1. The pH Scale

2. [H + ]pHExample Acids 1 X 10 0 0HCl 1 x 10 -1 1Stomach acid 1 x 10 -2 2Lemon juice 1 x 10 -3 3Vinegar 1 x 10 -4 4Soda 1 x 10 -5 5Rainwater 1 x 10 -6 6Milk Neutral1 x 10 -7 7Pure water Bases 1 x 10 -8 8Egg whites 1 x 10 -9 9Baking soda 1 x 10 -10 10Tums ® antacid 1 x 10 -11 11Ammonia 1 x 10 -12 12Mineral lime - Ca(OH) 2 1 x 10 -13 13Drano ® 1 x 10 -14 14NaOH

3. Practice  P.239 # 1 – 3

4. The pH Scale In acidic solutions, [H + ] > 1.0  10 -7, so pH < 7.00 pH = - log [ H + ] [ H + ] = antilog –pH

5. 1) Find the pH value of 0.0400 M HCl solution. HCl (g) + H 2 O (l) → H 3 O + (aq) + Cl - (aq) 0.04 M 0.04 M pH = -log [H 3 O + ] pH = -log [0.0400 M] pH = 1.398 2) Calculate the pH of a 0.025 mol/L solution of HNO 3(aq). HNO 3(aq) → H + (aq) + NO 3 – (aq) 0.025 mol/L pH = - log (0.025) = 1.60 pH = -log [H 3 O + ]

6. 3) Find the concentration of H 3 O + if the pH of the solution is 6.47 [ H 3 O + ] = 10 -pH [ H 3 O + ] = 10 -6.47 [ H 3 O + ] = 3.4 x 10 -7 M

7. Practice  P.242 # 4 – 8

8. The pOH Scale In basic solutions, [H+] 7.00 pOH = - log [ OH - ] [ OH _ ] = antilog –pOH

9. KOH (s) ↔ K + (aq) + OH - (aq) 0.10 mol/L pOH = -log (1.0 x 10 -1 ) pOH = 1.00 1) Calculate the pOH of a.10 mol/L KOH (aq) solution. pOH = -log (OH - )

10. 2) Calculate the pOH of a 0.050 mol/L solution of Mg(OH) 2(aq). Mg(OH) 2 (s) → Mg +2 (aq) + 2 OH - (aq) 0.050mol/L 0.10 mol/L pOH = -log (0.10 mol/L) pOH = 1.00 pOH = -log (OH - )

11. Practice  P.243 # 9 - 11 Section 6.2 Questions  P.244 # 1 - 12