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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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