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Ch. 15 & 16 - Acids & Bases II. pH (p. 481 - 491) C. Johannesson.

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Presentation on theme: "Ch. 15 & 16 - Acids & Bases II. pH (p. 481 - 491) C. Johannesson."— Presentation transcript:

1 Ch. 15 & 16 - Acids & Bases II. pH (p ) C. Johannesson

2 A. Auto-Ionization of Water
Water molecules react with other water molecules to IONIZE (make ions). + - H2O + H2O H3O+ + OH- base acid LPChem:Wz

3 H2O + H2O H3O+ + OH- A. Ionization of Water
“Pure” water contains hydronium and hydroxide ions. Note the equilibrium arrow! Auto-ionization only happens to one out of every quarter-billion molecules, so water doesn’t contain enough ions to be an electrolyte. LPChem:Wz

4 H2O + H2O H3O+ + OH- Kw = [H3O+][OH-] = 1.0  10-14
A. Ionization of Water H2O + H2O H3O+ + OH- The product of the concentrations of H3O+ and OH- in aqueous solution will always equal the equilibrium constant, Kw. Kw = [H3O+][OH-] = 1.0  10-14 C. Johannesson

5 HCl is a strong acid, [HCl] = [H3O+]
A. Ionization of Water Find the hydroxide ion concentration of 3.0  10-2 M HCl. [H3O+][OH-] = 1.0  10-14 [3.0  10-2][OH-] = 1.0  10-14 [OH-] = 3.3  M Acidic or basic? HCl is a strong acid, [HCl] = [H3O+] [H3O+] > 1x10-7 Or [H3O+] > [OH-]  acidic Acidic C. Johannesson

6 Write answers in Scientific Notation!
A. Ionization of Water Find the hydronium ion concentration of 2.5  10-5 M NaOH. [H3O+][OH-] = 1.0  10-14 [H3O+][2.5  10-5]= 1.0  10-14 [H3O+] = 4.0  M Acidic or basic? NaOH is a strong base, [NaOH] = [OH-] Write answers in Scientific Notation! Basic [H3O+] < 1x10-7 Or , [H3O+] < [OH-] C. Johannesson

7 pouvoir hydrogène (Fr.)
B. pH Scale 14 7 INCREASING ACIDITY INCREASING BASICITY NEUTRAL pH = -log[H3O+] pouvoir hydrogène (Fr.) “hydrogen power” C. Johannesson

8 pH of Common Substances
B. pH Scale pH of Common Substances C. Johannesson

9 pH = -log[H3O+] pOH = -log[OH-] pH + pOH = 14
B. pH Scale pH = -log[H3O+] pOH = -log[OH-] pH + pOH = 14 C. Johannesson

10 B. pH Calculations [H+] pH [OH-] pOH -log[H3O+] 10^(–pH) 14 = pH + pOH
Kw = [H3O+][OH-] 14 = pH + pOH -log[OH-] [OH-] pOH 10^(–pOH) LPChem:Wz

11 B. pH Scale What is the pH of 0.050 M HNO3? pH = -log[H3O+]
strong acid  [HNO3] = [H3O+] pH = -log[H3O+] pH = -log[0.050] pH = 1.30 Acidic or basic? __  2 sig figs __  2 sig figs after the decimal Acidic pH < 7 C. Johannesson

12 B. pH Scale What is the pH of a 1x10-9 M HNO3 solution? Basic
HNO3 is a strong acid Therefore [HNO3] = [H3O+] pH = -log[H3O+] pH = -log[1x10-9] = 9.0 Basic LPChem:Wz

13 B. pH Scale What is the [H3O+] of a solution with pH = 7.45?
pH = -log[H3O+] Therefore 10^ -pH = [H3O+] 10^ = [H3O+] = 3.6 x 10-8 M LPChem:Wz

14 B. pH Scale What is the molarity of HBr in a solution that has a pOH of 9.6? Given: Find: HBr is a: [HBr] = pOH Molarity of HBr Strong Acid [H3O+] How do we solve from pOH to [H3O+] ? C. Johannesson

15 B. pH Calculations [H+] pH [OH-] pOH -log[H3O+] 10^(–pH) 14 = pH + pOH
Kw = [H3O+][OH-] 14 = pH + pOH -log[OH-] [OH-] pOH 10^(–pOH) LPChem:Wz

16 B. pH Scale 10^ -pH = [H3O+] 10^ -4.4 = [H3O+]
What is the molarity of HBr in a solution that has a pOH of 9.6? pH + pOH = 14 pH = 14 pH = 4.4 pH = -log[H3O+] or 10^ -pH = [H3O+] 10^ -4.4 = [H3O+] = 4.0  10-5 M HBr Acidic C. Johannesson

17 B. pH Scale What is the OH- concentration of a solution whose pH is 1.94? Given: Find: pH [OH-] How do we solve from pH to [OH-] ? C. Johannesson

18 B. pH Calculations [H+] pH [OH-] pOH -log[H3O+] 10^(–pH) 14 = pH + pOH
Kw = [H3O+][OH-] 14 = pH + pOH -log[OH-] [OH-] pOH 10^(–pOH) LPChem:Wz

19 B. pH Scale 10^ -pOH = [OH-] 10^ -12.06 = [OH-]
What is the OH- concentration of a solution whose pH is 1.94? pH + pOH = 14 pOH = 14 pOH = 12.06 pOH = -log[OH-] or 10^ -pOH = [OH-] 10^ = [OH-] = 8.7  M OH- Acidic LPChem:Wz

20 B. pH Scale What is the pH of a 0.055 M KOH solution? Given: [OH-]
Find: [OH-] pH How do we solve from [OH-] to pH? C. Johannesson

21 B. pH Calculations [H+] pH [OH-] pOH -log[H3O+] 10^(–pH) 14 = pH + pOH
Kw = [H3O+][OH-] 14 = pH + pOH -log[OH-] [OH-] pOH 10^(–pOH) LPChem:Wz

22 B. pH Scale pH + pOH = 14 What is the pH of a 0.055 M KOH solution?
KOH is a strong base Therefore [KOH] = [OH-] pOH = -log[OH-] pOH = -log[0.055] pOH = 1.26 pH + pOH = 14 pH = 14 pH = 12.74 Basic LPChem:Wz


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