1. Concentrations of Solutions
Behavior of solutions depend on compound itself and on how much is present, i.e. on the concentration.
Two solutions can contain the same compounds but behave quite different because the proportions of those compounds are different.

2. Concentrations of Solutions
Concentration of a solution: the more solute in a given volume of solvent, the more concentrated
1 tsp salt (NaCl)/cup of water vs
3 Tbsp salt/cup water

3. Units of molarity are: mol/L = M
Molarity is one way to measure the concentration of a solution.
A 1.00 molar (1.00 M) solution contains 1.00 mol solute in every 1 liter of solution.
Units of molarity are: mol/L = M
moles of solute
volume of solution in liters
Molarity (M) =

4. Preparing a 1.0 Molar Solution
One liter of a 1.00 M NaCl solution
need 1.00 mol of NaCl
weigh out 58.5 g NaCl (1.00 mole) and
add water to make 1.00 liter (total volume) of solution.
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5. Molarity Practice
What is the molar NaCl concentration if you have 0.5 mol NaCl in 1.00 L of solution?
What is the molar NaCl concentration if you have 0.5 mol NaCl in 0.50 L of solution?
0.5 mol NaCl/1.00 L = 0.5 mol/L = 0.5 M
0.5 mol NaCl/0.50 L = 0.5/0.50 mol/L = 1 mol/L = 1 M

6. Molarity Practice
What is the molar NaCl concentration if you have 10.0 g of NaCl in 1.00 L of solution?
Have grams not mols!
Grams  mol
Need molar mass
NaCl: = g/mol
So: 10.0g x (1 mol/35.45g)=0.282 mol NaCl
Molar concentration: mol/1.00 L = M

7. Molarity – Moles - Volume
moles of solute
volume of solution in liters
Molarity (M) =
mol
Volume (L)
Molarity (M) =
Have mol and vol  molarity
Have molarity and vol  mol of solute
Have molarity and mol of solute  volume
AND: mol of solute  grams of solute

8. Practice 0.10 mol HCl mol HCl = 0.10 M HCl x 2.5 L = x 2.5 L 1 L
How many moles of HCl are present in 2.5 L of 0.10 M HCl?
Given: 2.5 L of soln
0.10M HCl
Find: mol HCl
Use: mol = molarity x volume
= 0.10 mol/1 L HCl
0.10 mol HCl
mol HCl =
0.10 M HCl x 2.5 L =
x 2.5 L
1 L
= 0.25 mol HCl

9. Practice
What volume of a 0.10 M NaOH solution is needed to provide 0.50 mol of NaOH?
Given: mol NaOH
0.10 M NaOH
Find: vol soln
Use: vol soln = mol solute / molarity
= 0.10 mol NaOH / 1L
0.50 mol NaOH
0.50 mol NaOH
Vol soln = =
0.10 M NaOH
0.10 mol NaOH 1L
0.50 mol NaOH
X 1L = =
5 L
0.10 mol NaOH

10. More Practice
How many grams of CuSO4 are needed to prepare mL of 1.00 M CuSO4?
Given: mL soln
1.00 M CuSO4
Find: g CuSO4
Use: mol CuSO4 = molarity x volume
Molarity = mol / 1L
Vol = mL

11. Concentration of Solutions Interconverting Molarity, Moles, and Volume
g CuSO4 = mL soln x 1 L x 1.00 mol
1000 mL 1 L soln
x g CuSO4
1 mol
= 39.9 g CuSO4

12. Steps involved in preparing solutions from pure solids

13. Steps involved in preparing solutions from pure solids
Calculate the amount of solid required
Weigh out the solid
Place in an appropriate volumetric flask
Fill flask about half full with water and mix.
Fill to the mark with water and invert to mix.
You should be able to describe this process (including calculating the mass of solid to use) for any solution I specify.

14. Dilutions
Many laboratory chemicals such as acids are purchased as concentrated solutions (stock solutions).
e.g. 12 M HCl
12 M H2SO4
More dilute solutions are prepared by taking a certain quantity of the stock solution and diluting it with water.

