1. Chapter 12 Solutions
12.5
Molarity and Dilution

2. Molarity (M) Molarity (M) is a concentration term for solutions
the moles of solute in 1 L of solution
moles of solute
liter of solution

3. Preparing a 6.0 M Solution A 6.00 M NaOH solution is prepared
by weighing out 60.0 g of NaOH (1.50 mol) and
adding water to make L of a 6.00 MNaOH solution

4. Preparation of Solutions

5. Preparation of Solutions
A solution of a desired concentration can be prepared by diluting a small volume of a more-concentrated solution, a stock solution, with additional solvent.
– Calculate the number of moles of solute desired in the final volume of the more-dilute solution and then calculate the volume of the stock solution that contains the amount of solute.
– Diluting a given quantity of stock solution with solvent does not change the number of moles of solute present.
– The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is
(Vs) (M s) = moles of solute = (Vd) (M d).

6. Preparation of Solutions

7. Calculating Molarity

8. Example of Calculating Molarity
What is the molarity of L of a NaOH solution if
it contains 6.00 g of NaOH?
STEP 1 State the given and needed quantities.
Given g of NaOH in L of solution
Need molarity (M)
STEP 2 Write a plan to calculate molarity.
molarity (M) = moles of solute
liters of solution
grams of NaOH moles of NaOH molarity

9. Example of Calculating of Molarity (continued)
STEP 3 Write equalities and conversion factors needed.
1 mol of NaOH = g of NaOH
1 mol NaOH and g NaOH
40.01 g NaOH mol NaOH
STEP 4 Set up problem to calculate molarity.
6.00 g NaOH x 1 mol NaOH = mol of NaOH
40.01 g NaOH
0.150 mol NaOH = mol
0.500 L solution 1 L
= M NaOH solution

10. Learning Check
What is the molarity of 325 mL of a solution containing g of NaHCO3?
A M NaHCO3 solution
B M NaHCO3 solution
C M NaHCO3 solution

11. Solution STEP 1 State the given and needed quantities.
Given g of NaHCO3 in L of solution
Need molarity (M)
STEP 2 Write a plan to calculate molarity.
molarity (M) = moles of solute
liters of solution
grams of NaHCO moles of NaHCO molarity

12. Solution (continued)
STEP 3 Write equalities and conversion factors needed.
1 mol of NaHCO3 = g of NaHCO3
1 mol NaHCO3 and g NaHCO3
84.01 g NaOH mol NaHCO3
STEP 4 Set up problem to calculate molarity.
46.8 g NaHCO3 x 1 mol NaHCO3 = mol of NaHCO g NaHCO3
0.557 mol NaHCO3 = M NaHCO3 solution
0.325 L

13. Molarity Conversion Factors
The units of molarity are used as conversion factors in calculations with solutions.

14. Example of Using Molarity in Calculations
How many grams of KCl are needed to prepare
0.125 L of a M KCl solution?
STEP 1 State the given and needed quantities.
Given L of a M KCl solution
Need grams of KCl
STEP 2 Write a plan to calculate mass or volume.
liters of KCl solution moles of KCl grams of KCl

15. Example of Using Molarity in Calculations (continued)
STEP 3 Write equalities and conversion factors needed.
1 mol of KCl = g of KCl
1 mol KCl and g KCl
74.55 g KCl mol KCl
1 L of KCl solution = mol of KCl
1 L KCl solution and mol KCl
0.720 mol KCl L KCl solution

16. Example of Using Molarity in Calculations (continued)
STEP 4 Set up problem to calculate mass or volume.
0.125 L x mol KCl x g KCl = 6.71 g of KCl
1 L mol KCl

17. Learning Check How many grams of AlCl3 are needed to prepare
37.8 mL of a M AlCl3 solution?
A g of AlCl3
B g of AlCl3
C g of AlCl3

