Dilution: Definition and Ten Examples

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Problems #1 - 10

Problems #11 - 25

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:

M1V1 = M2V2

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 NaNO3 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:

M1V1 + M2V2 = M3V3

Where the '1' refers to one starting solution, '2' to the other starting solution and '3' refers to the mixed solution (hints: V3 is the total volume after mixing and M3 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 M1V1 + M2V2 = M3V3, 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 M1V1 + M2V2 = M3V3, 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:

M2V1 = M2V2

(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:

AuNO3 ---> Au+ + NO3¯
Al(NO3)3 ---> Al3+ + 3NO3¯

(10.00 mL) x (0.200 mmol/mL AuNO3) x (1 mol NO3¯ / 1 mol AuNO3) = 2.00 mmol NO3¯ from AuNO3

(15.00 mL) x (0.180 mmol/mL Al(NO3)3) x (3 mol NO3¯ / 1 mol Al(NO3)3) = 8.10 mmol NO3¯ from Al(NO3)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 NO3¯

Example #10: How many mL do you have to add to 48.3 mL of 11.6 mg/dL glucose solution to obtain 3.19 mg/dL glucose?

Solution:

1) Determine volume of new solution:

C1V1 = C2V2 <--- where C stands for a non-molarity concentration

(48.3 mL) (11.6 mg/dL) = (3.19 mg/dL) (V2)

V2 = (48.3 mL x 11.6 mg/dL) / (3.19 mg/dL) = 175.6 mL total

2) Supposing additive volumes:

(175.6 mL total) − (48.3 mL initially) = 127.3 mL to be added

Bonus Example: 1.00 L each of two aqueous solutions of sucrose, C12H22O11, are mixed:

(A) One solution is 0.1487 M and has a density of 1.018 g/mL

(B) The other solution is 10.00%(w/w) and has a density of 1.038 g/mL.

Calculate the mole percent of sucrose in the solution that results from the mixing. (Note that mole percent is different from mole fraction.)

Solution:

1) We need to know the moles of solute and the moles of solvent for each solution.

(A) This solution has a concentration of 0.1487 mole/L and 1.00 L of it weighs 1018 g.
(0.1487 mol/L) (1.00 L) = 0.1487 mol (moles of solute)

(0.1487 mol) (342.2948 g/mol) = 50.9 g (mass of solute)

1018 g − 50.9 g = 967.1 g (mass of solvent)

967.1 g / 18.015 g/mol = 53.683 mol (moles of solvent)

(B) 1.00 L of the solution weighs 1038 g. 10% by mass is solute.

(1038 g) (0.1000) = 103.8 g (mass of solute)

103.8 g / 342.2948 g/mol = 0.30325 mol (moles of solute)

1038 − 103.8 = 934.2 g (mass of solvent)

934.2 g / 18.015 g/mol = 51.8568 mol (moles of solvent)

2) Calculate the mole fraction of the sucrose first.

total moles sucrose ---> 0.1487 + 0.30325 = 0.45195 mol

total moles ---> 53.683 + 51.8568 + 0.1487 + 0.30325 = 105.99175 mol

mole fraction ---> 0.45195 mol / 105.99175 mol = 0.004264

3) Mole percent of C12H22O11:

(0.004264) (100) = 0.4264%

Problems #1 - 10

Problems #11 - 25

Problems #26 - 35

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