1. Unit: Acids, Bases, and Solutions
Day 2 - Notes
Unit: Acids, Bases, and Solutions
Units of Concentration: Molarity

2. After today you will be able to…
Define molarity in terms of its mathematical formula
Calculate moles, liters, or molarity of a given solution
Explain how to make a solution

3. Recall, concentration is a measure of the amount of solute dissolved in a given quantity of solvent.

4. Molarity
In chemistry, the most important unit of concentration is molarity.
Molarity: (M) is the number of moles of solute dissolved in one liter of solution.

5. Molarity
Molarity=
Moles of solute
Liters of solution
Important: The volume is the total volume of resulting solution, not the solvent alone.

6. Steps for making a 0.5M solution
Add 0.5mol of solute to a 1.0L volumetric flask half-filled with distilled water

7. Steps for making a 0.5M solution
2. Swirl the flask to dissolve the solute. 3. Fill the flask to the mark etched on the side of the flask (1-L mark).

8. Performing Calculations With Molarity
Example: What is the molarity of a solution that contains 0.25 moles of NaCl in 0.75L of solution? M=
mol= L= ?
0.25 mol NaCl
0.75L
0.25 mol NaCl M =
0.75L
Molarity =
0.33 mol/L or M

9. Performing Calculations With Molarity
Example: What volume of a 1.08M KI solution would contain moles of KI? M=
mol= L=
1.08 M
0.642 mol KI ?
0.642 mol KI
1.08 M = L
Volume =
0.594L

10. Performing Calculations With Molarity
Example: How many grams of CaBr2 are dissolved in 0.455L of a 0.39M CaBr2 solution? M=
mol= L=
1 Ca=40.08
0.39 M CaBr2
2 Br=79.90(2)
mol
199.88g
0.39 M = ?
0.455L
0.455L
mol =
0.18 mol CaBr2
199.88gCaBr2
0.18 mol CaBr2 x =
35g CaBr2
1 mol CaBr2

11. Molality Molality (m) = moles solute/kg of solvent molality  molarity
Example problem: What is the molality of a solution with 0.70 mol NaCl and 2.2 kg of water?
m = mol solute = 0.70 mol
kg solvent 2.2 kg
answer = 0.32 m solution

12. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (1)
after dilution (2) =
M1V1
M2V2 =

13. How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
M1V1 = M2V2
M1 = 4.00
M2 = 0.200
V2 = 0.06 L
V1 = ? L
V1 =
M2V2 M1 =
0.200 x 0.06
4.00
= L = 3 mL
3 mL of acid
+ 57 mL of water
= 60 mL of solution

14. Another Dilution Problem
If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mL, what would be the resulting concentration?
M1*V1 = M2*V2
(6.5M) * (32 mL) = M2 * (500.0 mL)
6.5 M * 32 mL
M2 =
500 mL
M2 = M

15. moles of solute before dilution = moles of solute after dilution
Concentration of Solutions
Dilution is the process of preparing a less concentrated solution from a more concentrated one.
moles of solute before dilution = moles of solute after dilution

16. (2.00 M CuCl2)(Lc) = (0.100 M CuCl2)(0.2500 L)
Concentration of Solutions
In an experiment, a student needs mL of a M CuCl2 solution. A stock solution of 2.00 M CuCl2 is available.
How much of the stock solution is needed?
Solution: Use the relationship that moles of solute before dilution = moles of solute after dilution.
(2.00 M CuCl2)(Lc) = (0.100 M CuCl2)( L)
Lc = L or 12.5 mL
To make the solution:
Pipet 12.5 mL of stock solution into a mL volumetric flask.
Carefully dilute to the calibration mark.
Mc × Lc = Md × Ld

17. Mc × mLc = Md × mLd Concentration of Solutions
Because most volumes measured in the laboratory are in milliliters rather than liters, it is worth pointing out that the equation can be written as
Mc × mLc = Md × mLd