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Acid-Base Strength: Ka, Kb, Kw

Acid-Base Strength: Ka, Kb, Kw. Mrs. Kay Chemistry 12 Chapter 15 Pages: 583-584,587-597. Relative Strengths Of Binary Acids. H –X The greater the tendency for the transfer of a proton from HX to H 2 O, the more the forward reaction is favored and the stronger the acid.

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Acid-Base Strength: Ka, Kb, Kw

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  1. Acid-Base Strength:Ka, Kb, Kw Mrs. Kay Chemistry 12 Chapter 15 Pages: 583-584,587-597

  2. Relative Strengths Of Binary Acids H –X The greater the tendency for the transfer of a proton from HX to H2O, the more the forward reaction is favored and the stronger the acid. in a periodic group: • The weaker the bond, the stronger the acid. • The larger the resultant anion’s radius, the stronger is the acid. • The strengths of binary acids increase from top to bottom in a group of the periodic table.

  3. Relative Strengths Of Binary Acids H –X in a periodic group: Bond dissociation energy: the weaker the bond, the stronger the acid. Bond dissociation energy 569 > 431 > 368 > 297 (kJ/mol) HF HCl HBr HI Acid strength Ka 6.6x10-4 < ~106 < ~108 < ~109 Anion radius: the larger the anion’s radius, the stronger the acid. Anion radius (ppm) 136 < 181 < 195 < 216 (kJ/mol) HF HCl HBr HI Acid strength Ka 6.6x10-4 < ~106 < ~108 < ~109 The strength of binary acids increase from top to bottom in a group of the periodic table.

  4. Relative Strengths Of Binary Acids H –X in a period: • The larger the electronegativity difference between H and X, the more easily the proton is removed and the stronger is the acid.  EN 0.4 < 0.9 < 1.4 < 1.9 Acid strength CH4 NH3 H2O HF • The strengths of binary acids increase from left to right across a period of the periodic table.

  5. Representative Trends In Strengths of Binary Acids

  6. The Acid dissociation constant, Ka

  7. Ka • A weak acid only ionizes to a small extent and comes to a state of chemical equilibrium. • We can determine how much it ionizes by calculating the equilibrium constant for this reaction, the ionization constant, Ka. • The larger the Ka the more acid ions are found in solution and the stronger the acid because the more easily it donates a proton. • The reverse is true for the smaller the Ka

  8. HCOOH (aq)+ H2O (l) < -- > H3O+(aq) + HCOO-(aq) • Ka = [H3O+][HCOO-] [HCOOH] • Notice how the Ka ignores the water since we are dealing with dilute solutions of acids, water is considered a constant, and when multiplied by both sides it is cancelled out.

  9. Equilibrium In Solutions Of Weak Acids And Weak Bases weak acid: HA + H2O º H3O+ + A- [H3O+][A-] Ka = [HA] weak base: B + H2O º HB+ + OH- [HB+][OH-] Kb = [B] You need to be able to write acid and base ionization equations!!!

  10. Practice: • A solution of a weak acid, “HA”, is made up to be 0.15 M. Its pH was found to be 2.96. Calculate the value of Ka. • Steps to follow: • Write balanced equation • Calculate [H+] using 10-pH • Set up chart for equilibrium (ICE or i Δ f) • Solve using Ka expression

  11. Answer • HA (aq) + H2O (l) < -- >H3O+(aq) + A-(aq) • [H3O+]= 10 -2.96 = 0.0011 M • Set up table: i Δ f

  12. Ka = [H3O+][ A-] [HA] = [0.0011][0.0011] [0.139] = 8.7 x 10 -6

  13. Percent Dissociation The fraction of acid molecules that dissociate compared with the initial concentration of the acid. Percent Dissociation = [H3O+] x 100% [HA i] For the previous question: Percent Dissociation = [0.0011] x 100% =0.73 % [0.15]

  14. Practice: • The ionization constant, Ka, for a hypothetical weak acid, HA, at 25°C is 2.2 x 10-4. a) Calculate the [H3O+] of a 0.20 M solution of HA. b) Calculate the percent ionization of HA. c) Calculate the [A-]. d) What initial concentration of HA is needed to produce a [H3O+] of 5.0 x 10-3 M?

  15. Answer a) HA(aq) + H2O(l) < -- >H3O+(aq) + A-(aq) Ka = [H3O+][A-] = 2.2 x 10-4 [HA] 2.2 x 10-4 = (x)(x) 0.20 M x2 = (2.2 x 10-4) (0.20 M) x = 0.0066 M The [H3O+] is 0.0066 M. Because it is a 1:1 ratio they are both the same concentration (x)

  16. b) % ionization = 0.0066 M x 100% = 3.3% 0.20 M c) From the stoichiometry of the reaction, [H3O+] = [A-] Therefore, [A-] = 0.0066 M

  17. Found d) from table set up d) 2.2 x 10-4 = (5.0 x 10-3 M)(5.0 x 10-3 M) x – 5.0 x 10-3 M • 2.2 x 10-4(x – 5.0 x 10-3M) = 2.5 x 10-5 • 2.2 x 10-4x – 1.1 x 10-6 = 2.5 x 10-5 • x = (2.5 x 10-5 + 1.1 x 10-6) / 2.2 x 10-4 • x= 0.12 M • The initial concentration of HA required is 0.12 M.

  18. Acid And Base Ionization Constants weak acid: CH3COOH + H2O º H3O+ + CH3COO- [H3O+][CH3COO-] Acid ionization constant: Ka = [CH3COOH] weak base: NH3 + H2O º NH4+ + OH- [NH4+][OH-] Base ionization constant: Kb = [NH3] Acid and base ionization constants are the measure of the strengths of acids and bases.

  19. Kb • When using weak bases, the same rules apply as with weak acids, except you are solving for pOH and using [OH-]

  20. Another relationship • Useful to know: Ka x Kb = Kw = 1.0 x 10-14

  21. Buffer Solutions • A buffer solution is a solution that changes pH only slightly when small amounts of a strong acid or a strong base are added. • A buffer contains a weak acid with its salt (conjugate base) or a weak base with its salt (conjugate acid) CH3COOH/CH3COONa NH3/NH4Cl

  22. Depicting Buffer Action

  23. How A Buffer Solution Works • The acid component of the buffer can neutralize small added amounts of OH-, and the basic component can neutralize small added amounts of H3O+. • Pure water does not buffer at all.

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