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pH = - log [H 3 O + ] [H 3 O + ] = 10 - pH mol/L For pure water at 25 o C

pH = - log [H 3 O + ] [H 3 O + ] = 10 - pH mol/L For pure water at 25 o C pH = - log (1.0 x 10 -7 ) = 7.00 For a change in pH by 1, H 3 O + concentration changes by 10 Higher pH, lower H 3 O + concentration pH of pure water is 7 pH of an acidic solution is less than 7

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pH = - log [H 3 O + ] [H 3 O + ] = 10 - pH mol/L For pure water at 25 o C

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  1. pH = - log [H3O+] [H3O+] = 10- pH mol/L For pure water at 25oC pH = - log (1.0 x 10-7) = 7.00 For a change in pH by 1, H3O+ concentration changes by 10 Higher pH, lower H3O+ concentration pH of pure water is 7 pH of an acidic solution is less than 7 pH of a basic solution is greater than 7

  2. pOH = - log [OH-] pKw = - log Kw pKw = 14.00 at 298 K [H3O+ (aq)] [OH- (aq)] = Kw - log[H3O+ (aq)] - log[OH- (aq)] = - log Kw pH + pOH = pKw At 298 K pH + pOH = 14.00

  3. Strengths of Acids and Bases The pH of 0.10 M HCl(aq) will be recorded as close to 1 The pH of a 0.10 M solution of CH3COOH(aq) solution is recorded as ~ 3. H3O+(aq) concentration in 0.10 M HCl(aq) is greater than that in 0.10 M CH3COOH(aq) HCl(aq) + H2O(l)  H3O+(aq) + Cl- (aq) CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO- (aq)

  4. [CH3COO- (aq)] [H3O+(aq)] Ka = [CH3COOH(aq)] [OH- (aq)] [NH4+(aq)] Kb = [NH3(aq)] Ka: acidity constant or acid dissociation constant CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO- (aq) At 298 K, Ka for CH3COOH(aq)= 1.8 x 10-5 NH3 (aq) + H2O(l)  NH4+(aq) + OH- (aq) Kb: basicity constant or base dissociation constant At 298 K, Kb for NH3(aq)= 1.8 x 10-5

  5. The proton donor strength of an acid is measured by the value of Ka; higher Ka, stronger the acid The proton acceptor strength of a base is measured by Kb; higher Kb, stronger the base pKa = - log Ka pKb = - log Kb The larger the pK values, weaker the acid or base

  6. [OH- (aq)] [H3O+(aq)] [NH4+(aq)] [NH3(aq)] Kb = Ka = [NH3(aq)] [NH4+( aq)] Relationship between conjugate acid/base pairs NH3 (aq) + H2O(l)  NH4+(aq) + OH- (aq) Ka Kb = [H3O+(aq)] [OH- (aq)] = Kw Or pKa + pKb = pKw NH4+(aq) + H2O(l)  H3O+(aq) + NH3 (aq)

  7. HClO2(aq)/ ClO2-(aq) The stronger the acid/base, the weaker its conjugate base/acid CH3COOH(aq)/ CH3COO-(aq) HOCl(aq)/ OCl-(aq) NH4+(aq) / NH3(aq) CH3NH3+(aq) / CH3NH2(aq) pKa - pink pKb - blue

  8. Using tabulated Ka and Kb values determine which species is stronger as an acid or base 1) as acid HF(aq) or HIO(aq) 2)as base C6H5NH2(aq) or (CH3)3N(aq) 3) as acid C6H5NH3+(aq) or (CH3)3NH+(aq)

  9. Molecular Structure and Acid Strength The more polar or weaker the H-A bond, the stronger the acid Effect of bond strength HF < HCl < HBr < HI H - I bond is weakest H2O < H2S < H2Se < H2Te H-Te bond weakest

  10. For an acid HA, greater the electronegativity of A, stronger the acid electronegativity difference N-H 0.8 F-H 1.8 HF is an acid in water, NH3 is a base

  11. Solutions of Weak Acids/Bases For a strong acid and base; assume that deprotonation/protonation reactions go to completion HCl(aq) + H2O(l)  H3O+ (aq) + Cl- (aq) pH = - log [H3O+ (aq)] Knowing the concentration of HCl, can determine pH For weak acids/bases, set up equilibrium table to determine the H3O+ (aq) / OH- (aq) concentration at equilibrium, knowing the value of Ka/Kb.