15. Dilutions
A given volume of a stock solution contains a specific number of moles of solute.
e.g.: 25 mL of 6.0 M HCl contains 0.15 mol HCl
(How do you know this???)
If 25 mL of 6.0 M HCl is diluted with 25 mL of water, the number of moles of HCl present does not change.
Still contains 0.15 mol HCl
25 mL HCl
25 mL H2O
0.15 mol
50 mL
0.15 mol + =

16. moles solute = moles solute
Dilutions
moles solute = moles solute
before dilution after dilution
Although the number of moles of solute does not change, the volume of solution does change.
The concentration of the solution will change since
moles solute
Molarity =
Volume of solution

17. Dilution Calculation Mc x Vc = Md x Vd
When a solution is diluted, the concentration of the new solution can be found using:
Mc x Vc = Md x Vd
where Mc= initial concentration (mol/L)= more concentrated
Vc = initial volume of more conc. solution
Md = final concentration (mol/L) in dilution
Vd = final volume of diluted solution

18. Dilution Calculation Find: Md Use
What is the concentration of a solution prepared by diluting 25.0 mL of 6.00 M HCl to a total volume of 50.0 mL?
Given: Vc = 25.0 mL
Mc = 6.00 M
Vd = 50.0 mL
Find: Md
Use
Note: Vcand Vd do not have to be in liters, but they must be in the same units.
Vcx Mc= Vdx Md
Solve for Md

19. Dilution Make a diluted solution once you know Vc and Vd
Use a pipet to deliver a volume of the concentrated solution to a new volumetric flask.
Add solvent to the line on the neck of the new flask.
Mix well.

20. Practice
How many mL of 5.0 M K2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution?
If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting solution?
Vc = ?
Mc = 5.0M
Vd = 250 mL
Md = 0.10M
Md = ?
Vc = 10.0 mL
Mc = 10.0M
Vd = 250 mL

21. Solution Stoichiometry
Remember: reactions occur on a mole to mole basis.
For pure reactants, we measure reactants using mass
For reactants that are added to a reaction as aqueous solutions, we measure the reactants using volume of solution.

22. Solution Stoichiometry
grams A
Molar
mass
moles A
Molarity A
Vol
Soln A
Molar ratio
Molar
mass
grams B
moles B
Molarity B
Vol Soln B

23. Solution Stoichiometry Practice
If 25.0 mL of 2.5 M NaOH are needed to neutralize (i.e. react completely with) a solution of H3PO4, how many moles of H3PO4 were present in the solution?
Given: mL 2.5 M NaOH
balanced eqn: 3 mol NaOH/1 mol H3PO4
Find: moles of H3PO4
3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)

24. Approach moles NaOH Vol NaOH Soln moles H3PO4 Molarity NaOH
2.5 M (=mol/L)
Molar ratio
0.025 L NaOH soln
3 mol NaOH/1 mol H3PO4
moles
H3PO4
25.0 mL NaOH soln 1L
2.5 mol
Mol NaOH = 25.0 mL x x
= mol NaOH
1000 mL
1 L

25. More practice
What mass of aluminum hydroxide is needed to neutralize 12.5 mL of 0.50 M sulfuric acid?

26. Solution Stoichiometry
Solution stoichiometry can be used to determine the concentration of aqueous solutions used in reactions.
Concentration of an acid can be determined using a process called titration.
reacting a known volume of the acid with a known volume of a standard base solution (i.e. a base whose concentration is known)

27. Titration

28. Practice
If mL of 2.5 M NaOH are needed to neutralize 50.0 mL of an H3PO4 solution, what is the concentration (molarity) of the H3PO4 solution?
Given: mL 2.5 M NaOH
50.0 mL of H3PO4 sol’n
Find: molarity (mol/L) H3PO4
3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)

29. Strategy: M = moles L
To find the concentration of H3PO4 soln, we need both # moles and volume of H3PO4.
Since volume is given, we can simply find moles and plug into the equation for M.

30. Plan
Molarity
NaOH
moles
NaOH
Vol
NaOH Soln
Molar ratio
moles
H3PO4

31. Mol H3PO4 = mL x 1 L
1000 mL
x 2.50 mol NaOH x 1 mol H3PO4
1 L mol NaOH
= mol H3PO4
We’re not done….we need molarity.

32. Molarity of H3PO4 Molarity = moles L = 0.0296 mol H3PO4 x 1000 mL
50.0 mL L
= M H3PO4