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Equilibrium In Solutions Of Weak Acids And Weak Bases

Equilibrium In Solutions Of Weak Acids And Weak Bases. weak acid: HA + H 2 O  H 3 O + + A - [H 3 O + ][A - ] K a = [HA] weak base: B + H 2 O  HB + + OH - [HB + ][OH - ] K b = [B]. Some Acid-Base Equilibrium Calculations.

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Equilibrium In Solutions Of Weak Acids And Weak Bases

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  1. http:\\asadipour.kmu.ac.ir........57 slides

  2. Equilibrium In Solutions Of Weak Acids And Weak Bases weak acid: HA + H2O H3O+ + A- [H3O+][A-] Ka = [HA] weak base: B + H2O  HB+ + OH- [HB+][OH-] Kb = [B] http:\\asadipour.kmu.ac.ir........57 slides

  3. Some Acid-Base Equilibrium Calculations • cHA≈[HA] [H3O+][A-] [H3O+][A-] Ka = --------------------= ---------------- cHA–[H3O+]cHA • cHA> [HA] Analytical C> Equilibrium C • - the calculations can be simplified. • - When Macid/KaorMbase/Kb > 100, • - When Ka or Kb<1×10-4 (In usual Conc.) http:\\asadipour.kmu.ac.ir........57 slides

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  5. An Example 1.Determine the concentrations of H3O+, CH3COOH and CH3COO-, and the pH of 1.00 M CH3COOH solution. Ka = 1.8 x 10-5. 2. What is the pH of a solution that is 0.200 M in methylamine, CH3NH2? Kb = 4.2 x 10-4. http:\\asadipour.kmu.ac.ir........57 slides

  6. Are Salts Neutral, Acidic or Basic? Salts are ionic compounds formed in the reaction between an acid and a base. 1. NaCl Na+ is from NaOH , a strong base Cl- is from HCl, a strong acid H2O NaCl (s) → Na+ (aq) + Cl- (aq) Na+ and Cl- ions do not react with water. The solution is neutral. http:\\asadipour.kmu.ac.ir........57 slides

  7. Are Salts Neutral, Acidic or Basic? 2. KCN K+ is from KOH , a strong base CN- is from HCN, a weak acid H2O KCN (s) → K+ (aq) + CN- (aq) K+ ions do not react with water, but CN- ions do. CN- + H2O  HCN + OH-hydrolysis The OH- ions are produced, so the solution is basic. http:\\asadipour.kmu.ac.ir........57 slides

  8. Are Salts Neutral, Acidic or Basic? 3. NH4Cl NH4+ is from NH3 , a weak base Cl- is from HCl, a strong acid H2O NH4Cl (s) → NH4+ (aq) + Cl- (aq) Cl- ions do not react with water, but NH4+ ions do. NH4+ + H2O  H3O+ + NH3 hydrolysis The H3O+ ions are produced, so the solution is acdic. http:\\asadipour.kmu.ac.ir........57 slides

  9. Are Salts Neutral, Acidic or Basic? 3. NH4CN NH4+ is from NH3 , a weak base CN- is from HCN, a weak acid H2O NH4CN (s) → NH4+ (aq) + CN- (aq) NH4+ + H2O  H3O+ + NH3 Kahydrolysis CN- + H2O  HCN + OH- Kbhydrolysis (Ka>Kb ,Acidic)’’’(Ka<Kb,Basic)‘’’(Ka=Kb,Nutral) http:\\asadipour.kmu.ac.ir........57 slides

  10. Ions As Acids And Bases Certain and anion ions can cause an aqueous solution to become acidic or basic due to hydrolysis. • Salts of strong acids and strong bases form neutral solutions. • Salts of weak acids and strong bases form basic solutions. • Salts of strong acids and weak bases form acidic solutions. • Salts of weak acids and weak bases form solutions that are acidic in some cases, neutral or basic in others. http:\\asadipour.kmu.ac.ir........57 slides

  11. Strong Acids And Strong Bases Strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4 Strong bases: Group IA and IIA hydroxides http:\\asadipour.kmu.ac.ir........57 slides

  12. An Example Indicate whether the solutions (a) Na2S and (b) KClO4 are acidic, basic or neutral. http:\\asadipour.kmu.ac.ir........57 slides

  13. The pH Of a Salt Solution What is the pH of 0.1M NaCN solution? What is the pH of 0.1M NH4Cl solution? What is the pH of 0.1M NH4CN solution? Ka of HCN=1.0×10-9. Kb for NH3=1.0×10-5 Ka xKb = Kw so, Kb = Kw/Ka http:\\asadipour.kmu.ac.ir........57 slides

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  15. Common Ion Effect Illustrated CH3COOH  CH3COO- + H+ blue-violet: pH > 4.6 yellow: pH < 3.0 ((1.00 M CH3COOH)) ((1.00 M CH3COOH + 1.00 M CH3COONa)) http:\\asadipour.kmu.ac.ir........57 slides

