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Acid-Base Equilibrium: Ka and Kb Review Questions

Practice solving acid-base equilibrium problems involving Ka and Kb values, calculating pH for different weak acids and bases. Importance of Ka and Kb in determining acid and base strength. Interpretation of dissociation constants for weak acids. Understanding the stability of conjugate bases. Study examples for better comprehension.

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Acid-Base Equilibrium: Ka and Kb Review Questions

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  1. Acid-Base Eqm (2): Ka and Kb Acid-Base Eqm (2): Ka and Kb p.01 Review Questions Calculate the pH : Q.1 10-8 HCl(aq) [100% ionized] [H3O+] = (10-7 + 10-8) M = 1.110-7 M,  pH = -log(1.110-7) = 6.96 Q.2 0.01M NaOH(aq) [100% ionized] [OH-] = 0.01M, pOH = -log(0.01) = -2  pH = 14 – 2 = 12 C. Y. Yeung (CHW, 2009)

  2. Acid Dissociation Constant (Ka) p.02 weak acid HA + H2O [H3O+(aq)][A-(aq)] Kc = [HA(aq)][H2O(l)] [H3O+(aq)][A-(aq)] Kc[H2O(l)] = [HA(aq)] [H3O+(aq)][A-(aq)] Ka = [HA(aq)] [temperature dependent!!] A-+ H3O+

  3. p.03 at start HCOO-+ H3O+ HCOOH + H2O at eqm 0 1.00 0 x2 Ka = = 1.7810-4 1 – x [H3O+(aq)][A-(aq)] Ka = [HA(aq)] Q: Ka of methanoic acid (HCOOH) at 298K is 1.7810-4 mol dm-3. Calculate the pH value of 1.00M HCOOH. x x 1.00 – x x2 = 1.7810-4(x << 1, Ka is small) x = 0.0133 solved eqn without approximation, x = 0.01325, pH = 1.88  pH = -log(0.0133) = 1.88

  4. p.04 x2 = 1.7810-6(x << 1, Ka is small) x = 1.33 10-3  pH = -log(1.33 10-3) = 2.87 x2 Ka = = 1.7810-4 0.01 – x Q: Ka of methanoic acid (HCOOH) at 298K is 1.7810-4 mol dm-3. Calculate the pH value of 0.01 M HCOOH. solved eqn without approximation, x = 1.24210-3, pH = 2.91 Note: Different conc. of HA has different “% of dissociation” For 1.00M HCOOH, % of dissociation = (0.01325/1)100% = 1.33 % For 0.01M HCOOH, % of dissociation = (1.24210-3/0.01)100% = 12.4 % *** i.e. [HA] decreases, % of dissociation increases.

  5. For Polyprotic Acid: Ka1, Ka2, ….. p.05 Ka1 Ka2 HSO4-+ H3O+ SO42-+ H3O+ H2SO4 + H2O HSO4- + H2O [Note: Ka1 > Ka2 > … ]  WHY? Note: It is more difficult to remove a H+ from the negatively charged HSO4-. H3O+ is already produced in the first dissociation, it would suppress the formation of H3O+ from the second dissociation. [COMMON ION EFFECT (p.148)] Overall Ka of Polyprotic acid = Ka1 Ka2  …

  6. p.06 at start SO42-+ H3O+ HSO4- + H2O at eqm 0.1 0.10 0 x(0.1+x) Ka2 = = 0.01 0.1 – x Q: Assuming that the first ionization of H2SO4 is 100%, and the Ka2 of sulphuric acid is 0.01M. Find the pH of 0.10M H2SO4. Ka2 = 0.01M x 0.1+x 0.10 – x x = 8.4410-3  pH = -log (0.1 + 8.4410-3) = 0.965

  7. p.07 Importance of Ka ?? 1. Indicates the strength of acid: (a) larger Ka stronger acid (b) smaller pKa  stronger acid (ref.: p. 141) 2. Indicates the relative stability of conjugate bases: e.g. Ka of ethanoic acid = 1.8010-5 mol dm-3, Ka of butanoic acid = 1.4810-5 mol dm-3. i.e. CH3CH2CH2COO- is less stable than CH3COO-.

