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How Much Does the pH of a Buffer Change When an Acid or Base Is Added?

How Much Does the pH of a Buffer Change When an Acid or Base Is Added?. though buffers do resist change in pH when acid or base are added to them, their pH does change calculating the new pH after adding acid or base requires breaking the problem into 2 parts

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How Much Does the pH of a Buffer Change When an Acid or Base Is Added?

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  1. How Much Does the pH of a Buffer Change When an Acid or Base Is Added? • though buffers do resist change in pH when acid or base are added to them, their pH does change • calculating the new pH after adding acid or base requires breaking the problem into 2 parts • a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other • added acid reacts with the A− to make more HA • added base reacts with the HA to make more A− • an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]

  2. Compare the effect on pH of adding 0.010 molNaOH to a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to adding 0.010 molNaOH to 1.00 L of pure water? Pure Water Buffer HC2H3O2 + H2O C2H3O2-+ H3O+ pKa for HC2H3O2 = 4.745

  3. H2O(l) + NH3(aq) NH4+(aq) + OH−(aq) Basic Buffers B:(aq) + H2O(l) H:B+(aq) + OH−(aq) • buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, (H:B+)Cl−

  4. What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl? NH3 + H2O NH4+ + OH−

  5. Buffering Effectiveness • a good buffer should be able to neutralize moderate amounts of added acid or base • however, there is a limit to how much can be added before the pH changes significantly • the buffering capacity is the amount of acid or base a buffer can neutralize • the buffering range is the pH range the buffer can be effective • the effectiveness of a buffer depends on two factors • (1) the relative amounts of acid and base • (2) the absolute concentrations of acid and base

  6. Effect of Relative Amounts of Acid and Conjugate Base a buffer is most effective with equal concentrations of acid and base pH change after adding 0.010M NaOH Buffer 2 0.18 mol HA & 0.020 mol A- Initial pH = 4.05 Buffer 1 0.100 mol HA & 0.100 mol A- Initial pH = 5.00 pKa (HA) = 5.00 HA + OH−A- + H2O after adding 0.010 mol NaOH pH = 5.09 after adding 0.010 mol NaOH pH = 4.25

  7. Effect of Absolute Concentrations of Acid and Conjugate Base a buffer is most effective when the concentrations of acid and base are largest pH change after adding 0.010M NaOH Buffer 1 0.50 mol HA & 0.50 mol A- Initial pH = 5.00 Buffer 2 0.050 mol HA & 0.050 mol A- Initial pH = 5.00 pKa (HA) = 5.00 HA + OH−A- + H2O after adding 0.010 mol NaOH pH = 5.02 after adding 0.010 mol NaOH pH = 5.18

  8. Effectiveness of Buffers • a buffer will be most effective when the [base]:[acid] = 1 • equal concentrations of acid and base • effective when 0.1 < [base]/[acid] < 10 • a buffer will be most effective when the [acid] and the [base] are large

  9. Buffering Range • A buffer will be effective when 0.1 < [base]:[acid] < 10 • substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum pH at which the buffer will be effective Highest pH Lowest pH therefore, the effective pH range of a buffer is pKa ± 1 when choosing an acid to make a buffer, choose one whose is pKa is closest to the pH of the buffer

  10. Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous Acid, HClO2pKa = 1.95 Nitrous Acid, HNO2pKa = 3.34 Formic Acid, HCHO2pKa= 3.74 Hypochlorous Acid, HClOpKa = 7.54

  11. Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous Acid, HClO2pKa = 1.95 Nitrous Acid, HNO2pKa = 3.34 Formic Acid, HCHO2pKa= 3.74 Hypochlorous Acid, HClOpKa = 7.54 The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range.

  12. What ratio of NaCHO2 : HCHO2 would be required to make a buffer with pH 4.25? Formic Acid, HCHO2, pKa = 3.74 to make the buffer with pH 4.25, you would use 3.24 times as much NaCHO2 as HCHO2

  13. Buffering Capacity • buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness • the buffering capacity increases with increasing absolute concentration of the buffer components • as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves • buffers that need to work mainly with added acid generally have [base] > [acid] • buffers that need to work mainly with added base generally have [acid] > [base]

  14. Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer

  15. Titration • in an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete • when the reaction is complete we have reached the endpoint of the titration • an indicator may be added to determine the endpoint • an indicator is a chemical that changes color when the pH changes • when the moles of H3O+ = moles of OH−, the titration has reached its equivalence point

  16. Titration

  17. Titration Curve • a plot of pH vs. amount of added titrant • the inflection point of the curve is the equivalence point of the titration • prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH • the pH of the equivalence point depends on the pH of the salt solution • equivalence point of neutral salt, pH = 7 • equivalence point of acidic salt, pH < 7 • equivalence point of basic salt, pH > 7 • beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH

