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Acid Base Chemistry

Acid Base Chemistry. Autoionization of water and pH. Water is amphoteric 2H 2 O (l) H 3 O + ( aq ) + OH - ( aq ) Ion product constant for water: K w K w = [H 3 O + ][OH - ] = 1.0x10 -14 In pure water: [ H 3 O + ]=[ OH - ] K w = 1 x 10 -14 at 25 ⁰ C

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Acid Base Chemistry

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  1. Acid Base Chemistry

  2. Autoionization of water and pH • Water is amphoteric 2H2O(l) H3O+(aq) + OH-(aq) Ion product constant for water: Kw Kw = [H3O+][OH-] = 1.0x10-14 In pure water: [H3O+]=[OH-] Kw = 1 x 10-14 at 25⁰C Kw = 1 x 10-12 at 95⁰C Temperature affects K – so only a scale of 14 at standard lab conditions

  3. Neutral solution: [H3O+]= [OH-] = 1.0x10-7 Acidic solution: [H3O+] > [OH-] Basic solution: [H3O+] < [OH-]

  4. USING KW IN CALCULATIONS • THE [OH-] IN A CERTAIN HOUSEHOLD AMMONIA CLEANING SOLUTION IS 0.0025 M. CALCULATE THE CONCENTRATION OF H+ IONS.

  5. Acid-Base definitions

  6. Acid Strength • Strong acid—completely ionizes, equilibrium lies far to the right. • Weak Acid—partially ionizes, equilibrium lies to the left. • NOTe*--do not confuse with dilute or concentrated acids.

  7. Strong acids • Memorize this list of strong acids: • HCl • HBr • HI • HNO3 • HClO4 • H2SO4 Generally, the more oxygen atoms, the stronger the acid. Why? All strengths are relative to solvent used…mostly in water.

  8. WEAK ACIDS • Degree of ionization depends on attraction between anion (conjugate base) and the hydrogen ion, relative to attractions of these ions to water. • Acid ionization constant (Ka) • HA(aq) + H2O(l) H3O+(aq) + A-(aq) • The smaller the constant, the weaker the acid.

  9. Write the simple ionization reaction for the following acids • Hydrochloric acid • Acetic Acid • The ammonium ion • The anilinium ion (C6H5NH3+)

  10. Conjugate acid-Base pairs Conjugate acid-base pairs consists of two substances related to each other by the transfer of a proton. HNO2(aq) + H2O(l) NO2-(aq) + H3O+(aq) NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Discuss relative strengths of conjugate acid base pairs…

  11. Relative strengths in aqueous solutions • Strong acid = weak conjugate base • Weak acid = strong conjugate base • Weak Base = Strong conjugate acid • Strong Base = weak conjugate acid • Can use relative strengths to predict which way equilibrium lies

  12. pRACTICE • Identify the conjugate acid-base pairs in the following reactionS: • HBr + NH3 NH4+ + Br- • HCO3- + HCl H2CO3 + Cl- • Predict the direction of the following reaction in aqueous solution: HNO2(aq) + CN-(aq) HCN(aq) + NO2-(aq)

  13. pH Scale • Way to specify the acidity of a solution. • pH = -log[H3O+] • Example: A solution with [H3O+] = 1.0x10-3 M • pH = -log[1.0x10-3] = 3.00 Note* Sig figs: only numbers to the right of the decimal point are significant.

  14. Relation of pH to acidity

  15. Calculating pH • Calculate the pH of each solution at 25oC and indicate whether the solution is acidic or basic. • [H3O+] = 1.8x10-4 M • [OH-] = 1.3x10-2 M

  16. Calculating [H3O+] from pH • Calculate the [H3O+] of a solution with a pH of 4.80 (use correct sig figs in your answer)

  17. Finding the [H3O+] and pH of strong and weak acids • For strong acids, [H3O+] = concentration of strong acid. • For weak acids, must use Ka to find concentrations at equilibrium. • Remember ICE tables??

  18. Practice • Find the pH of a 0.100 M solution of HCN. • Ka for HCN is 4.9x10-10

  19. Practice • Find the pH of a 0.200 M HNO2 solution. • Ka for HNO2 = 4.6x10-4

  20. Finding equilibrium constant from pH • A 0.100 M weak acid solution has a pH of 4.35. Find the Ka for the acid.

  21. A little different… • Find the molarity of acetic acid if the pH of the solution is 4.15.

  22. Diprotic and Polyprotic Acids • Acids that may yield more than one hydrogen ion per molecule. • Ex: Sulfuric acid • Ionize in a stepwise manner • Oxalic acid is a poisonous substance used chiefly as a bleaching and cleansing agent. Calculate the concentrations of all species preset at equilibrium in a 0.20 M solution. • Ka1 = 6.5x10-2 • Ka2 = 6.1x10-5

  23. Mixtures of Acids • When two or more acids are mixed together, you must calculate them in order from strongest to weakest • Based on Ka

  24. Mixtures of Acids • If both are strong…. • Basically finding the new molarity of H+ and taking the pH • Ex – calculate the pH of a mixture if 50.0 mL of 1.00 M HCl and 100.0 mL of 3.00 M HNO3 are mixed.

