1. Concentration

2. Concentration of Solution
Molarity
Moles of solute
Liter of solution
Mol L
(M)
Solvent Solute

3. How to mix a Standard Solution
Wash bottle
Volume marker
(calibration mark)
Weighed
amount
of solute
Use a VOLUMETRIC FLASK to make a standard solution of known concentration
Step 1> add the weighed amount of solute in the volumetric flask
Step 2> add distilled water (about half of final volume)
Step 3> cap volumetric flask, and shake to dissolve solute completely
Step 4> add distilled water to volume marker (calibration mark)
The solution process may be exothermic (release heat). This may cause the liquid to show a larger volume than is real. Allow the solution to return to ambient (room) temperature and check volume again.
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480

4. Reading a pipette 4.48 - 4.50 mL 4.86 - 4.87 mL 5.00 mL
Identify each volume to two decimal places
(values tell you how much you have expelled) mL mL
5.00 mL

5. Molarity Example Problem 1
NaCl
12.6 g of NaCl are dissolved in water making 344mL of solution. Calculate the molar concentration.

6. Molarity Example Problem 2
NaCl
How many moles of NaCl are contained in 250.mL of solution with a concentration of 1.25 M?
Volume x concentration = moles solute

7. Molarity Example Problem 3
NaCl
What volume of solution will contain 15 g of NaCl if the solution concentration is 0.75 M?
moles solute ÷ concentration = volume solution

8. Diluting Solutions C1V1 = C2V2
Often once you have made a stock solution, you need to dilute it to a working concentration.
To determine how to dilute the stock solution, use the formula:
Great! Now that you’ve made your one molar stock solution of sodium chloride, you may need to dilute it to a different concentration as a working solution. The formula C1V1 = C2V2 is used to figure out this dilution. C2 is the concentration of the new solution you want to make and V2 is the volume of that solution you want to make. C1 is the concentration of your stock solution. The question is “how much of that stock do you need to make your dilution”? That’s why you will usually be solving for V1.
C1 – concentration of stock
C2 - concentration of diluted solution
V1 – volume needed of stock
V2 – final volume of dilution
C1V1 = C2V2

9. Diluting Solutions C1 V1 = C2 V2 (5) V1 = (0.4)(100) V1= 8 ml
Example 5:
How many milliliters of a 5 M stock solution of NaCl are
needed to prepare 100 ml of a 0.4 M solution?
C1 V1 = C2 V2
(5) V1 = (0.4)(100)
V1= 8 ml
If you want to make 100 milliliters (that’s your V2) of a 0.4 M sodium chloride solution (that’s your C2). The concentration of the stock solution is 5M (that’s your C1) How much of that stock solution will you need (that’s your V1)? So to solve for V1, let’s plug our numbers in to the formula and see. Solving for V1 means we have to multiply 0.4 by 100 and divide by 5. That gives us an answer of 8 milliliters. That means we will use 8 milliliters of our stock solution. Can you guess how much water we add to that 8 mls to dilute it? That’s right! We wanted a final volume of 100 mls, so means we will add 92 mls of water to produce our final product of a 0.05 M solution of sodium chloride. Alright!

10. Dilution GIVEN: C1 = 15.8M V1 = ? C2 = 6.0M V2 = 250 mL WORK:
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
GIVEN:
C1 = 15.8M
V1 = ?
C2 = 6.0M
V2 = 250 mL
WORK:
C1 V1 = C2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3

11. 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?
C1*V1 = C2*V2
(6.5M) * (32 mL) = C2 * (500.0 mL)
6.5 M * 32 mL
C2 =
500 mL
C2 = M