  18. Titration Curve:Unknown Strong Base Added to Strong Acid

  19. added 30.0 mL NaOH 0.00050 mol NaOH pH = 11.96 added 35.0 mL NaOH 0.00100 mol NaOH pH = 12.22 25.0 mL 0.100 M HCl 0.00250 mol HCl pH = 1.00 added 5.0 mL NaOH 0.00200 mol HCl pH = 1.18 added 10.0 mL NaOH 0.00150 mol HCl pH = 1.37 added 25.0 mL NaOH equivalence point pH = 7.00 Adding NaOH to HCl added 15.0 mL NaOH 0.00100 mol HCl pH = 1.60 added 40.0 mL NaOH 0.00150 mol NaOH pH = 12.36 added 20.0 mL NaOH 0.00050 mol HCl pH = 1.95 added 50.0 mL NaOH 0.00250 mol NaOH pH = 12.52

  20. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH Ka = 1.8 x 10-4 • HCHO2(aq) + NaOH(aq)NaCHO2(aq) + H2O(aq) • What is the Initial pH before adding any NaOH?

  21. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH • HCHO2(aq) + NaOH(aq)NaCHO2(aq) + H2O(aq) • initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3 • What is the pH after adding 5.0 mL 0.100M NaOH?

  22. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH • HCHO2(aq) + NaOH(aq)NaCHO2(aq) + H2O(aq):Kb = 5.6 x 10-11 • initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3 • at equivalence (25 mL of NaOH added [OH-] = 1.7 x 10-6 M

  23. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH • HCHO2(aq) + NaOH(aq)NaCHO2(aq) + H2O(aq) • Added 30 mL NaOH (after equivalence point) 5.0 x 10-4 mol NaOH xs

  24. added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 initial HCHO2 solution 0.00250 mol HCHO2 pH = 2.37 added 5.0 mL NaOH 0.00200 mol HCHO2 pH = 3.14 added 10.0 mL NaOH 0.00150 mol HCHO2 pH = 3.56 added 25.0 mL NaOH equivalence point 0.00250 mol CHO2− [CHO2−]init = 0.0500 M [OH−]eq = 1.7 x 10-6 pH = 8.23 Adding NaOH to HCHO2 added 12.5 mL NaOH 0.00125 mol HCHO2 pH = 3.74 = pKa half-neutralization added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 15.0 mL NaOH 0.00100 mol HCHO2 pH = 3.92 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52 added 20.0 mL NaOH 0.00050 mol HCHO2 pH = 4.34

  25. Titrating Weak Acid with a Strong Base • the initial pH is that of the weak acid solution • calculate like a weak acid equilibrium problem • before the equivalence point, the solution becomes a buffer • calculate mol HAinit and mol A−init using reaction stoichiometry • calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−init • half-neutralization pH = pKa

  26. Titrating Weak Acid with a Strong Base • at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established • mol A− = original mole HA • [A−]init = mol A−/total liters • calculate like a weak base equilibrium problem • beyond equivalence point, the OH is in excess • [OH−] = mol MOH xs/total liters • [H3O+][OH−]=1 x 10-14

  27. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the equivalence point HNO2 + KOH NO2-+ H2O

  28. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH HNO2 + KOH NO2-+ H2O 0.00300 0.00100

  29. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH HNO2 + H2O NO2-+ H3O+ Look up Ka= 4.6 x 10-4

  30. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point HNO2 + H2O NO2-+ H3O+ at half-equivalence, moles KOH = ½ mole HNO2 0.00200 0.00200

  31. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point HNO2 + H2O NO2-+ H3O+ Look up Ka = 4.6 x 10-4

  32. Titration Curve of a Weak Base with a Strong Acid

  33. Titration of a Polyprotic Acid • if Ka1 >> Ka2, there will be two equivalence points in the titration • the closer the Ka’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH

  34. Monitoring pH During a Titration • the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+] • using a probe that specifically measures just H3O+ • the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve • if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with anindicator

  35. Monitoring pH During a Titration

  36. Indicators • many dyes change color depending on the pH of the solution • these dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution HInd(aq) + H2O(l)Ind-(aq) + H3O+(aq) • the color of the solution depends on the relative concentrations of Ind-:HInd≈ 1, the color will be mix of the colors of Ind- and HInd • when Ind-:HInd> 10, the color will be mix of the colors of Ind- • when Ind-:HInd< 0.1, the color will be mix of the colors of HInd

  37. Phenolphthalein

  38. Methyl Red

  39. Monitoring a Titration with an Indicator • for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point • an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH • pKa of HInd≈ pH at equivalence point

  40. Acid-Base Indicators

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