The document describes an acid-base titration experiment involving titrating a weak acid with a strong base. It provides 3 key points:
1) The document outlines the experimental procedure, reagents, and purpose which is to determine the concentration of an unknown acid through titration with a strong base such as NaOH.
2) It discusses acid-base titration curves and indicators, and how indicators work through acid-base equilibria. Buffer solutions are also introduced.
3) Several examples of calculations are shown for strong acid-strong base and strong acid-weak base titrations, including determining concentrations and pH at various points in the titration. Step-by-step workings are provided.
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