1. Solutions in which water is the dissolving medium
Aqueous Solution
Solutions in which water is the dissolving medium

2. Solutions are defined as homogeneous mixtures of two or more pure substances.
The solvent is present in greatest abundance.
All other substances are solutes.

3. Dissociation NaCl (s) → Na+ (aq) + Cl-(aq)
When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them.
This process is called dissociation.
NaCl (s) → Na+ (aq) + Cl-(aq)

4. An electrolyte is a substances that dissociates into ions when dissolved in water. e.g. ionic compunds
A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.
e.g. molecular compounds.

5. A strong electrolyte dissociates completely when dissolved in water.
A weak electrolyte only dissociates partially when dissolved in water

6. Strong Electrolytes Are…
Soluble ionic compounds
Strong acids
Strong bases

7. Concentration of aqueous solutions
The amount of solute dissolved in a certain volume of water.
The concentration of an aqueous solution is often expressed in terms of Molarity .
moles of solute
volume of solution in liters
Molarity (M) =

8. How many grams of sodium sulfate are required to make 0. 350 L of 0
How many grams of sodium sulfate are required to make L of M Na2SO4?

9. To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute.
The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

10. Procedure for preparation of 0.25 L of 1.00 M solution of CuSO4
1. Find moles: M × V
2. Convert moles to grams

11. Dilution One can also dilute a more concentrated solution by
Using a pipet to deliver a volume of the solution to a new volumetric flask, and
Adding solvent to the line on the neck of the new flask.

13. Dilution
The molarity of the new solution can be determined from the equation
Mc  Vc = Md  Vd
Where Mc and Md are the molarity of the concentrated and dilute solutions, respectively, and Vc and Vd are the volumes of the two solutions.

14. To prepare 250 ml of 0.10 M CuSO4 by diluting a stock solution containing 1.0 M CuSO4. Calculate the volume of the more concentrated solution that must be diluted.
How many milliliters of 3.0 M H2SO4 are required to make 450 mL of 0.10 M H2SO4?

15. Acids
Arrhenius defined acids as substances that increase the concentration of H+ when dissolved in water.
Brønsted and Lowry defined them as proton donors.

16. Acids There are only seven strong acids: Hydrochloric (HCl)
Hydrobromic (HBr)
Hydroiodic (HI)
Nitric (HNO3)
Sulfuric (H2SO4)
Chloric (HClO3)
Perchloric (HClO4)

17. Bases
Arrhenius defined bases as substances that increase the concentration of OH− when dissolved in water.
Brønsted and Lowry defined them as proton acceptors.

18. Bases Alkali metals Calcium Strontium Barium
The strong bases are the soluble metal salts of hydroxide ion:
Alkali metals
Calcium
Strontium
Barium

19. Acid and Base Strength
Strong acids are completely dissociated in water.
Weak acids only dissociate partially in water.

20. Acid-Base Reactions
In an acid-base reaction, the acid donates a proton (H+) to the base.

21. Neutralization Reactions
Generally, when solutions of an acid and a base are combined, the products are a salt and water.
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

22. HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
When a strong acid reacts with a strong base, the net ionic equation is…
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq) 
Na+ (aq) + Cl- (aq) + H2O (l)
H+ (aq) + OH- (aq)  H2O (l)

23. Using Molarities in Stoichiometric Calculations

24. How many moles of water form when 25. 0 mL of 1
How many moles of water form when 25.0 mL of 1.00 M HNO3 is completely neutralized by NaOH?
How many moles of NaOH are needed to neutralize 20.0mL of 0.150M H2SO4 solution?

25. Titration
In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).
© 2009, Prentice-Hall, Inc.

26. Titration is a process in which:
A solution of known concentration (a standard solution) reacts with a solution of unknown concentration.
Titration is used to determine the concentration of the unknown.
The point at which stoichiometrically equivalent quantities reacted together is called the equivalence point.
A pH meter, conductivity meter, or an indicator( by color change) can be used to show the end point of the titration.

27. Titration
A pH meter, conductivity meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

28. HCl (aq) + NaOH (aq) → NaCl (aq) + H20 (l)
1. Place HCl (20.00 mL) in a flask 2. Fill the burette with standard NaOH (1.00 M) 3. Add 2-3 drops of indicator (e.g. phenolphthalein which is colorless in acidic solutions). 4. Add NaOH from burette dropwise until solution changes color to pale pink; this is called end point.

29. 5. Titration is stopped at the end point; titration volume is recorded
5. Titration is stopped at the end point; titration volume is recorded. At the equivalence point: # of moles NaOH = # mole of HCl Vbase * Mbase = Vacid * Macid(unknow

30. Normality equivalents of solute Volume of solution (Litres)
An equivalent is defined according to the reaction in Acid-Base reactions:
An equivalent of an acid is the quantity that supplies 1 mol of H+ while an equivalent of base is the quantity reacting with 1 mol of H+
e.g M H2SO4 = 2N, 1M HCl= 1N
1 M Al(OH)3=3N, 1M NaOH=1N