  16. The Common Ion Effect Calculate the pH of 0.10 M CH3COOH solution. Ka of CH3COOH=1.0×10-5 Calculate the pH of 0.10 M CH3COONa solution. Calculate the pH of 0.10 M CH3COOH/ 0.10 M CH3COONa solution. http:\\asadipour.kmu.ac.ir........57 slides

  17. Depicting Buffer Action http:\\asadipour.kmu.ac.ir........57 slides

  18. 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 CH3COOH  CH3COO- Acidic buffer NH3  NH4+Alkalin buffer http:\\asadipour.kmu.ac.ir........57 slides

  19. 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+. • CH3COOH  CH3COO- + H+ http:\\asadipour.kmu.ac.ir........57 slides

  20. Ionization constant of an acid Taking log of the equation on both sides, http:\\asadipour.kmu.ac.ir........57 slides

  21. Ionization constant of an acid Multiplying both sides of the equation by -1 Henderson-Hasselbach equation http:\\asadipour.kmu.ac.ir........57 slides

  22. Henderson-Hasselbalch Equation For Buff Solutions [conjugate base] pH = pKa + log [conjugate acid] If [conjugate acid] = [conjugate base], pH = pKa Requirement: • [B] / [A] between 0.10 and 10 http:\\asadipour.kmu.ac.ir........57 slides

  23. Buffer Capacity • There is a limit to the capacity of a buffer solution to neutralize added acid or base, and this limit is reached before all of one of the buffer components has been consumed. • In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize. • As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal. [B]=[A] • [Buffer]=[Acid]+[Base] http:\\asadipour.kmu.ac.ir........57 slides

  24. Buffer Capacity • [Buffer]=[Acid]+[Base] • [Acid]↑ & [Base]↑Capacity↑ • In equimolarbuffersis is important • Capacity↑ http:\\asadipour.kmu.ac.ir........57 slides

  25. Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH3 (pKb=5)and 1.0 M NH4Cl. • What is the pH of this buffer? http:\\asadipour.kmu.ac.ir........57 slides

  26. Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH3 (pKb=5)and 1.0 M NH4Cl. • What is the pH of this buffer? • If 5 mmolNaOHis added to 0.500 L of this solution, what will be the pH? http:\\asadipour.kmu.ac.ir........57 slides

  27. Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH3 (pKb=5)and 1.0 M NH4Cl. • What is the pH of this buffer? • If 5 mmolNaOH is added to 0.500 L of this solution, what will be the pH? • If 5 mmolHClis added to 0.500 L of this solution, what will be the pH? http:\\asadipour.kmu.ac.ir........57 slides

  28. Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH3 (pKb=5)and 1.0 M NH4Cl. • What is the pH of this buffer? • If 5 mmolNaOH is added to 0.500 L of this solution, what will be the pH? • If 5 mmolHCl is added to 0.500 L of this solution, what will be the pH? • If 5 mmol NH4Cl is added to 0.500 L of this solution, what will be the pH? http:\\asadipour.kmu.ac.ir........57 slides

  29. Calculations in Buffer Solutions 2) What concentration of acetate ion in 500 ml of 0.500 M CH3COOH (pKa=5) produces a buffer solution with pH = 4.00? http:\\asadipour.kmu.ac.ir........57 slides

  30. Calculations in Buffer Solutions 2) What concentration of acetate ion in500 ml of 0.500 M CH3COOH (pKa=5) produces a buffer solution with pH = 4.00? How many mg? http:\\asadipour.kmu.ac.ir........57 slides

  31. Acid-Base Indicators • An acid-base indicator is a weak acid having one color and the conjugate base of the acid having a different color. One of the “colors” may be colorless. HIn + H2OH3O+ + In- • Acid-base indicators are often used for applications in which a precise pH reading isn’t necessary. • A common indicator used in chemistry laboratories is Phenolphetalein. http:\\asadipour.kmu.ac.ir........57 slides

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  34. Neutralization Reactions • Neutralization is the reaction of an acid and a base. • Titration is a common technique for conducting a neutralization. • At the equivalence point in a titration, the acid and base have been brought together in exact stoichiometric proportions. • The point in the titration at which the indicator changes color is called the end point. • The indicator endpoint and the equivalence point for a neutralization reaction can be best matched by plotting a titration curve, a graph of pH versus volume of titrant. • In a typical titration, 50 mL or less of titrant that is 1 M or less is used. http:\\asadipour.kmu.ac.ir........57 slides

  35. Drawing titration Curve ForStrong Acid - Strong Base HCl + NaOH→ NaCl +H2O Calculate the pH at the some points and draw the curve. 4 essential points. 1)initial point 2)equivalence point 3)before the equivalence point 4)beyond the equivalence point http:\\asadipour.kmu.ac.ir........57 slides

  36. Drawing titration Curve ForStrong Acid - Strong Base HCl + NaOH→ NaCl +H2O 4 questions. 1)What are the present compounds? 2)Which of them is effective on pH? 3)How much are the concentrations? 4)What is the relationship between their Conc. And pH? http:\\asadipour.kmu.ac.ir........57 slides