18. Solution STEP 1 State the given and needed quantities.
Given mL of a M AlCl3 solution
Need grams of AlCl3
STEP 2 Write a plan to calculate mass or volume.
milliliters of AlCl3 solution liters of AlCl3 solution
moles of AlCl grams of AlCl3

19. Solution (continued)
STEP 3 Write equalities and conversion factors needed.
1 mol of AlCl3 = g of AlCl3
1 mol AlCl and g AlCl3
g AlCl mol AlCl3
1000 mL of AlCl3 solution = 1 L of AlCl3 solution
1000 mL AlCl3 solution and L AlCl3 solution
1 L AlCl3 solution mL AlCl3 solution
1 L of AlCl3 solution = mol of AlCl3
1 L AlCl3 solution and mol AlCl3
0.150 mol AlCl L AlCl3 solution

20. Solution (continued)
STEP 4 Set up problem to calculate mass or volume.
37.8 mL x L x mol x g
1000 mL L mol
= g of AlCl3 (B)

21. Learning Check How many milliliters of a 2.00 M HNO3 solution
contain 24.0 g of HNO3?
A mL of HNO3 solution
B mL of HNO3 solution
C mL of HNO3 solution

22. Solution STEP 1 State the given and needed quantities.
Given g of HNO3; 2.00 M HNO3 solution
Need milliliters of HNO3 solution
STEP 2 Write a plan to calculate mass or volume.
g of solution moles of HNO mL of HNO3

23. Solution (continued)
STEP 3 Write equalities and conversion factors needed.
1 mol of HNO3 = g of HNO3
1 mol HNO and g HNO3
63.02 g HNO mol HNO3
1000 mL of HNO3 = mol of HNO3
1000 mL HNO3 and mol HNO3
2.00 mol HNO mL HNO3

24. Solution (continued)
STEP 4 Set up problem to calculate mass or volume.
24.0 g HNO3 x 1 mol HNO3 x mL
63.02 g HNO mol HNO3
= mL of HNO3 solution (C)

25. Dilution In a dilution, water is added volume increases
concentration decreases

26. NO3-
Na+
Moles = 1.0
Volume = 1.0 L
Molarity = 1.0 M
NaNO3 solution

27. Solution volume is doubled Moles of solute remain the same
NO3-
Na+
Moles = 1.0
Volume = 2.0 L
Molarity = 0.50 M
Solution volume is doubled
Moles of solute remain the same
Solution concentration is halved

28. Comparing Initial and Diluted Solutions
In the initial and diluted solution,
the moles of solute are the same
the concentrations and volumes are related by the equation
M1V1 = M2V2
initial diluted

29. Calculating Dilution Quantities

30. Example of Dilution Calculations
What is the final molarity of the solution when
0.180 L of M KOH is diluted to L?
STEP 1 Prepare a table of the initial and diluted volumes and concentrations.
Initial Solution Diluted Solution
M1 = M M2 = ?
V1 = L V2 = L

31. Example of Dilution Calculations (continued)
STEP 2 Solve the dilution expression for the unknown quantity.
M1V1 = M2V2
V V2
M = M1V1 V2
STEP 3 Set up the problem by placing known quantities in the dilution expression.
M2 = M1V1 = (0.600 M)(0.180 L) = M
V L

32. Learning Check What is the final volume, in milliliters, if 15.0 mL
of a 1.80 M KOH solution is diluted to give a
0.300 M KOH solution?
A mL of M KOH solution
B mL of M KOH solution
C mL of M KOH solution

33. Solution
STEP 1 Prepare a table of the initial and diluted volumes and concentrations.
Initial Solution Diluted Solution
M1= M V1 = mL
M2= M V2 = ?

34. Solution (continued)
STEP 2 Solve the dilution expression for the unknown quantity.
M1V1 = M2V2
M M2
V2 = M1V1 M2
STEP 3 Set up the problem by placing known quantities in the dilution expression.
V2 = M1V1 = (1.80 M)(15.0 mL)
M M = mL (C )