1.  Strong Acids- do dissociate completely  Weak Acids – do not dissociate completely The strong acids are: HNO 3 H 2 SO 4 HClHIO 4 HClO 4 HBr HClO 3 HI

2.  HCl  H + + Cl - ◦ Single arrow in the equation because all of the HCl particles dissociate, this contributes large amounts of H + to the solution  HC 2 H 3 O 2  H + + C 2 H 3 O 2 - ◦ Double arrow, very few of the HC 2 H 3 O 2 particles fall apart, contributing a smaller amount of H + to the solution

3. Acids - donate hydrogen ions (protons) Bases - accept hydrogen ions (protons) NH 3 + H 2 O  NH 4 + + OH - W hen you look at the reaction from left to right, NH 3 is a base because it is accepting a hydrogen, and H 2 O is an acid because it is donating a hydrogen.

4. NH 3 + H 2 O  NH 4 + + OH -  Reading the reaction from right to left, NH 4 + is an acid because it is donating a hydrogen and OH - is a base because it is accepting a hydrogen.  These are conjugate acid - base pairs.

5. NH 3 + H 2 O  NH 4 + + OH - Base Acid Conjugate Conjugate Acid Base Water can act as an acid or a base, it is amphoteric

6. Weak acids have a Ka = acid dissociation constant  This is used to compare the strength of weak acids  HC 2 H 3 O 2 Ka = 1.8 x 10 -5  HF Ka = 7.2 x 10 -4  HC 2 H 3 O 2 is a weaker acids because Ka is smaller

7. Water falls apart to a small extent H 2 O  H + + OH -  [H + ] = 1.0 x 10 -7 M  [OH - ] = 1.0 x 10 -7 M  K w = [H + ] [OH - ]  K w = 1.0 x 10 -14  To calculate [H + ] use [H + ] =1.0 x 10 -7 M/[OH - ]  To calculate [ OH - ] use [ OH - ] =1.0 x 10 -7 M/[H + ]

8. pH = -log[H + ] pH + pOH = 14 pOH =-log[OH - ] Example : [H + ] = 1.0 x 10 -4 M pH = - log (1.0 x 10 -4 ) pH = 4 pOH = 14 - 4 =10

9.  The opposite of log is antilog, this function is located above log on your calculator.  [H + ] = 10 ^ -pH  [OH - ] = 10 ^ -pOH  pH = 6  10 ^-6 = 1.0 x 10 -6