  37. Drawing titration Curve ForStrong Acid - Strong Base HCl + NaOH→ NaCl +H2O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. • initial pH. (Before the addition of any NaOH) . Answer Q1. There are:HCl & H2O Answer Q2.HCl Answer Q3. [HCl] Answer Q4. pH=-log[H+] http:\\asadipour.kmu.ac.ir........57 slides

  38. Drawing titration Curve ForStrong Acid - Strong Base HCl + NaOH→ NaCl +H2O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. b)equivalence point. Answer Q1. There are:NaCl & H2O Answer Q2. H2O Answer Q3. Answer Q4. pH=7 http:\\asadipour.kmu.ac.ir........57 slides

  39. Drawing titration Curve ForStrong Acid - Strong Base HCl + NaOH→ NaCl +H2O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. c)before the equivalence point. Answer Q1. There are:HCl,NaCl & H2O Answer Q2.HCl Answer Q3. Answer Q4. [H+]=NpH=-log[H+] http:\\asadipour.kmu.ac.ir........57 slides

  40. Drawing titration Curve ForStrong Acid - Strong Base HCl + NaOH→ NaCl +H2O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. d)after the equivalence point. Answer Q1. There are:NaOH,NaCl & H2O Answer Q2.NaOH Answer Q3. Answer Q4. [OH-]=N pOH=-log[OH-] pH=14-pOH http:\\asadipour.kmu.ac.ir........57 slides

  41. Titration Curve ForStrong Acid - Strong Base pH is low at the beginning. pH changes slowly until just before equivalence point. pH changes sharply around equivalence point. pH = 7.0 at equivalence point. Further beyond equivalence point, pH changes slowly. Any indicator whose color changes in pH range of 4 – 10 can be used in titration. http:\\asadipour.kmu.ac.ir........57 slides

  42. Drawing titration Curve Forweak acid- Strong Base CH3COOH + NaOH→ CH3COO- + Na+ +H2O Calculate the pH at the some points and draw the curve. Ka=1×10-5 5 essential points. 1)initial point 2)equivalence point 3)beyond the initial point 4)before the equivalence point 5)beyond the equivalence point http:\\asadipour.kmu.ac.ir........57 slides

  43. Drawing titration Curve Forweak acid- Strong Base CH3COOH + NaOH→ CH3COO- + Na+ +H2O 4 questions. 1)What are the present compounds? 2)Which of them is effective on pH? 3)How much are the concentrations? 4)What is the relationship between their Conc. And pH? http:\\asadipour.kmu.ac.ir........57 slides

  44. Drawing titration Curve Forweak acid- Strong Base CH3COOH + NaOH→ CH3COO- + Na+ +H2O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH. • initial pH. (Before the addition of any NaOH) . Answer Q1. There are: CH3COOH & H2O Answer Q2. CH3OOH Answer Q3. CH3OOH Answer Q4. pH=-log[H+] http:\\asadipour.kmu.ac.ir........57 slides

  45. Drawing titration Curve Forweak acid- Strong Base CH3COOH + NaOH→ CH3COO- + Na+ +H2O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH. b)equivalence point. Answer Q1. There are:CH3COO- , Na+ & H2O Answer Q2.CH3COO- Answer Q3. Answer Q4. pOH=-log[OH-] Ka×Kb=Kw http:\\asadipour.kmu.ac.ir........57 slides

  46. Drawing titration Curve Forweak acid- Strong Base CH3COOH + NaOH→ CH3COO- + Na+ +H2O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH. c)beyond the initial point. Answer Q1. There are: CH3COOH,CH3COO- ,Na+ & H2O Answer Q2. CH3COOH,CH3COO- Answer Q3. Answer Q4. http:\\asadipour.kmu.ac.ir........57 slides

  47. Drawing titration Curve Forweak acid- Strong Base CH3COOH + NaOH→ CH3COO- + Na+ +H2O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH. d)before the equivalence point. Answer Q1. There are: CH3COOH,CH3COO- ,Na+ & H2O Answer Q2. CH3COOH,CH3COO- Answer Q3. Answer Q4. http:\\asadipour.kmu.ac.ir........57 slides

  48. Drawing titration Curve Forweak acid- Strong Base CH3COOH + NaOH→ CH3COO- + Na+ +H2O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH. e)after the equivalence point. Answer Q1. There are:NaOH,CH3COO- , Na+ & H2O Answer Q2.NaOH Answer Q3. Answer Q4. [OH-]=N pOH=-log[OH-] pH=14-pOH http:\\asadipour.kmu.ac.ir........57 slides

  49. Titration Curve ForWeak Acid - Strong Base The initial pH is higher because weak acid is partially ionized. At the half-neutralization point, pH = pKa. pH >7 at equivalence point because the anion of the weak acid hydrolyzes. The steep portion of titration curve around equivalence point has a smaller pH range. The choice of indicators for the titration is more limited. http:\\asadipour.kmu.ac.ir........57 slides

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