1. Acid/Base Indicators Substance that changes color in the presence of an acid or a base – Red or Blue Litmus – Phenolphthalein (phth) – Bromothymol blue (blue food coloring) – Red cabbage juice Episode 1102

2. Strong acids: – Dissociate completely in water – Ex: HCl Weak Acids: – Dissociate partially in water – Ex: HC 2 H 3 O 2 or vinegar Episode 1102

3. Strong bases: – Dissociate completely in water – Ex: NaOH Weak Bases: – Dissociate partially in water – Ex: NH 4 OH or ammonia solution Episode 1102

4. Acid/Base Concentration pH = - log [H+] 0 -----ACID-----7-----BASE-----14 Episode 1102

5. Determine the pH of a solution of HCl that has a molarity of 1 x 10 -4 M. HCl is a strong acid- know it completely dissociates. HCl  H + + Cl - 1 mole HCl = 1 mol H + 1 x 10 -4 M HCl = 1 x10 -4 M H + pH = -log (1x 10 -4 ) pH = 4 Episode 1102

6. Calculate the pH for a solution of HNO 3 with a molarity of 1 x 10 -3 M. Strong acid – completely dissociates HNO 3  H + + NO 3 - 1 mol HNO 3 = 1 mol H + 1 x 10 -3 M HNO 3 = 1 x 10 -3 M H + pH = -log (1 x 10 -3 ) pH = 3 Episode 1102

7. Calculate the pH for a solution of H 2 SO 4 with a molarity of 1 x 10 -4 M. Strong acid – completely dissociates H 2 SO 4  2H + + SO 4 -2 1 mol H 2 SO 4 = 2 moles H + 1 x 10 -4 M H 2 SO 4 = 2(1 x 10 -4 M H + ) 2(1 x 10 -4 M H + ) = 2 x 10 -4 M H + pH = -log (2 x 10 -4 ) pH = 3.7 Episode 1102

8. Self Ionization of Water [H+][OH-] = 1 x 10 -14 Origin of pH scale – (-log[H+]) + (-log[OH-]) = -log (1 x 10 -14 ) – pH + pOH = 14 Episode 1102

9. Calculate the pH of a solution of NaOH with a molarity of 3.0 x 10 -2 M. Notice NaOH – acid or base? Strong bases – completely dissociates NaOH  Na + + OH - 1 mol NaOH = 1 mol OH - 3.0 x 10 -2 M NaOH = 3.0 x 10 -2 M OH - [H + ][OH - ] = 1 x 10 -14 [H + ](3.0 x 10 -2 ) = 1 x 10 -14 [H + ] = 3.3 x 10 -13 pH = -log [H + ] pH = -log (3.3 x 10 -13 ) pH = 12.5 Episode 1102

10. Find the pH for a solution of Ca(OH) 2 with a molarity of 1 x 10 -4 M. Notice Ca(OH) 2 – acid or base? Strong bases – completely dissociates Ca(OH) 2  Ca +2 + 2OH - 1 mol Ca(OH) 2 = 2 mol OH - 1 x 10 -4 M Ca(OH) 2 = 2(1 x 10 -4 M OH - ) [OH-] = 2 x 10 -4 M OH - [H + ][OH - ] = 1 x 10 -14 [H + ](2 x 10 -4 ) = 1 x 10 -14 [H + ] = 5 x 10 -11 pH = -log [H + ] pH = -log (5 x 10 -11 ) pH = 10.3 Episode 1102

11. Calculate both the hydrogen ion concentration and the hydroxide concentration for an aqueous solution that has a pH of 4.0. pH = -log [H + ] 4.0 = -log [H + ] Log is a function of the number 10, so …  -4.0 = log [H + ]  10 -4 = [H + ] or 1 x 10 -4 M [H + ]  [H + ][OH - ] = 1 x 10 -14  (1 x 10 -4 )[OH - ] = 1 x 10 -14  [OH - ] = 1 x 10 -10 M Episode 1102