Solution and concentration

1. Solution Concentration

2. Review
A solution is a homogeneous mixture.
The solvent is the major component of the
solution.
The solute is the minor component and
active ingredient.
A saturated solution holds the maximum
amount of solute that is theoretically
possible for a given temperature.

3. How would you describe this picture?

4. Solution Concentration
Is one glass of tea stronger than the other?
– What’s true about the “stronger” glass of tea?
– How much tea does it have in it compared to
the other glass?

5. Solution Concentration
Concentration – a ratio comparing the
amount of solute to the amount of solution.
Many ways of expressing concentration:
– Molarity (M) is the one we will be dealing
with

6. Concentrated vs. Dilute
The words “concentrated” and “dilute” are
opposites.
EX: The dark tea is more concentrated than
the light tea.
EX: The light tea is more dilute than the
dark tea.

7. Concentrated vs. Dilute
Concentrated
solution
Dilute
solution
= solute particles

8. Molarity
Molarity (M)
– UNITS: mol/L or Molar (M)
– Example: 0.500 mol/L = 0.500 M
solution
of
Liters
solute
of
moles
Molarity 

9. Molarity
 What is the Molar concentration of a sol’n if 20.0 grams
of KNO3 (MM = 101.11 g/mol) is dissolved in enough
water to make 800. mL?
– Convert g of KNO3 to mol of KNO3
– Convert mL to L
3
3
3
3 KNO
mol
0.198
KNO
g
101.11
KNO
mol
1
x
KNO
g
20.0 
L
0.800
mL
1
L
0.001
x
mL
800. 

10. Molarity
What is the Molar concentration of a sol’n if
0.198 mol KNO3 is dissolved in enough water
to make 0.800 L?
M
0.248
L
0.800
KNO
mol
0.198
Molarity 3



11. Molarity
What is the Molar concentration of a sol’n if 10.5 grams of
glucose (MM = 180.18 g/mol) is dissolved in enough water
to make 20.0 mL of sol’n?
– Convert g of glucose to mol of glucose.
– Convert mL to L.
glucose
mol
0.0583
glucose
g
180.18
glucose
mol
1
x
glucose
g
10.5 
L
0.0200
mL
1
L
0.001
x
mL
20.0 

12. Molarity
What is the Molar concentration of a sol’n if 0.0583 mol of
glucose is dissolved in enough water to make 0.0200 L of
sol’n?
M
2.92
L
0.0200
glucose
mol
0.0583
Molarity 


13. Calculating Grams
 How many grams of KI (MM = 166.00 g/mol) are needed to prepare
25.0 mL of a 0.750 M solution?
– Convert mL to L.
– Solve for moles.
– moles of KI = 0.750 M x 0.0250 L = 0.0188 mol KI
L
0.0250
mL
1
L
0.001
x
mL
25.0 
L
0.0250
KI
of
moles
x
M
0.750

14. Calculating Grams
 How many grams of KI (MM = 166.00 g/mol) are needed to prepare
25.0 mL of a 0.750 M solution?
– Convert 0.0188 mol KI to grams.
KI
g
3.12
KI
mol
1
KI
g
166.00
x
KI
mol
0.0188 

15. Calculating Grams
How many grams of HNO3 (MM = 63.02
g/mol) are present in 50.0 mL of a 1.50 M
sol’n?
– Convert mL to L.
• 50.0 mL = 0.0500 L
– Solve for moles:
• moles = (1.50 M)(0.0500 L) = 0.0750 mol HNO3
– Convert 0.0750 mol HNO3 to grams:
• 0.0750 mol HNO3 = 4.73 g HNO3

16. Dilution
Dilute (verb) - to add solvent to a solution.
– Decreases sol'n concentration.
– M1V1 = M2V2
• M1 = initial conc.
• V1 = initial volume
• M2 = final conc.
• V2 = final volume
– Assumes no solute is added.

17. Dilution
Stock Solution
Impractically
High Concentration
Usable
Solution
Add H2O
Question for Consideration: Why do you
think chemical supply companies typically sell
acids (and other solutions) in extremely high
concentrations when it would be safer to ship
more dilute solutions?

18. Dilution
 To what volume should 40.0 mL of 18 M H2SO4
be diluted if a concentration of 3.0 M is desired?
– What do we want to know?
• V2
– What do we already know?
• M1 = 18 M
• V1 = 40.0 mL
• M2 = 3.0 M
– (18 M)(40.0 mL) = (3.0 M)V2
– 720 M*mL = (3.0 M)V2
– V2 = 240 mL

19. Dilution
 You are asked to prepare 500. mL of 0.250 M HCl,
starting with a 12.0 Molar stock sol'n. How much
stock should you use?
– What do we want to know?
• V1
– What do we already know?
• M1 = 12.0 M
• M2 = 0.250 M
• V2 = 500. mL
– (12.0 M) V1 = (0.250 M)(500. mL)
– (12.0 M) V1 = 125 M*mL
– V1 = 10.4 mL

20.  To how much water should you add 20.0 mL of
5.00 M HNO3 to dilute it to 1.00 M?
– What do we want to know?
• How much water to add. (V2 - V1)
– What do we already know?
• M1 = 5.00 M
• V1 = 20.0 mL
• M2 = 1.00 M
– (5.00 M)(20.0 mL) = (1.00 M) V2
– 100. M*mL = (1.00 M) V2
– V2 = 100. mL
– Water added = 100. mL - 20.0 mL = 80. mL
Dilution

21. Colligative Properties
Colligative properties are properties of
solutions that are affected by the number of
particles but not the identity of the solute