1. Dr. Fred Omega Garces Chemistry 100 Miramar College
9.01c-PART2 Solutions
The Chemistry of Matter in Water
Dr. Fred Omega Garces
Chemistry 100
Miramar College

2. Dilution Process
Dilution: When the number of Solute:Solvent ratio decreases because the Volume of solvent is increased.
1Can OJ concentrated is diluted to prepare 4-can volume of OJ

3. Dilute Solutions Dilution Analysis:
Suppose 10 ml of the 3.4•10-2 M HNO3 solution is diluted to 50.0mL. What is the new concentration ?
10 ml 3.4•10-2 M HNO3 in 50 ml Volume. What is Conc?
100 mL of 3.4•10-2 M HNO3
Given:
C1 =
C2 =
V1 =
V2 =
Goal
Methodology
M1 • V1 = M2 • V2 or
C1 • V1 = C2 • V2 1

4. Dilute Solutions Dilution Analysis:
Suppose 10 ml of the 3.4•10-2 M HNO3 solution is diluted to 50.0mL. What is the new concentration ?
10 ml 3.4•10-2 M HNO3 in 50 ml Volume. What is Conc?
100 mL of 3.4•10-2 M HNO3
C1 • V1 = C2 • V2
Before Dilution:
10 mL Aliquot:
10ml of 0.034•10 M
Moles in aliquot-
Moles = 3.4•10-2 mol • L L
= •10 mol =3.4•10-4 mol
After Dilution:
50 mL Volume:
50 mL vol. contains 3.4•10-4 mol
Moles of new solution-
3.4•10-4 mol = Mnew • Vol new
= Mnew • L 6.8•10-3 M = Mnew 1

5. Dilution Example C1 • V1 = C2 • V2 C1 = ? V1 = ? C2 = ? V2 = ?
Suppose you have M sucrose stock solution. How do you prepare 250 mL of M sucrose solution ?
Dilution Example
Example (100 RS): There is a bottle of M sucrose stock solution in the lab. Give precise directions to your assistant to prepare mL of a M sucrose solution.
Concentration M Sucrose
240mL
250.0 mL of M sucrose
C1 • V1 = C2 • V2
Try to solve this Problem
C1 = ?
V1 = ?
C2 = ?
V2 = ?

6. Dilution Example in Analytical Chemistry
10.0 g MnSO4•H2O is placed in a 1-L volumetric flask. What volume is necessary to prepare 250mL of 0.050% solution? Assume dsoln 1 g/ml
1 Recognize the problem and state it clearly
Making the observation
i.e., there is a ball of fire in the sky

7. Solution Stoichiometry
How is concentration used in stoichiometry problems?

8. Solution Stoichiometry: Titration
From Previous problem:
Suppose ml of the 6.8•10-3 M HNO3 is titrated with NaOH. What volume of M NaOH is necessary to neutralize the acid solution.
1 Recognize the problem and state it clearly
Making the observation
i.e., there is a ball of fire in the sky

9. Solution Stoichiometry: Titration
From Previous problem:
Suppose ml of the 6.8•10-3 M HNO3 is titrated with NaOH. What volume of M NaOH is necessary to neutralize the acid solution.
Reaction: HNO3 (aq) + NaOH (aq)  H2O (l) + NaNO3 (aq)
6.8 e-3 M e-3M
100.0 ml V ?
F g D
Moles HNO3 = L x (6.8e-3 mol / 1 L) = 6.8 e-4 mol HNO3
D g 4
At neutralization, moles HNO3 = moles NaOH = 6.8 e-4 mol NaOH
4 g 6
Volume NaOH = 6.8 e-4 mol NaOH x (1 L / 1.00e-3 mol NaOH) = L
Answer:
Vol NaOH = mL
1 Recognize the problem and state it clearly
Making the observation
i.e., there is a ball of fire in the sky

10. Solution Stoichiometry: Titration
End of Chapter: Practice problems
8.64
8.111
1 Recognize the problem and state it clearly
Making the observation
i.e., there is a ball of fire in the sky

11. Solution Stoichiometry: Titration
End of Chapter: Practice problems
8.64
8.111
1 Recognize the problem and state it clearly
Making the observation
i.e., there is a ball of fire in the sky

12. Solution Stoichiometry: Titration
End of Chapter: Practice problems
8.64
8.111
1 Recognize the problem and state it clearly
Making the observation
i.e., there is a ball of fire in the sky

13. Solution Stoichiometry: Titration
End of Chapter: Practice problems
8.101
8.118
8.101

14. Solution Stoichiometry: Titration
End of Chapter: Practice problems
8.101
8.118
8.101

15. Solution Stoichiometry: Titration
End of Chapter: Practice problems
8.101
8.118
8.101

16. Solution at a Glance Solutions can be describe by the following:
Solvent
The component of a solution present in the greatest quantity
Solute
The component of solution present in the lesser quantity
Solution
A homogeneous mixture of two or more substances in which each substance retains its chemical identity
Concentration of a Solution
The amount of solute in a specific amount of solution.
Molarity
(M)
moles of solute
Liters of solution