Go to dilution problems #1 - 10

Go to dilution problems #11 - 25

Go to dilution problems #26 - 35

To dilute a solution means to add more solvent without the addition of more solute. Of course, the resulting solution is thoroughly mixed so as to ensure that all parts of the solution are identical.

The fact that the solute amount stays constant allows us to develop calculation techniques.

First, we write:

moles solute before dilution = moles solute after dilution

From rearranging the equation that defines molarity, we know that the moles of solute equals the molarity times the volume. (Calculating the moles of solute from molarity times volume will be very useful in other areas of chemistry, particularly acid base. Remember that the volume must be in liters.)

So we can substitute MV (molarity times volume) into the above equation, like this:

M_{1}V_{1}= M_{2}V_{2}

The "sub one" refers to the situation before dilution and the "sub two" refers to after dilution.

This equation does not have an official name like Boyle's Law, so we will just call it the dilution equation.

**Example #1:** 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make?

**Solution:**

Using the dilution equation, we write:

(1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x)x = 100. mL

Notice that the volumes need not be converted to liters. Any old volume measurement is fine, just so long as the same one is used on each side. (However, as mentioned above, if you are calculating how many moles of solute are present, you need to have the volume in liters.)

**Example #2:** 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results?

Placing the proper values into the dilution equation gives:

(2.500 mol/L) (100.0 mL) = (0.5500 mol/L) (x)x = 454.5454545 mL (oops, my fingers got stuck typing.) (Bad attempt at humor, really bad!)

x = 454.5 mL

Sometimes the problem might ask how much more water must be added. In this last case, the answer is 454.5 - 100.0 = 354.5 mL.

Go ahead and answer the question, if your teacher asks it, but it is bad technique in the lab to just measure out the "proper" amount of water to add and then add it. This is because the volumes (the soltion and the diluting water) are not necessarily additive. The only volume of importance is the final solution's volume. You add enough water to get to the final solution volume without caring how much the actual volume of water you added is.

**Example #3:** A stock solution of 1.00 M NaCl is available. How many milliliters are needed to make 100.0 mL of 0.750 M

(0.750 mol/L) (100.0 mL) = (1.00 mol/L) (x)x = 75.0 mL

**Example #4:** What volume of 0.250 M KCl is needed to make 100.0 mL of 0.100 M solution?

(0.100 mol/L) (100.0 mL) = (0.250 mol/L) (x)Please go ahead and solve for x.

**Example #5:** 2.00 L of 0.800 M NaNO_{3} must be prepared from a solution known to be 1.50 M in concentration. How many mL are required?

(1.50 mol/L) (x) = (0.800 mol/L) (2.00 L)x = 1.067 L

Divide the liters by 1000 to get mL, the answer is 1070 mL (notice that it is rounded off to three sig figs)

These next two are a bit harder and involve slightly more calculation than the discussion above. Here's a summary of the steps:

1) calculate total moles

2) calculate total volume

3) divide moles by volume to get molarity

You can also think of it this way:

M_{1}V_{1}+ M_{2}V_{2}= M_{3}V_{3}

Where the '1' refers to one starting solution, '2' to the other starting solution and '3' refers to the mixed solution (hints: V_{3} is the total volume after mixing and M_{3} is almost always the unknown).

**Example #6:** Calculate the final concentration if 2.00 L of 3.00 M NaCl and 4.00 L of 1.50 M NaCl are mixed. Assume there is no volume contraction upon mixing.

Here are the two mole calculations:

x = (3.00 mol/L) (2.00 L)

x = (1.50 mol/L) (4.00 L)

I hope it is obvious that you add the two answers to get the total moles.

The total volume calculation is 2.00 + 4.00 = 6.00 L.

Divide total moles by total volume to get the final answer. The answer is 2.00 M.

Using M_{1}V_{1} + M_{2}V_{2} = M_{3}V_{3}, we have this:

(3.00 mol/L) (2.00 L) + (1.50 mol/L) (4.00 L) = (x) (6.00 L)

6.00 mol + 6.00 mol = (x) (6.00 L)

x = 12.0 mol / 6.00 L = 2.00 mol/L

**Example #7:** Calculate the final concentration if 2.00 L of 3.00 M NaCl, 4.00 L of 1.50 M NaCl and 4.00 L of water are mixed. Assume there is no volume contraction upon mixing.

The solution to this problem is almost exactly the same as 10a. The only "problem child" appears to be the 4.00 L of water. Hint: the water contributes to the final volume, but NOT to the total moles. The ChemTeam gets a final answer of 1.20 M in this problem.

Using M_{1}V_{1} + M_{2}V_{2} = M_{3}V_{3}, we have this:

(3.00 mol/L) (2.00 L) + (1.50 mol/L) (4.00 L) = (x) (10.0 L)

6.00 mol + 6.00 mol = (x) (10.0 L)

x = 12.0 mol / 10.0 L = 1.20 M

**Example #8:** What final volume should you dilute 51.0 mL of a 4.05 M KI solution so that 22.0 mL of the diluted solution contains 3.10 g of KI

**Solution:**

1) Let's determine the molarity of 22.0 mL containing 3.10 g of KI:

MV = grams / molar mass(x) (0.0220 L) = 3.10 g / 165.998 g/mol

x = 0.84886 M

2) Let's dilute to the final volume:

M_{2}V_{1}= M_{2}V_{2}(4.05 mol/L) (51.0 mL) = (0.84886 mol/L) (y)

y = 243 mL (to three sig figs)

**Example #9:** If you mix 10.00 mL of 0.200 M gold(I) nitrate with 15.00 mL of 0.180 M aluminum nitrate, what is the resulting nitrate concentration?

**Solution:**

1) Supposing complete ionization of both salts:

AuNO_{3}---> Au^{+}+ NO_{3}¯

Al(NO_{3})_{3}---> Al^{3+}+ 3NO_{3}¯(10.00 mL) x (0.200 mmol/mL AuNO

_{3}) x (1 mol NO_{3}¯ / 1 mol AuNO_{3}) = 2.00 mmol NO_{3}¯ from AuNO_{3}(15.00 mL) x (0.180 mmol/mL Al(NO

_{3})_{3}) x (3 mol NO_{3}¯ / 1 mol Al(NO_{3})_{3}) = 8.10 mmol NO_{3}¯ from Al(NO_{3})_{3}Note the alternate definition of molarity, that being mmol/mL.

2) Supposing additive volumes:

(2.00 mmol + 8.10 mmol) / (10.00 mL + 15.00 mL) = 0.404 mmol/mL = 0.404 mol/L NO_{3}¯

Go to dilution problems #1 - 10

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