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Acid Base Equilibria

Acid Base Equilibria. Complications: Beyond pH of strong acids or strong bases HCl(aq) H + (aq) + Cl - (aq) K a > 1 HF(aq) H + (aq) + F - (aq) K a = 7.2 x 10 -4. Weak conjugate base. Strong acid. Weak acid. stronger conjugate base. Conjugate Bases. Bronsted-Lowry Model

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Acid Base Equilibria

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  1. Acid Base Equilibria Complications: Beyond pH of strong acids or strong bases HCl(aq) H+(aq) + Cl- (aq) Ka > 1 HF(aq) H+(aq) + F- (aq) Ka = 7.2 x 10-4 Weak conjugate base Strong acid Weak acid stronger conjugate base Mullis

  2. Conjugate Bases • Bronsted-Lowry Model • Conjugate base remains after acid loses its proton • Strongest conjugate base = best at getting available protons • Factors affecting conjugate base strength: • Electronegativity S- > Cl- • Size if acids were binary F- > I- • Number of oxygens if acids were oxyacids ClO- > ClO2- Mullis

  3. pH of Mixture of Weak Acids • Find pH of a solution that contains 1.00M HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2 (Ka = 4.0 x 10-4). • Because these are weak acids, dissociation is not complete and [H+] ≠ [HNO2]. • You must use ICE table strategy to solve pH problems with weak acids or bases. Mullis

  4. How to Determine [H+] in an Acid Mixture • HCN H+ + CN-Ka = 6.2 x 10-10 • HNO2 H+ + NO2-Ka = 4.0 x 10-4 • H2O H+ + OH-Ka = 1.0 x 10-14 • Compare species and find the most likely contributor of protons. Here it is by far HNO2: Ka is 6 orders of magnitude greater than the closest possible H+ contributor. Mullis

  5. Small: reactants favored Change must be very small relative to initial conc.: Ignore x in this term Equilibrium Method to Find [H+] • Ka = x2 = 4.0 x 10-4 (5.00 –x) x2 = 20. x 10-4 x = 4.5 x 10-2 Mullis

  6. Accuracy check • When dealing with weak acids and bases, Ka values are typically valid to a ±5% accuracy. Verify that our assumption in ignoring x in the reduction of [HNO2]0 was valid: • x x 100% = 4.5 x 10-2 (100%) = 0.90 % [HNO2] 5.00 • Since 0.90% is less than 5%, the assumption is valid---in other words, our answer for x is good. Mullis

  7. What was the Original Question? • Find pH of this solution: • [H+] = 4.5 x 10-2 • pH = -log[H+] = -log(4.5 x 10-2) = 1.35 • Number of decimals in pH = Number of significant figures in concentration (2 here) Mullis

  8. Find [CN-] in that problem. • [H+] = 4.5 x 10-2 M • HCN H+ + CN-Ka = 6.2 x 10-10 • 6.2 x 10-10 = [4.5 x 10-2 ][CN-] [HCN] Remember our assumption that since Ka is small, we ignore the small reduction in initial concentration of product? Be consistent and make that assumption here, too. So: • 6.2 x 10-10 = [4.5 x 10-2 ][CN-] 1.00 [CN-] = 1.4 x 10-8 M (That’s a small number!) Mullis

  9. % DissociationEx: HNO2 H+ + NO2- [H+] x 100% = % dissociation [HNO2] Or… [NO2-] x 100% = % dissociation [HNO2] If you know % dissociation and original concentration of acid [HNO2], can solve for [H+] or [NO2-] (or x, the change in concentration). Mullis

  10. % Dissociation and Dilution More Concentrated Diluted Less % dissociation More % dissociation (more chance of recombination) If Q = K for this reaction Q < Ka for this one, so equilibrium shifts right Mullis

  11. Ka from % DissociationA 0.100 M solution of lactic acid (HC3H5O3) is 3.7% dissociated. Calculate the value of Ka. % dissociation = x x (100%) Ka = x2 0.10 (0.100 –x) 3.7% (0.10) = x Ka = 0.00372 x = 0.0037 M (0.100 - .0037) Ka = 1.4 x 10-4 Mullis

  12. -OH phenol Organic Acids(Acid = proton donor) Mullis

  13. Bases without OH- • B (aq) + H2O BH+ (aq) + OH- (aq) • NH3 (aq) + H2O NH4+ (aq) + OH- (aq) • Kb = [BH+][OH-] [B] base conjugate base acid conjugate acid Mullis

  14. OH . . N HO CHCH2NHCH3 C2H5 HO H N . . H pyridine ethylamine Amines RxNH(3-x) • Ammonia with one or more of the H atoms replaced by another group. • Many organic molecules involve amines. adrenaline Mullis

  15. Find the pH of 1.0M methylamine (Kb=4.38x10-4). CH3NH2 + H2O CH3NH3+ + OH- Kb = x2 = 4.38x10-4x2≈ 4.38x10-4 1.0-x x≈ 2.1x10-2 [OH-] = x = 2.1x10-2 M pOH=-log[2.1x10-2 ] pOH = 1.68 pH= 14 -1.68= 12.32 Mullis

  16. PolyProtic Acids • Lose one proton at a time: H3PO4 H2PO4-  HPO42- PO43- • The reactions written to express this loss have successively smaller Ka values: • Ka1 > Ka2 > Ka3 • …..makes sense: The loss of a second proton occurs less readily than the loss of the first one. In fact, Ka1 is usually so much larger than the others, the loss of the first proton is the only reaction that significantly contributes [H+]. Mullis

