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Polyprotic Acids & Bases

Polyprotic Acids & Bases. A polyprotic acid can donate more than one H + Carbonic acid: H 2 CO 3 (aq); dissolved CO 2 in water Sulfuric acid: H 2 SO 4 (aq) Phosphoric acid: H 3 PO 4 (aq) A polyprotic base: can accept more than one proton Carbonate ion: CO 3 2- (aq)

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Polyprotic Acids & Bases

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  1. 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

  2. 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.

  3. Calculate the pH of 0.010 M H2SO4(aq) at 25oC. Sulfuric acid is the only common polyprotic acid where the first deprotonation step is complete. The second deprotonation step is much weaker and adds slightly to the H3O+(aq) concentration. For the first step assume all H2SO4(aq) deprotonates H2SO4 (aq) + H2O(l)  H3O+ (aq) + HSO4-(aq) From the first step [H3O+(aq)] = 0.010 M

  4. Second deprotonation HSO4- (aq) + H2O(l)  H3O+ (aq) + SO42- (aq) Ka2 = 0.012 HSO4- (aq) SO42- (aq) H3O+ (aq) Initial 0.010 0 0.010 Change -x + x 0.010 + x Equilibrium 0.010-x x 0.010 + x Ka2 = ([H3O+ (aq)])([SO42- (aq)]) / ([HSO4- (aq) ]) 0.012 = (0.010+x)(x) / (0.010-x) Solve the quadratic equation for x. Ka2 is large; cannot assume that x << 0.010 [H3O+ (aq)] = 1.4 x 10-2 M pH = 1.9

  5. Determine the pH of 0.20 M H2S(aq) at 25oC H2S (aq) + H2O(l)  H3O+ (aq) + HS- (aq) Ka1 = 1.3 x 10-7 HS- (aq) + H2O(l)  H3O+ (aq) + S2- (aq) Ka2 = 7.1 x 10-15 For the first deprotonation step determine [H3O+(aq)] using equilibrium tables. [H3O+(aq)] = 1.6 X 10-4 M Can assume that x << 0.20 since Ka1 is small Second deprotonation constant is very small, so ignore addition of H3O+(aq) due to second step. pH determined by first step alone. pH = 3.8

  6. Composition and pH For a solution of H2CO3(aq): at low pH the fully protonated species (H2CO3) dominates; at high pH the fully deprotonated form (CO32-) dominates; and at intermediate pH the intermediate species (HCO3-) dominates. LeChatelier’s principle at work 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

  7. H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq) Ka1 = [H3O+(aq)] [HCO3-(aq)] / [H2CO3(aq) ] HCO3-(aq) + H2O(l)  H3O+(aq) + CO32-(aq) Ka2 = [H3O+(aq)] [CO32-(aq)] / [HCO3-(aq) ] Define a(X): fraction of species X X a(X) = [H2CO3(aq) ] + [HCO3-(aq)] + [CO32-(aq)] The fraction of deprotonated species increases as the pH increases

  8. Determine the concentration of H2CO3(aq), HCO3-(aq), CO32-(aq), H3O+(aq) present and the pH at equilibrium in a solution that is initially 0.010 M in H2CO3. (Ka1 = 4.3 x 10-7, Ka2 = 4.8 x 10-11) Step 1 H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq) H2CO3(aq) H3O+(aq) HCO3-(aq) Initial 0.010 0 0 Change -x x x Equilibrium 0.010-x x x 4.3 x 10-7 = x2 / (0.010-x) Assume x << 0.010; x = 6.6 x 10-5 [H2CO3(aq)] ≈ 0.010 M; [H3O+(aq)] = 6.6 x 10-5 M; [HCO3-(aq)] = 6.6 x 10-5 M