  25. Mixtures of Acids • Calculate the pH of a solution that contains 1.00 M HCN (Ka=6.2 x 10-10) and 5.00 M HNO3. • Also calculate the [CN-] in this solution.

  26. Base solutions • Strong Bases: completely dissociate in water • Memorize these: LiOH, NaOH, KOH, Sr(OH)2, Ca(OH)2, Ba(OH)2 • Completely dissociate in water.

  27. Weak Bases • Analogous to weak acid • Can find pH, pOH, Kb, and [OH-]

  28. Different bases at the same concentration: base with larger Kb is the stronger base Stronger base will have: higher [OH–] higher pH higher % dissociation Same base at 2 different concentrations: Kb is the same, so base strength is the same solution with higher concentration will have: higher [OH–] higher pH lower % dissocation

  29. Salt HYdrolysis • Salts dissolving in water can produce acidic, neutral, or basic solutions. • All depends on conjugate acids or bases being formed when the ions interact with water molecules.

  30. Summary • Salts derived from: • Strong Acid, strong base = neutral (NaOH) • Weak acid, strong base = basic (NaCH3COO) • Strong acid, weak base = acidic (NH4Cl)

  31. Practice • Predict whether an aqueous solution of the following salts is acidic, basic, or neutral. Prove with appropriate equations. • NaC2H3O2 • NH4NO3 • Al2(SO4)3

  32. Practice • Calculate the pH of a 0.30 M solution of NaF. The Ka of HF is 7.2x10-4.

  33. Practice Find the pH of a 0.100 M NaCHO2 solution.

  34. Polyprotic acids • Ionizes in successive steps • Example: Sulfurous acid • H2SO3(aq)H+(aq) + HSO3-(aq) Ka1 =1.6x10-2 • HSO3-(aq) H+(aq) + SO32-(aq) Ka2 = 6.4x10-8 Note*--Ka2 is much smaller than Ka1.

  35. Finding pH of polyprotic acids First ionization step is much larger than any sequential step. Inhibits any more formation of H3O+ in the second step. Find the pH of a 0.100 M ascorbic acid (H2C6H6O6). Ka1 = 8.0x10-5 Ka2 = 1.6x10-12

  36. Buffers • Resist pH change by neutralizing added acid or base. • Contains either: • Weak acid and its conjugate base • Weak base and its conjugate acid • Example: Carbonic acid in blood

  37. Concept check • Which one of the following solutions is a buffer? A. 0.100 M HNO3 AND 0.100 M HCl B. 0.100 M HNO3 AND 0.100 M NaNO3 C. 0.100 M HNO2 AND 0.100 M NaCl D. 0.100 M HNO2 AND 0.100 M NaNO2

  38. Calculating pH of a buffer Solution • Consider a solution of sodium acetate and acetic acid. • Common Ion effect: Same ions in buffer solution cause acid to ionize less (Le Châtelier’s principle) Practice: A buffer solution contains 0.100 M acetic acid and 0.100 M sodium acetate. Determine the [H3O+], pH, and % ionization in this solution. For HC2H3O2, Ka = 1.8 x 10–5

  39. Henderson-Hasselbalch equation • pH = pKa + log • Calculate the pH of a buffer solution that is 0.050 M in benzoic acid (HC7H5O2) and 0.150 M sodium benzoate (NaC7H5O2). For benzoic acid, Ka = 6.5x10-5

  40. Calculating pH Changes in Buffer solutions • Two parts: • Stoichiometry calculation: calculate how the addition changes the relative amounts of acids and bases. • Equilibrium calculation: calculate pH based on new amounts of acid and conjugate base.

  41. Calculating the pH change in a buffer solution • A 1.0 L buffer solution contains 0.100 M acetic acid and 0.100 M sodium acetate. The value for Ka for acetic acid 1.8x10-5. Calculate the pH of the buffer solution. Calculate the new pH after addition of 0.010 mol of solid NaOH to the buffer. For comparison, calculate the pH after adding 0.010 mol of solid NaOH to 1.0 L of pure water.

  42. Buffer Effectiveness • Most effective with the following characteristics: • Relative concentrations of weak acid and conjugate base do not differ by more than a factor of 10. • Concentrations of acid and conjugate base are high. • Range of the system is one pH unit on either side of the pKa

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