Chapter 15 - Acids and Bases Behavior of Weak Acids and Bases

6) Calculate [H+] and ph for 1.0 10 8 M HCl. HCl H + + Cl - For a strong acid, [H+] = 1.0 10 8 M, ph = 8.0, BUT THIS DOES NOT MAKE SENSE!!!

THERE IS ANOTHER EQUILIBRIUM TO CONSIDER H2O + H2O H3O + + OH - KW = [H3O+][OH-] = 1.0 x 10-14 R 2 H I C 1.00 x 10-8 0 + x + x E x + 1.00 x 10-8 x IF x REPRESENTS [H+] AND [OH-] FROM AUTOIONIZATION (x + 1.00 x 10-8 ) (x) = 1.0 x 10-14 solve quadratic x 2 + (1.00 x 10-8 )(x)-1.0 x 10-14 = 0 x = 9.5 x 10-8 = [OH-] [H+] = 9.5 x 10-8 + 1.0 x 10-8 = 10.5 x 10-8 = 1.05 x 10-7 ph = 6.98

Polyprotic Acids

7) Calculate the ph of 0.034 M carbonic acid solution (Ka1 = 4.3 10 7 and Ka2 = 4.7 10 11 ). R I C E H2 0.034 0 0 0 +x +x 0.034 - x x x Ka1 = 4.3 x 10-7 = [HCO3 - ][H+] [HA] [x][x] [0.034] 1.5 x 10-8 = x 2 1.2 x 10-4 = x = [H+] 1.2 x 10-4 = x = [HCO3 - ] 0.034 = [H2CO3]

7) Calculate the ph of 0.034 M carbonic acid solution (Ka1 = 4.3 10 7 and Ka2 = 4.7 10 11 ). R I C E HCO 1.2 x 10-4 0 0 1.2 x 10-4 - x +x +x x 1.2 x 10-4 + x Ka2 = [CO3 2- ][H+] [HCO3 - ] [x][x +1.2 x 10-4 ] 4.7 x 10-11 = X [1.2 x 10-4 ] 1.2 x 10-4 [x][1.2 x 10-4 ] 4.7 x 10-11 = [1.2 x 10-4 ] 4.7 x 10-11 = x = [CO3 2- ] 1.2 x 10-4 = [HCO3 - ] ph = -log [H+] = -log (1.2 x 10-4 + 4.7 x 10-11 ) = 3.92

Polyprotic Acids

Acid Strength and Molecular Structure

Strength of Binary Acids Depends on Two Factors H? H+?- 1) How strong is the bond between H and? 2) How well? can stabilize a negative charge

Strength of Binary Acids The more polarized the H X bond is, the more acidic the bond. Acid Strength C-H < N-H < O-H << F-H Increasing Polarization of H-X Bond (Increasing Electronegativity Difference between H and X)

Strength of Binary Acids The weaker the H X bond is, the more acidic the bond. Acid Strength H-F < H-Cl < H-Br < H-I 567 431 366 299 kj/mol Increasing Strength of H-X Bond Strength

Strength of Binary Acids 6A 7A Binary acid strength increases across a period and increases down a column. H2O H2S HF HCl H2Se HBr H2Te HI

Strengths of Oxy Acids H-O-Y The more electronegative the Y atom, the stronger the acid. The more oxygens attached to Y, the stronger the acid.

Structure and Acid Strength H-O bond becomes weaker

Electronegativity and Acid Strength H-O bond becomes weaker

ph of Salt Solutions

Acid-Base Properties of Salts

1) Write balanced equilibrium equation for basic anion or acidic cation. CO 3 2 (aq) + H 2 O(l) HCO 3 (aq) + OH (aq) 2) Determine K for equilibrium (in this case, K b1 ). 3) Solve equilibrium problem for [H + ], [OH ]. HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) A (aq) + H 2 O(l) OH (aq) + HA(aq) K a K b H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) K w Kw = Ka Kb

