1. Strong Acids and Bases -- these are strong electro- lytes that exist entirely as ions in aqueous solution -- memorize the names and formulas of the seven strong acids......and the eight strong, hydroxide bases... the hydroxides of... “strong base cations” Li, Na, K, Rb, Cs, Ca, Sr, Ba hydrochloric, HCl hydrobromic, HBr hydroiodic, HI chloric, HClO 3 perchloric, HClO 4 nitric, HNO 3 sulfuric, H 2 SO 4

2. Strong electrolytes are often written using a one-sided arrow. e.g., HNO 3 (aq) H + (aq) + NO 3 – (aq) Find the pH of a 0.012 M perchloric acid solution. HClO 4 H + + ClO 4 – init. [ ] final [ ] 0.012 ~0 0 0.012 0 o o o o o p p p p p pH = –log [H + ] = –log (0.012) = 1.92 (“Who cares?”)

3. Equipment for the production of nitric acid, formerly part of a U.S. military munitions plant in Joliet, IL. If a nitric acid solution has pH = 2.66, find the solution’s concentration. [H + ] = 10 –pH = 10 –2.66 = 2.2 x 10 –3 M HNO 3 H + + NO 3 – So orig. [HNO 3 ] was also 2.2 x 10 –3 M.

4. ** NOTE: Ca(OH) 2, Sr(OH) 2, and Ba(OH) 2 are strong electrolytes, but have limited solubilities. This means that if you put a big chunk of any of these into water, little of the chunk will dissolve, BUT the little that does dissolve will float around as X 2+ (aq) and OH – (aq), NOT as X(OH) 2 (aq). Barium hydroxide is used in the titration of weak acids, particularly organic acids, because a clear solution of Ba(OH) 2 – unlike NaOH or KOH – will NOT contain any carbonates; carbonates would introduce errors into an analytical titration because the CO 3 2– ion acts as a weak base.

5. pH = 11.83 Find the pH of a 0.0034 M solution of calcium hydroxide. Ca(OH) 2 Ca 2+ + 2 OH – init. [ ] final [ ] 0.0034 ~0 0 0.0034 0 0.0068 pOH = –log [OH – ] = –log (0.0068) = 2.17 (“w.c?”) pH + pOH = 14 2.17

6. Metal oxides, hydrides, and nitrides (containing O 2–, H –, and N 3–, respectively) also react with water to produce strongly basic solutions. e.g., --O 2– (aq) + H 2 O(l) 2 OH – (aq) --H – (aq) + H 2 O(l) H 2 (g) + OH – (aq) --N 3– (aq) + 3 H 2 O(l) NH 3 (aq) + 3 OH – (aq) OH!