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  • 1. Teacher Prep
    • Distilled water
    • Buffer solution
    • Universal indicator
  • 2. Acid-Base Titrations
  • 3. Titrations Purpose: To determine the unknown concentration of the acid.
  • 4. Titration Curve
  • 5. Indicator Ranges
  • 6. Indicator Ranges
  • 7. Indicators
    • How do indicators work?
    • Through equilibrium!
    Phenolphthalein + H +
  • 8. Basic Calculations (Gr. 11)
    • Example #1
    • If it takes 54.0 mL of 0.1 M NaOH to neutralize 125.0 mL of an HCl solution. What is the concentration of the HCl?
    NaOH (aq) + HCl (aq)  NaCl (aq) + H 2 O (l) n NaOH =CV n NaOH =(0.1M)(0.0540L) n NaOH =5.4x10 -3 M NaOH = n HCl C HCl = n V C HCl = 5.4x10 -3 0.1250L C HCl = 0.0432M .: [HCl] = 0.0432M
  • 9. Example #2
    • What is the pH of the final solution where 30.0 mL of 0.1 M NaOH is mixed with 18.0 mL of 0.5 M HCl?
    • Do you have a good guess as to whether the final solution is acidic or basic?
    NaOH (aq) + HCl (aq)  NaCl (aq) + H 2 O (l) n NaOH =CV n NaOH =(0.1M)(0.0300L) n NaOH =0.00300mol n HCl =CV n HCl =(0.5M)(0.0180L) n HCl =0.00900mol n HCl leftover =0.00900mol-0.00300mol n HCl leftover =0.00600mol HCl leftover pH = -log [H + ] pH = -log [0.125M] pH = 0.9 .: the pH is 0.9 C= n HCl leftover = 0.00600mol = 0.125M V 0.048L
  • 10. Example #2
    • What is happening at the chemical level?
    • NaOH  Na+ + OH-
    • HCl  H+ + Cl-
    • The OH - and H + ions come together to produce H 2 O. The remaining ions will determine the final pH of the solution.
  • 11. STRONG-WEAK TITRATIONS
  • 12. Strong-Weak Titration Curves
  • 13. Buffers
    • Buffer solutions contain a mixture of an acid and its conjugate base or a base with its conjugate acid.
    • Due to the presence of the conjugate acid-base pair, the addition of more acid or base will not cause the pH to drastically change.
  • 14. Buffers
    • Buffers are important in biological systems where large changes in pH could be detrimental to the organism.
    • Almost all organic substances in our body act as a buffer.
    • H 2 O + CO 2 <===> H 2 CO 3 <===> H + + HCO 3 2-
  • 15. Buffers
  • 16. Strong-Weak Titrations
    • Example #3
    • What happens when you add a strong base to a weak acid?
    • CH 3 COOH <===> CH 3 COO - + H +
    • NaOH  Na + + OH -
  • 17. Strong-Weak Titrations
    • Two-step process:
    • All OH - ions from the strong base will react with H + to produce H 2 O.
    • Equilibrium is shifted to the right.
    • pH is then calculated from the new equilibrium based on the remaining [H + ].
  • 18. Strong Weak Titrations
    • Example #4
    • What happens when you add a strong acid to a weak base?
    • NH 3 + H 2 O <===> NH 4 + + OH -
    • HCl  H + + Cl -
  • 19. Strong-Weak Titrations
    • Two-step process:
    • All H + ions from the strong acid will react with OH - to produce H 2 O.
    • Equilibrium is shifted to the right.
    • pH is then calculated from the new equilibrium based on the remaining [H + ].
