NO, you must use a 250 mL volumetric flask and dilute to the line. Actually, you will add more than 202 mL water.

 



Dilutions

M(who)V(what) = huh?

(I hate M1V1 = M2V2 because most people memorize this equation,
but do not know when to use it!)

 

10.0 mL of a 0.20 M NaCl solution is diluted to 100.0 mL. What is the concentration of the final solution?

Instead of using the "MV abomination" why not analyze the problem. To find the concentration of the final solution you need to know two things...
# of moles

volume of solution
You know the final volume is 100.0 mL (that is what the problem says).

Now you need to know the number of moles of NaCl present in the solution. How can you find the number of moles?

All the NaCl that is in the new solution came from the old solution. So, find the number of moles of of NaCl that are in the original 10.0 mL solution.

So, molarity is just another conversion factor, like molar mass, or density.

 

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