1) Calculate the ph of a 0.22 M acetic acid solution (Ka of acetic acid = 1.8 10 5 ) R CH Ka = x 2 0.22-x = 1.8 x 10-5 I C 0.22 0.0 0.0 - x + x + x x 2 = 3.96 x 10-6 [H+] = x = 1.99 x 10-3 E 0.22 -x x x ph = 2.70

2) Calculate the ph of a 0.120 M solution of NH3 (Kb = 1.76 10 5 ). NH3 (aq) + H2O NH4 + (aq) + OH-(aq) R NH3 +H2O NH4 + + OH- Kb = x 2 0.120-x = 1.76 x 10-5 I 0.120 0.0 0.0 x 2 = 2.11 x 10-6 C - x + x + x [OH-] = x = 1.45 x 10-3 E 0.120 -x x x poh = -log(1.45 x 10 3 ) = 2.84, ph = 11.16

Acid-Base Properties of Salts (M+B-) In order to determine the acidic or basic properties of a salt, one must consider the effect of the components of a salt on the autoionization equilibrium for water: H2O + H2O H3O + + OH - As acid is added, the solution becomes more acidic. As base is added, the solution becomes more basic. As H3O + is removed, the solution becomes more basic. As OH- is removed, the solution becomes more acidic.

Acid-Base Properties of Salts What happens when you dissolve NaCl in water? Which equilibria must you consider? Na + + OH - NaOH NaCl Cl - + H3O + HCl + H2O However, neither equilibrium is important. Na + + OH - NaOH Cl - + H3O + HCl + H2O If the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, the salt will form a neutral solution.

Acid-Base Properties of Salts What happens when you dissolve NaF in water? Which equilibria must you consider? Na + + OH - NaOH NaF F - + H3O + HF + H2O In this case, only one equilibrium is important. Na + + OH - NaOH F - + H3O + HF + H2O If the salt cation is the counterion of a strong base and the anion is the conjugate base of a weak acid, the salt will form a basic solution.

Acid-Base Properties of Salts What happens when you dissolve NH4Cl in water? Which equilibria must you consider? NH4 + + OH - NH3 + H2O NH4Cl Cl - + H3O + HCl + H2O Once again, only one equilibrium is important. NH4 + + OH - NH3 + H2O Cl - + H3O + HCl + H2O If the salt cation is the counterion of a weak base and the anion is the conjugate base of a strong acid, the salt will form an acidic solution.

Acid-Base Properties of Salts What happens when you dissolve NH4F in water? Which equilibria must you consider? NH4 + + OH - NH3 + H2O NH4F F - + H3O + HF + H2O In this case, both equilibria are important. NH4 + + OH - NH3 + H2O F - + H3O + HF + H2O NH4 + is a stronger acid (Ka = 5.6 x 10-10 )than F- acting as a base (Kb = 1.5 x 10-11 )and the salt will form an acidic solution.

3) Calculate the ph of a 0.13 M of sodium acetate. Kb = Kw/Ka = 10-14 /(1.8 x 10-5 ) = 5.6 x 10-10 R CH I C 0.13 0.0 0.0 - x + x + x E 0.13- x x x [x][x] [0.13-x] [x][x] [0.13] = 5.6 10 10 = 5.6 10 10 poh = -log(8.5 10 6 ) = 5.07, ph = 8.93 x = 8.5 10 6 = [OH-]

4) Calculate the ph of a 0.120 M solution of NH4Cl NH4 + + Cl- NH3 + HCl Ka = Kw/Kb = 10-14 /(1.76 x 10-5 ) = 5.68 x 10-10 R x 2 NH4 + + H2O NH3 + H3O + Ka = 0.120-x = 5.68 x 10-10 I C E 0.120 0.0 0.0 - x + x + x 0.120 -x x x x 2 = 6.82 x 10-11 [H+] = x = 8.26 x 10-6 ph = -log(8.26 x 10 6 ) = 5.08

Acid-Base Properties of Salts If the salt cation is a highly charged metal ion, and the anion is the conjugate base of a strong acid, the salt will form an acidic solution. Al(NO3)3

Acidic Behavior of Hydrated Al3+

Ka Values of Hydrated Metal Cations