  9. HCO3-(aq) + H2O(l)  H3O+(aq) + CO32-(aq) Ka2 = [H3O+(aq)] [CO32-(aq)] / [HCO3-(aq) ] HCO3-(aq) H3O+(aq) CO32-(aq) Initial 6.6 x 10-5 6.6 x 10-5 0 Change -y 6.6 x 10-5 +y y Equilibrium 6.6 x 10-5 - y 6.6 x 10-5 + y y 4.8 x 10-11 = (6.6 x 10-5 +y) y/ (6.6 x 10-5 - y) Assume y << 6.6 x 10-5 4.8 x 10-11 = (6.6 x 10-5 y/ (6.6 x 10-5 ); y = 4.8 x 10-11 At equilibrium: [H2CO3(aq)] ≈ 0.010 M; [H3O+(aq)] = 6.6 x 10-5 M; [HCO3-(aq)] = 6.6 x 10-5 M; [CO32-(aq)] = 4.8 x 10-11 M pH = 4.18

  10. Buffers Buffer solutions : resists change in pH even with addition of small amounts of acid or base. Buffer solutions are mixed solutions: mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. Human blood has a pH maintained at pH = 7.4 due to a combination of carbonate, phosphate and protein buffers. The ocean is buffered to a pH of ~ 8.4 by buffering that depends the presence of hydrogen carbonates and silicates.

  11. Buffer Action An acid buffer is an aqueous solution of a weak acid and its conjugate base. It buffers solutions on the acid side of neutral (pH < 7). Example: solution of CH3COOH(aq) + CH3COONa(aq) CH3COOH / CH3COO- A base buffer is an aqueous solution of a weak base and its conjugate acid. It buffers solutions on the basic side of neutral (pH > 7). Example: NH3(aq) + NH4Cl(aq) NH3/ NH4+

  12. Buffer solution of CH3COOH(aq) / CH3COO- (aq) If a small amount of strong acid is added: H3O+(aq) + CH3COO-(aq)  CH3COOH(aq) + H2O(l) K for this reaction = 1/Ka(CH3COOH(aq)) = 5.5 x 104 The CH3COO-(aq) acts as a “sink” for the added protons, and the pH remains unchanged. If a small amount of strong base is added OH-(aq) + CH3COOH(aq)  CH3COO-(aq) + H2O(l) K for this reaction = 1/Kb(CH3COO-(aq)) = 1.8 x 109 The CH3COOH(aq) acts as a “sink” for the added OH-, and the pH remains unchanged

  13. [HA(aq)] [H3O+(aq)] = Ka [A-(aq)] [HA(aq)] - log [H3O+(aq)] = - log Ka - log [A-(aq)] [A- (aq)] pH = pKa + log [HA(aq)] Designing a buffer Make a solution with particular pH so that it buffers about this pH. Consider a solution of a weak acid and its conjugate base HA(aq) + H2O(l)  H3O+(aq) + A-(aq) Henderson-Hasselbach equation Note: for a solution of a weak base/conjugate acid, use Ka of the conjugate acid

  14. An optimal buffer is one in which the weak acid and its conjugate base have equal concentrations Select a weak acid that has its pKa as close as possible to the desired pH Having chosen the weak acid, use the Henderson Hasselbach equation to determine the ratio of [A-(aq)] and [HA(aq)] that will form solution that buffers around the desired pH

  15. Calculate the pH of a buffer solution that is 0.040 M CH3COONa(aq) and 0.080 M CH3COOH (aq). pKa(CH3COOH(aq)) = 4.75 CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq) CH3COOH(aq) H3O+(aq) CH3COO-(aq) Initial 0.080 M 0 0.040M Change -x x 0.040 + x Equilibrium 0.080 - x x 0.040 + x Ka = (0.040 + x) x / (0.080 - x) Assume x << 0.040 x = 3.6 x 10-5 pH = 4.44

