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Tang 07 titrations 2

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Tang 07 titrations 2

  1. 1. Teacher Prep <ul><li>Distilled water </li></ul><ul><li>Buffer solution </li></ul><ul><li>Universal indicator </li></ul>
  2. 2. Acid-Base Titrations
  3. 3. TITRATIONS
  4. 4. TITRATIONS Purpose: To determine the unknown concentration of the acid.
  5. 5. Titration Curve TITRATIONS
  6. 6. Indicator Ranges TITRATIONS
  7. 7. Indicator Ranges TITRATIONS
  8. 8. Indicators <ul><li>How do indicators work? </li></ul><ul><li>Through equilibrium! </li></ul>Phenolphthalein TITRATIONS + H +
  9. 9. Basic Calculations (Gr. 11) <ul><li>Example #1 </li></ul><ul><li>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? </li></ul>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 TITRATIONS
  10. 10. Example #2 <ul><li>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? </li></ul><ul><li>Do you have a good guess as to whether the final solution is acidic or basic? </li></ul>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 TITRATIONS
  11. 11. Example #2 <ul><li>What is happening at the chemical level? </li></ul><ul><li>NaOH  Na+ + OH- </li></ul><ul><li>HCl  H+ + Cl- </li></ul><ul><li>The OH - and H + ions come together to produce H 2 O. The remaining ions will determine the final pH of the solution. </li></ul>TITRATIONS
  12. 12. STRONG-WEAK TITRATIONS
  13. 13. Strong-Weak Titration Curves STRONG-WEAK TITRATIONS
  14. 14. Buffers <ul><li>Buffer solutions contain a mixture of an acid and its conjugate base or a base with its conjugate acid. </li></ul><ul><li>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. </li></ul>STRONG-WEAK TITRATIONS
  15. 15. Buffers <ul><li>Buffers are important in biological systems where large changes in pH could be detrimental to the organism. </li></ul><ul><li>Almost all organic substances in our body act as a buffer. </li></ul><ul><li>H 2 O + CO 2 <===> H 2 CO 3 <===> H + + HCO 3 2- </li></ul>STRONG-WEAK TITRATIONS
  16. 16. Buffers STRONG-WEAK TITRATIONS
  17. 17. Strong-Weak Titrations <ul><li>Example #3 </li></ul><ul><li>What happens when you add a strong base to a weak acid? </li></ul><ul><li>CH 3 COOH <===> CH 3 COO - + H + </li></ul><ul><li>NaOH  Na + + OH - </li></ul>STRONG-WEAK TITRATIONS
  18. 18. Strong-Weak Titrations <ul><li>Two-step process: </li></ul><ul><li>All OH - ions from the strong base will react with H + to produce H 2 O. </li></ul><ul><li>Equilibrium is shifted to the right. </li></ul><ul><li>pH is then calculated from the new equilibrium based on the remaining [H + ]. </li></ul>STRONG-WEAK TITRATIONS
  19. 19. Strong Weak Titrations <ul><li>Example #4 </li></ul><ul><li>What happens when you add a strong acid to a weak base? </li></ul><ul><li>NH 3 + H 2 O <===> NH 4 + + OH - </li></ul><ul><li>HCl  H + + Cl - </li></ul>STRONG-WEAK TITRATIONS
  20. 20. Strong-Weak Titrations <ul><li>Two-step process: </li></ul><ul><li>All H + ions from the strong acid will react with OH - to produce H 2 O. </li></ul><ul><li>Equilibrium is shifted to the right. </li></ul><ul><li>pH is then calculated from the new equilibrium based on the remaining [H + ]. </li></ul>STRONG-WEAK TITRATIONS
  21. 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) 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 TITRATIONS
  22. 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) 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 TITRATIONS
  23. 23. 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 TITRATIONS
  24. 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) 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 STRONG-WEAK TITRATIONS
  25. 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) 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 STRONG-WEAK TITRATIONS
  26. 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=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) STRONG-WEAK TITRATIONS
  27. 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 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 STRONG-WEAK TITRATIONS
  28. 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)  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? STRONG-WEAK TITRATIONS
  29. 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 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 STRONG-WEAK TITRATIONS
  30. 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) 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 STRONG-WEAK TITRATIONS
  31. 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 =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 STRONG-WEAK TITRATIONS
  32. 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) 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 STRONG-WEAK TITRATIONS
  33. 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 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 STRONG-WEAK TITRATIONS
  34. 34. 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 STRONG-WEAK TITRATIONS
  35. 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) 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 STRONG-WEAK TITRATIONS
  36. 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) 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 STRONG-WEAK TITRATIONS
  37. 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) 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 STRONG-WEAK TITRATIONS
  38. 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 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 STRONG-WEAK TITRATIONS
  39. 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 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) STRONG-WEAK TITRATIONS
  40. 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) 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 STRONG-WEAK TITRATIONS
  41. 41. 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 STRONG-WEAK TITRATIONS

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