  17. Sequential Loss of Protons H3PO4 H+ + H2PO4-Ka1 = 7.5 x 10-3 H2PO4- H+ + HPO42-Ka2 = 6.2 x 10-8 HPO42- H+ + PO43-Ka3 = 4.8 x 10-13 Mullis

  18. pH of a Polyprotic Acid • Find pH of 5.0M H3PO4 and the equilibrium concentrations of each species: H3PO4, H2PO4-, HPO42-, PO43- • Ka1 = [H+][H2PO4-] = x2 = 7.5 x 10-3 [H3PO4] 5.0-x (see previous problems’ RICE tables & pH calculations for details) x2 = (5.0)7.5 x 10-3 x ≈ 0.19 M= [H2PO4-] = [H+] pH = 0.72 [H3PO4] = 5.0 – x = 4.8 M Ka2 =[H+][HPO42-] = (0.19) [HPO42-] = 6.2 x 10-8 [H2PO4-] 0.19 [HPO42-]= 6.2 x 10-8 M Ka3 =[H+][PO43-] = (0.19) [PO43-] = 4.8 x 10-13 [HPO42-] 6.2 x 10-8 [PO43-]= 1.6 x 10-19 M Mullis

  19. Sulfuric acid: Relatively High Conc. • Find pH of 1.0 M H2SO4. • H2SO4 H+ + HSO4- Ka1 > 1 • HSO4- H+ + SO42- Ka2 = 1.2 x 10-2 • In this case, does the 2nd reaction contribute significantly to [H+]? • [H+] is at least 1.0 since the first step is total dissociation. Mullis

  20. From 1st dissociation step Sulfuric acid: Relatively High Conc. 1.2 x 10-2= [H+][SO42-] = (1.0 + x)(x) [HSO4-] (1.0 – x) Assume change is small compared to 1.0, so: 1.0(x) 1.0 X = 1.2 x 10-2, or 1.2% of 1.0 M. If this is added to 1.0, result is 1.0012M, but with 2 sig. figs., [H+] is 1.0 M. (pH = 0.00) Mullis

  21. From 1st dissociation step 0.01 M H2SO4 (Relatively Low Conc.) 1.2 x 10-2= [H+][SO42-] = (.01 + x)(x) [HSO4-] (.01 – x) If assume change is small compared to .01, .01(x) .01 x = 1.2 x 10-2, or more than 0.01 M! Therefore, cannot ignore the subtraction or addition of x—use the quadratic to find x = .0045. Total [H+] = 0.01(from 1st dissociation step) + .0045 = 0.0145. pH = -log(0.0145) = 1.84 Mullis

  22. Salts of Weak Acids or Bases • Cations of strong bases do not change pH of a solution. • (Ex: Na+ will not attract or contribute H+) • Salts in which cations do not change pH and the anion is the conjugate base of a weak acid will produce basic solutions. • Ex. KC2H3O2 dissociates to produce C2H3O2- and OH- • Salts in which anions are not a base and the cation is a conjugate acid of a weak base will produce acidic solutions. • Ex. NH4Cl dissociates to produce NH3 and H+. Mullis

  23. Finding Kb for Conjugates • Look up the Ka for the weak acid that makes the conjugate base. • Ex: NaC2H3O2 (aq) • C2H3O2- + H2O  HC2H3O2 + OH- • Ka(acetic acid)= 1.8 x 10-5 • KaKb = Kw • Kb = 1x10-14 = 5.6 x 10-10 1.8 x 10-5 Mullis

  24. Salt as a Weak Base • Find pH of a 0.30 M NaF solution. • Na+ does not change pH. • F- is the conjugate base of HF. Water has to be the proton donor for this solution. • F-(aq)+ H2O (aq)  HF (aq)+ OH- (aq) • Ka for HF is 7.2 x 10-4. • Kb = 1.0 x 10-14= 1.4 x 10-11 7.2 x 10-4 Mullis

  25. Salt as a Weak Base Kb = 1.4 x 10-11 = [HF][OH-] = x2 [F-] .30-x x2 =.30(1.4 x 10-11) x = 2.0 x 10-6(<5% of .30) [OH-] = 2.0x10-6 pOH = 5.69 pH= 14.00 – 5.69 = 8.31 Mullis

  26. Salt as a Weak Acid • Find pH of a 0.10 M NH4Cl solution. • Cl- does not change pH. • NH4+ produces H+ in water. It is the conjugate acid of NH3. • NH4+(aq)  NH3(aq)+ H+ (aq) • Kb for NH3 is 1.8 x 10-5. • Ka for NH+ = 1.0 x 10-14= 5.6 x 10-10 1.8 x 10-5 Mullis

  27. Salt as a Weak Acid Kb = 5.6 x 10-10 = [NH3][H+] = x2 [NH4+] .10-x x2 =.10(5.6 x 10-10 ) x = 7.5 x 10-6(<5% of .10) [H+] = 7.5 x 10-6 pH = -log(7.5 x 10-6 ) = 5.13 Mullis

  28. Acid-Base Properties of Salts Mullis

  29. Practice • A 0.15 M solution of a weak acid is 3.0% dissociated. Calculate Ka. • What are the major species present in a 0.150 M NH3 solution? Calculate the [OH-] and the pH of this soln. • Calculate the pH of a 5.0 x 10-3 M solution of H2SO4. • Sodium azide is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN3. The Ka value for HN3 (hydrazoic acid) is 1.9 x 10-5. • Calculate the pH of 0.10 M CH3NH3Cl. Kb of CH3NH3+ is 4.38 x 10-4. Mullis

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