  16. Suppose that a solution is made by dissolving 1.2 g NaOH(s) (0.030 moles) in 500 mL of the buffer solution in the previous problem. Calculate the pH of the resulting solution and the change in pH. Assume the volume of the solution to be constant. The OH-(aq) will react with CH3COOH(aq) CH3COOH(aq) + OH-(aq)  CH3COO-(aq) + H2O(l) Moles of OH- added = 0.030 moles Moles of CH3COOH(aq) = (0.500 L) (0.080 M) = 0.040 mol Amount of unreacted CH3COOH(aq) = 0.010 moles Molarity of CH3COOH(aq) = 0.020 M

  17. Moles of CH3COO-(aq) = initial amount + amount formed by reaction of OH- (aq) and CH3COOH(aq) = (0.040M x 0.500 L) + (0.030 moles) = 0.050 moles Molarity of CH3COO-(aq) = 0.10 M CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq) pH = pKa + log ([CH3COO-(aq) ] / [CH3COOH(aq) ] ) = 5.45 If the solution had contained HCl at pH = 4.4, addition of the NaOH would have raised the pH to 12.8

  18. Titrations Strong Acid - Strong Base H3O+(aq) + OH-(aq)  2H2O(l) pH changes slowly initially, changes rapidly through pH = 7 (equivalence point) and then changes slowly again If the analyte is a strong acid, pH increases as base is added

  19. If the analyte is a strong base, pH decreases as acid is added

  20. Analyte: 25.00 mL of 0.250 M NaOH(aq) Titrant: 0.340 M HCl(aq) Determine the pH of the solution when 5.00mL of titrant added Answer: pH = 13.18 Determine the amount of titrant that must be added to reach the equivalence point? Answer: 18.4 mL Determine the pH of the solution after the addition of 20.4 mL of titrant. Answer: pH = 1.82

  21. Strong Acid-Weak Base and Weak Acid - Strong Base

  22. CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l) Slow change in pH before equivalence point; solution is a buffer CH3COOH(aq)/CH3COO-(aq) At halfway point [HA] = [A-] pH = pKa At equivalence, pH determined by CH3COO-(aq)

  23. Changes in pH during a titration of a weak acid/base with a strong base/acid: Halfway to the stoichiometric point, the pH = pKa of the acid The pH is greater than 7 at the equivalence point of the titration of a weak acid and strong base The pH is less that 7 at the equivalence point of the titration of a weak base and strong acid Beyond the equivalence point, the excess strong acid or base will determine the pH of the solution

  24. Titration of 100.0 mL of 0.1000 M CH3COOH(aq) with 0.1000 M NaOH Before addition of NaOH: pH determined by CH3COOH(aq) CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq) Answer: pH = 2.88 Before the equivalence point: determine pH for a buffer Addition of 30.00 mL of NaOH(aq) The OH-(aq) reacts with the CH3COOH(aq). Determine concentration of CH3COOH(aq) and CH3COO- (aq) in solution after addition of the base. Answer: pH = 4.38 At half equivalence: [CH3COOH(aq)] = [CH3COO-(aq)] pH = pKa

  25. At equivalence: enough OH-(aq) added to react with all CH3COOH(aq). For this problem, equivalence is reached when 100.0mL of OH- is added; i.e. 0.01000 moles of OH-(aq) added Solution contains 0.01000 moles CH3COO-(aq) in 200.0 mL solution; [CH3COO-(aq)] = 0.05000 M pH determined by CH3COO-(aq) + H2O(l)  CH3COOH(aq) + OH- (aq) pH = 8.72 (note greater than 7.0) Beyond equivalence: pH determined by excess OH-(aq)

  26. Estimate the pH at the equivalence point of the titration of 25.00 mL of 0.100 M HCOOH(aq) with 0.150 M NaOH(aq) (Ka(HCOOH) = 1.8 x 10-4) At the equivalence point, enough NaOH(aq) has been added to react with all the HCOOH(aq) forming HCOO-(aq) The reaction: HCOO-(aq) + H2O(l)  HCOOH(aq) + OH- determines the pH at equivalence Answer: 8.26

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