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Chemistry 1011

Chemistry 1011. TOPIC Acids and Bases TEXT REFERENCE Masterton and Hurley Chapter 4.2 (Review), 13, 14.1, 15.1 (page 427), 21.2 (page589). 13.4 Weak Acids and Their Ionization Constants. YOU ARE EXPECTED TO BE ABLE TO:

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Chemistry 1011

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  1. Chemistry 1011 TOPIC Acids and Bases TEXT REFERENCE Masterton and Hurley Chapter 4.2 (Review), 13, 14.1, 15.1 (page 427), 21.2 (page589) Chemistry 1011 Slot 5

  2. 13.4 Weak Acids and Their Ionization Constants YOU ARE EXPECTED TO BE ABLE TO: • Distinguish strong acids and bases from weak acids and bases (Review Chapter 4) • Write an expression for the ionization constant, Ka, for a weak acid • Define pKa • Place a list of weak acids in order of strength given values for Ka or pKa • Calculate Ka, given the pH of a solution of a weak acid • Given Ka, calculate the percent ionization of a weak acid • Given Ka, calculate the [H+] and pH of a solution of a weak acid of known concentration Chemistry 1011 Slot 5

  3. Strong and Weak Acids and Bases • Strong acids and bases are fully ionized in dilute aqueous solution HCl(aq) H+(aq) + Cl-(aq) NaOH(aq) Na+(aq) + OH-(aq) • Weak acids and bases are only partly ionized in dilute aqueous solution CH3COOH(aq) CH3COO-(aq) + H+(aq) Chemistry 1011 Slot 5

  4. Strong and Weak Acids and Bases • Weak bases produce hydroxide ions by reacting with water molecules NH3(aq) + H2O(aq)NH4+(aq) + OH-(aq) cb ca ca cb • In this process, • the NH3 accepts a proton and is a Bronsted-Lowry base • the H2O donates a proton and is a Bronsted-Lowry acid • NH4+ is the conjugate acid of NH3 • OH- is the conjugate base of H2O Chemistry 1011 Slot 5

  5. Ionization Constants of Weak Acids • Weak acids are only partly ionized in dilute aqueous solution CH3COOH(aq) CH3COO-(aq) + H+(aq) • This can also be written CH3COOH(aq) + H2O(aq)CH3COO-(aq) + H3O+(aq) acetic acid acetate ion Chemistry 1011 Slot 5

  6. Ionization Constants of Weak Acids • The equilibrium constant can be written from either equation, ([H2O] is constant) Ka = [CH3COO-]x[H3O+] [CH3COOH] • In general, for HB(aq) H+(aq) + B-(aq) Ka = [H+]x[B-] [HB] Chemistry 1011 Slot 5

  7. Ionization Constants of Weak Acids • The equilibrium constant, Ka, is known as the ionization constant • For acetic acid Ka = [CH3COO-]x[H3O+] = 1.8 x 10-5 [CH3COOH] • Note the low value: [CH3COO-] and [H3O+] will be small Chemistry 1011 Slot 5

  8. pKa Values • The value of Ka can be used to compare the strength of different weak acids • Because of the very small values, it is sometimes useful to use pKa data for comparison pKa = -log10Ka Chemistry 1011 Slot 5

  9. pKa for Some Weak Acids • ACID Ka pKa Nitrous acid, HNO2 6.0 x 10-4 3.22 Formic acid, HCOOH 1.9 x 10-4 3.72 Acetic acid, CH3COOH 1.8 x 10-5 4.74 Carbonic acid, H2CO3 4.2 x 10-7 6.38 Chemistry 1011 Slot 5

  10. Calculating Ka from the pH Value CH3COOH(aq) CH3COO-(aq) + H+(aq) Ka = [CH3COO-]x[H+] [CH3COOH] • The pH of a 1.0 x 10-1 mol/L solution of acetic acid is 2.9. Calculate the value of Ka for the acid • (Ignore [H+] from the water - it is so small as to be insignificant) Chemistry 1011 Slot 5

  11. Calculating Ka from the pH Value pH = -log10[H+] = 2.9 [H+] = 1.3 x 10-3 Ka = [CH3COO-]x[H+] = (1.3 x 10-3)x(1.3 x 10-3) [CH3COOH] (1.0 x 10-1 - 1.3 x 10-3) = (1.3 x 10-3)2 9.9 x 10-2 Ka = 1.7 x 10-5 Chemistry 1011 Slot 5

  12. Percent Ionization • The % ionization of a weak acid HB is given by [H+](equilibrium) x 100% [HB](original) • For the acetic acid, the percent ionization is (1.3 x 10-3) x 100% = 1.3% (1.0 x 10-1) Chemistry 1011 Slot 5

  13. Calculating [H+] and pH from Ka • Given that Ka for acetic acid is 1.8 x 10-5, calculate the pH of a 0.20mol/L solution. CH3COOH(aq) CH3COO-(aq) + H+(aq) [original] 2.0 x 10-1 0.0 0.0 D[ ] -x +x +x [equilibrium] (2.0 x 10-1- x) x x Ka = [CH3COO-]x[H+] = x2 [CH3COOH] (2.0 x 10-1- x) Chemistry 1011 Slot 5

  14. Solving for x • If x is small compared to the initial concentration of the acid (% ionization is low), we can solve for x using an approximate method • If x is not small, the quadratic formula must be used • In this case, assume that x is small. Then (2.0 x 10-1 - x)  2.0 x 10-1 Chemistry 1011 Slot 5

  15. Solving for x Ka = 1.8 x 10-5 = x2 (2.0 x 10-1) x2 = (1.8 x 10-5)(2.0 x 10-1) = 3.6 x 10-6 x = 1.9 x 10-3 mol/L pH = 2.7 (Check that [H+] is << [CH3COOH] ) 1.9 x 10-3 mol/L<< 2.0 x 10-1 mol/L Chemistry 1011 Slot 5

  16. Solving for x Using the Quadratic Formula Ka = x2 = 1.8 x 10-5 (2.0 x 10-1 - x) x2 = (1.8 x 10-5) (2.0 x 10-1 - x) = (3.6 x 10-6) - (1.8 x 10-5 )x (x) x2 + (1.8 x 10-5 )x (x) - (3.6 x 10-6) = 0 • x = 1.9 x 10-3 mol/L • pH = 2.7 Chemistry 1011 Slot 5

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