  • 20. Calculations In a titration, 20.0 mL of 0.300 M HCl(aq) is titrated with 0.300 M NaOH(aq). What is the pH of the solution after the following volumes of NaOH(aq) have been added? a) 0.00 mL b) 10.0 mL c) 20.0 mL d) 30.0 mL NaOH (aq) + HCl (aq)  NaCl (aq) + H 2 O (l) a) pH = -log [H + ] pH = -log [0.300] pH = 0.5 b) n HCl =0.006mol n NaOH =CV n NaOH =(0.300M)(0.010L) n NaOH =0.003 mol n HCl leftover =0.006mol-0.003mol n HCl leftover =0.003mol pH = -log [H + ] pH = -log [0.1] pH = 1.00 C=n/V total C=0.003mol/0.030L C=0.1M
  • 21. Calculations In a titration, 20.0 mL of 0.300 M HCl(aq) is titrated with 0.300 M NaOH(aq). What is the pH of the solution after the following volumes of NaOH(aq) have been added? a) 0.00 mL b) 10.0 mL c) 20.0 mL d) 30.0 mL NaOH (aq) + HCl (aq)  NaCl (aq) + H 2 O (l) c) n HCl =0.006mol n NaOH =CV n NaOH =(0.300M)(0.020L) n NaOH =0.006 mol n HCl leftover =0mol pH = -log [H + ] pH = -log [1.0x10 -7 ] pH = 7 d) n HCl =0.006mol n NaOH =CV n NaOH =(0.300M)(0.030L) n NaOH =0.009 mol n NaOH leftover =0.009mol-0.006mol n NaOH leftover =0.003mol C=n/V total C=0.003mol/0.050L C=0.06M
  • 22. Calculations In a titration, 20.0 mL of 0.300 M HCl(aq) is titrated with 0.300 M NaOH(aq). What is the pH of the solution after the following volumes of NaOH(aq) have been added? a) 0.00 mL b) 10.0 mL c) 20.0 mL d) 30.0 mL NaOH (aq) + HCl (aq)  NaCl (aq) + H 2 O (l) d) pOH = -log [OH - ] pOH = -log [0.06] pOH = 1.22 pH = 12.8
  • 23. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL HC 2 H 3 O 2(aq) <===> H + (aq) + C 2 H 3 O 2 - (aq) a) I 0.300M 0 0 C -x +x +x E 0.300-x x x K a = [H + (aq) ][C 2 H 3 O 2 - (aq) ] [HC 2 H 3 O 2(aq) ] 1.8x10 -5 = [x][x] [0.300-x]  Assumption used 2.32379x10 -3 = x pH = -log [H + ] pH = -log [2.32x10 -3 ] pH = 2.63 .: pH = 2.63
  • 24. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL HC 2 H 3 O 2(aq) <===> H + (aq) + C 2 H 3 O 2 - (aq) b) NaOH  Na + (aq) + OH - (aq) V total =0.030L n NaOH =CV n NaOH =(0.3M)(0.010L) n NaOH = 0.003 mol  so 0.003 mol of HC 2 H 3 O 2 are used  so 0.003 mol of C 2 H 3 O 2 - are formed n HC 2 H 3 O 2 =(0.300M)(0.020L) n HC 2 H 3 O 2 =0.006mol  so 0.003 mol of HC 2 H 3 O 2 remain C=n/V total C=0.003mol/0.030L C=0.1M
  • 25. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL HC 2 H 3 O 2(aq) <===> H + (aq) + C 2 H 3 O 2 - (aq) C=0.1M I 0.1 0 0.1 C -x +x +x E 0.1-x x 0.1+x 1.8x10 -5 = [x][0.1+x] [0.1-x]  Assumption used 1.8x10 -5 = [x][0.1] [0.1] 1.8x10 -5 = x pH = -log [H + ] pH = -log [1.8x10 -5 ] pH = 4.74 .: pH = 4.74 b)
  • 26. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL HC 2 H 3 O 2(aq) <===> H + (aq) + C 2 H 3 O 2 - (aq) c) NaOH  Na + (aq) + OH - (aq) At equivalence point, the #mol acid = #mol base n HC 2 H 3 O 2 =(0.300M)(0.020L) = 0.006mol = n NaOH V= n NaOH C V= 0.006mol 0.3M V=0.02L  V total =0.040L Since n acid =n conjugate base , 0.006mol of C 2 H 3 O 2 - were formed
  • 27. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL C 2 H 3 O 2 - (aq) + H 2 O (l) <===> OH - (aq) + HC 2 H 3 O 2(aq) c)  V total =0.040L I 0.15 0 0 C -x +x +x E 0.15-x x x  C= n V C= 0.006mol 0.040L C=0.15M K b = [OH - (aq) ][HC 2 H 3 O 2(aq) ] [C 2 H 3 O 2 - (aq) ] K b = [x][x] [0.15-x] What is the K b value?