  8. Base Dissociation Constant (Kb) p.08 [temperature dependent!!] B + H2O x2 Kb = = 1.7810-5 1 – x [HB+(aq)][OH-(aq)] Kb = [B(aq)] HB++ OH- Q: Kb of NH3 at 298 K is 1.7810-5 mol dm-3. What is the pH value of 1M NH3 at 298K? x = 4.22 10-3  pOH = -log(4.22 10-3) = 2.37 pH = 14 – pOH = 11.6

  9. p.09 NH4++ OH- NH3 + H2O 0 1 1 at start x 1+x 1 – x at eqm x(1+x) Kb = = 1.7810-5 1 – x Q: Kb of NH3 at 298 K is 1.7810-5 mol dm-3. What is the pH value of a solution which is both 1M NH3 and 1M NH4Cl at 298K? x = 1.7810-5  pOH = -log(1.78 10-5) = 4.75 pH = 14 - 4.75 = 9.25

  10. p.10 Importance of Kb ?? 1. Indicates the strength of base: (a) larger Kb stronger base (b) smaller pKb  stronger base (ref.: p. 142) 2. Indicates the relative stability of conjugate acid: e.g. Kb of NH3 = 1.7410-5 mol dm-3, Kb of (CH3)2NH = 9,5510-4 mol dm-3. i.e. (CH3)2NH2+ is more stable than NH4+.

  11. p.11 A-+ H3O+ HA+ OH- A- + H2O HA + H2O [H3O+(aq)][A-(aq)] Ka = [HA(aq)] [HA(l)][OH-(aq)] Kb = [A-(aq)] [HA(l)][OH-(aq)] [HA(l)][OH-(aq)] [H3O+(aq)][A-(aq)] [H3O+(aq)][A-(aq)]   Ka Kb= [A-(aq)] [A-(aq)] [HA(aq)] [HA(aq)] Relationship between Kw, Ka and Kb ?? = Kw

  12. p.12 Next …. Explain the Strength of Organic Acids & Bases (Book 3A p. 113 – 133) Assignment Study all examples in p.142 - 149 p.152 Q.3 –12 [due date: 26/3(Thur)] Pre-Lab: Expt. 12 Determination of Ka

  13. p.13 A-+ H3O+ HA + H2O [H3O+(aq)][A-(aq)] Ka = [HA(aq)] [H3O+(aq)][A-(aq)] log log Ka = [HA(aq)] [A-(aq)] + log log [H3O+(aq)] pKa = [HA(aq)] [A-(aq)] + log pH pKa = [HA(aq)] Expt. 12 Dissociation Constant of Weak Acids

  14. p.14 A-+ H3O+ Step 5 0 mol 110-3 mol HA + H2O Step 6 (10cm3 NaOH added) 0 mol 110-3 mol Step 7 (another 10cm3 HA added) 110-3 mol 110-3 mol in the same volume, i.e. same conc.! at eqm 110-3 mol – x 110-3 mol + x [A-(aq)] + log pH pKa = [HA(aq)] measured by pH meter = pH + log (1) = pH

  15. Simple explanation on the Relative Acidity of Organic Acids p.15 The more stable is the conjugate base, the more acidic is the acid. 1. If the –ve charge of conjugate base is localized, it is more likely to recombine with the H3O+ to form the acid again.  the acid is less acidic! 2. If the –ve charge of conjugate base is delocalized, it is less likely to recombine with the H3O+ to form the acid again.  the acid is more acidic!

  16. Alkyl group (e.g. CH3) is e- releasing, halo group (e.g. Cl) is e- withdrawing. p.16 H H O O C C C C Cl CH3 - - O O H H e- withdrawing group e- releasing group more stable conjugate base,  ClCH2COOH is more acidic! *** Longer alkyl group, more e- realeasing. *** More halogen atoms, more e- withdrawing. (i.e. larger Ka, smaller pKa)

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