  12. [CH3COO- (aq)] [H3O+(aq)] Ka = [CH3COOH(aq)] Calculate the pH and percentage deprotonation of 0.10 M CH3COOH(aq) given that Ka is 1.8 x 10-5. CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO- (aq) CH3COOH(aq) CH3COO-(aq) H3O+(aq) Initial 0.10 0 0 Change - x x x Equilibrium 0.10 - x x x 1.8 x 10-5 = x2/(0.10 - x)

  13. 1.8 x 10-5 = x2/(0.10 - x) Since Ka is so small, assume that x << 0.10 1.8 x 10-5 ≈ x2/(0.10) x = 1.3 x 10-3 M [H3O+(aq)] = 1.3 x 10-3 M pH = 2.89 % deprotonation = 100% x ([CH3COO-(aq)]/[CH3COOH]initial) = 100% x (1.3 x 10-3 M)/(0.10 M) = 1.3 % Note: x < 5% of 0.10 , OK to make this approximation

  14. For a weak base B(aq) + H2O(l)  HB+ (aq) + OH- (aq) Use a similar approach to determine pOH knowing Kb, and then determine pH Determine the pH and percentage protonation of a 0.20 M aqueous solution of methylamine, CH3NH2. The Kb for CH3NH2 is 3.6 x 10-4. pH = 11.9 % protonation = 4.2%

  15. pH of Salt Solutions CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l) “neutralization” reaction If a 0.3M solution of CH3COOH(aq) is added to a 0.3M solution of NaOH, pH of resulting solution is not 7.0 but ~ 9.0 Solution of a salt is a solution of an acid (usually the cation) and a base (usually the anion), and the pH depends on their relative strength. CH3COO-(aq) determines the pH of the solution

  16. Ni(H2O)62+(aq) + H2O(l)  H3O+(aq) + Ni(H2O)5(OH)+(aq)

  17. [NH3(aq)] [H3O+(aq)] Ka = [NH4+ (aq) ] Estimate the pH of 0.15 M NH4Cl(aq). Kb (NH3(aq)) = 1.8 x 10-5 NH4+ (aq) is an acid and Cl- (aq) is neutral; expect pH < 7 NH4+ (aq) + H2O(l)  H3O+ (aq) + NH3(aq) NH4+(aq) NH3 (aq) H3O+(aq) Initial 0.15 0 0 Change -x x x Equilibrium 0.15-x x x

  18. Kw Ka (NH4+ (aq)) = x2 Kb (NH3(aq)) 0.15 - x 5.6 x 10-10 = Assume x << 0.15 x ≈ 9.2 x 10-6 (agrees with the assumption) pH = - log(9.2 x 10-6 ) = 5.04

  19. Polyprotic Acids & Bases A polyprotic acid can donate more than one H+ Carbonic acid: H2CO3(aq); dissolved CO2 in water Sulfuric acid: H2SO4(aq) Phosphoric acid: H3PO4(aq) A polyprotic base: can accept more than one proton Carbonate ion: CO32-(aq) Sulfate ion: SO42-(aq) Phophate ion: PO43-(aq) Treat each step of protonation or deprotonation sequentially

  20. H2CO3 (aq) + H2O(l)  H3O+(aq) + HCO3-(aq) Ka1 = 4.3 x 10-7 HCO3-(aq) + H2O(l)  H3O+(aq) + CO32-(aq) Ka2 = 4.8 x 10-11 Typically: Ka1 >> Ka2 >> Ka3 >>… Harder to loose a positively charged proton from a negatively charged ion, because of attraction between opposite charges.

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