  • 28. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL C 2 H 3 O 2 - (aq) + H 2 O (l) <===> OH - (aq) + HC 2 H 3 O 2(aq) c) I 0.15 0 0 C -x +x +x E 0.15-x x x 5.555556x10 -10 = [x][x] [0.15-x] K a x K b = K w K b = K w K a K b = 1.0x10 -14 1.8x10 -5 K b = 5.555556x10 -10  Assumption used 5.555556x10 -10 = [x][x] [0.15] 9.128709x10 -6 = x 9.1287x10 -6 9.1287x10 -6 0.15
  • 29. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL C 2 H 3 O 2 - (aq) + H 2 O (l) <===> OH - (aq) + HC 2 H 3 O 2(aq) c) I 0.15 0 0 C -x +x +x E 0.15-x x x 9.1287x10 -6 9.1287x10 -6 0.15 pOH = -log [OH - ] pOH = -log [9.128709x10 -6 ] pOH = 5.03959 pH = 14-5.03959 pH = 8.96 .: pH = 8.96
  • 30. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL C 2 H 3 O 2 - (aq) + H 2 O (l) <===> OH - (aq) + HC 2 H 3 O 2(aq) d) n NaOH =CV n NaOH =(0.3M)(0.030L) n NaOH =0.009mol n HC 2 H 3 O 2 =(0.300M)(0.020L) = 0.006mol  0.003mol of NaOH will remain after equivalence point has been reached
  • 31. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL C 2 H 3 O 2 - (aq) + H 2 O (l) <===> OH - (aq) + HC 2 H 3 O 2(aq) d) n NaOH remaining = 0.003mol  V total =0.050L  C = n NaOH V C = 0.003mol 0.050L C = 0.06M n C 2 H 3 O 2 - = 0.006mol C = n C 2 H 3 O 2 - V C = 0.006mol 0.050L C = 0.12M
  • 32. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL C 2 H 3 O 2 - (aq) + H 2 O (l) <===> OH - (aq) + HC 2 H 3 O 2(aq) d) I 0.12 0.06 0 C -x +x +x E 0.12-x 0.06+x x K b = 5.555556x10 -10 = [0.06+x][x] [0.12-x]  Assumption used 5.555556x10 -10 = [0.06][x] [0.12] 1.1111x10 -9 = x 1.1111x10 -9 0.06 0.12
  • 33. Calculations In a titration 20.0 mL of 0.300 M HC 2 H 3 O 2(aq) with 0.3 M NaOH (aq) , what is the pH of the solution after the following volumes of NaOH (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL C 2 H 3 O 2 - (aq) + H 2 O (l) <===> OH - (aq) + HC 2 H 3 O 2(aq) d) I 0.12 0.06 0 C -x +x +x E 0.12-x 0.06+x x 1.1111x10 -9 0.06 0.12 pOH = -log [OH - ] pOH = -log [0.06] pOH = 1.2218 pH = 14-1.2218 pH = 12.78 .: pH = 12.8
  • 34. Calculations In a titration 20.0 mL of 0.1 M NH 3(aq) is titrated with 0.1 M HCl (aq) . What is the pH of the resulting solution after the following volumes of HCl (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL NH 3(aq) + H 2 O (l) <===> OH - (aq) + NH 4 + (aq) a) I 0.1M 0 0 C -x +x +x E 0.1-x x x K b = [OH - (aq) ][NH 4 + (aq) ] [NH 3(aq) ] 1.8x10 -5 = [x][x] [0.1-x]  Use assumption 1.34164x10 -3 = x pOH = -log [OH - ] pOH = -log [1.34x10 -3 ] pOH = 2.87 pH = 14 - 2.87 pH = 11.13 .: pH = 11.13
  • 35. Calculations In a titration 20.0 mL of 0.1 M NH 3(aq) is titrated with 0.1 M HCl (aq) . What is the pH of the resulting solution after the following volumes of HCl (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL NH 3(aq) + H 2 O (l) <===> OH - (aq) + NH 4 + (aq) b) n NH 3 =CV n NH 3 =(0.1M)(0.020L) n NH 3 =0.002 mol n HCl =CV n HCl =(0.1M)(0.010L) HCl (aq)  H + (aq) + OH - (aq) n HCl =0.001mol n NH 4 + formed =0.001 mol n NH 3 remaining =0.001 mol V total =0.030L C=n/V total C=0.001mol/0.030L C=0.0333M
  • 36. Calculations In a titration 20.0 mL of 0.1 M NH 3(aq) is titrated with 0.1 M HCl (aq) . What is the pH of the resulting solution after the following volumes of HCl (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL NH 3(aq) + H 2 O (l) <===> OH - (aq) + NH 4 + (aq) b) I 0.0333 0 0.0333 C -x +x +x E 0.0333-x x 0.0333+x 1.8x10 -5 = [x][0.0333+x] [0.0333-x]  Use assumption 1.8x10 -5 = [x][0.0333] [0.0333] 1.8x10 -5 = x pOH = -log [OH - ] pOH = -log [1.8x10 -5 ] pOH = 4.74 .: pH = 9.26 pH = 14 - 4.74 pH = 9.26
  • 37. Calculations In a titration 20.0 mL of 0.1 M NH 3(aq) is titrated with 0.1 M HCl (aq) . What is the pH of the resulting solution after the following volumes of HCl (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL NH 3(aq) + H 2 O (l) <===> OH - (aq) + NH 4 + (aq) c) HCl (aq)  H + (aq) + OH - (aq) n NH 3 =CV n NH 3 =(0.1M)(0.020L) n NH 3 =0.002 mol = n HCl at equivalence point = n NH 4 + at equivalence point V= n HCl C V= 0.002mol 0.1M V=0.02L  V total =0.040L C= n NH 4 + V C= 0.002mol 0.040L C=0.05mol/L
  • 38. Calculations In a titration 20.0 mL of 0.1 M NH 3(aq) is titrated with 0.1 M HCl (aq) . What is the pH of the resulting solution after the following volumes of HCl (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL NH 4 + (aq) <===> NH 3(aq) + H + (aq) I 0.05 0 0 C -x +x +x E 0.05-x x x 5.555x10 -10 = [x][x] [0.05-x]  Use assumption 5.555x10 -5 = [x][x] [0.05] 5.27x10 -6 = x pH = -log [H + ] pH = -log [5.27x10 -6 ] pH = 5.28 .: pH = 5.28 5.27x10 -6 0.05 5.27x10 -6 c)
  • 39. Calculations In a titration 20.0 mL of 0.1 M NH 3(aq) is titrated with 0.1 M HCl (aq) . What is the pH of the resulting solution after the following volumes of HCl (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL NH 3(aq) + H 2 O (l) <===> OH - (aq) + NH 4 + (aq) d) n NH 3 =CV n NH 3 =(0.1M)(0.020L) n NH 3 =0.002 mol n HCl =CV n HCl =(0.1M)(0.030L) HCl (aq)  H + (aq) + OH - (aq) n HCl =0.003mol n HCl remaining =0.001 mol n NH 3 remaining = 0 mol V total =0.050L C=n/V total C=0.001mol/0.050L C=0.02M n NH 4 + formed = 0.002 mol
  • 40. Calculations In a titration 20.0 mL of 0.1 M NH 3(aq) is titrated with 0.1 M HCl (aq) . What is the pH of the resulting solution after the following volumes of HCl (aq) have been added? a) 0.00 mL b) 10.0 mL c) equivalence pt d) 30.0 mL NH 3(aq) + H 2 O (l) <===> OH - (aq) + NH 4 + (aq) d) Since we discovered before that the conjugate acid/base formed does not significantly affect pH when a strong acid/base is present, then we can solve for the pH by using the [HCl] HCl (aq)  H + (aq) + OH - (aq) C HCl =0.02M pH = -log [H + ] pH = -log [0.02] pH = 1.69897 .: pH = 1.70