Titration a solution of known concentration, called a standard solution
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1 Acid-Base Titrations Titration is a form of analysis in which we measure the volume of material of known concentration sufficient to react with the substance being analyzed.
2 Titration a solution of known concentration, called a standard solution is added gradually to another solution of unknown concentration until the chemical reaction between the two solutions is complete equivalence point the exact number of moles required for the reaction detected by a color change in an acidbase indicator or electronically
3 Acid-Base Titrations How does the ph change during the course of the following kinds of titration? strong acid-strong base weak acid-strong base weak base-strong acid
4 Titrations Involving a Strong Acid and a Strong Base
5 Titrations Involving a Strong Acid and a Strong Base titrate 25.0 ml of M HCl with M NaOH Note: mmol mol ml and L are equivalent
6 Titrations Involving a Strong Acid and a Strong Base What is the ph after the addition of 10.0 ml of M NaOH to 25.0 ml of M HCl? HCl (aq) NaOH(aq) H 2 O (l ) NaCl (aq) H (aq) OH - (aq) H 2 O (l )
7 Titrations Involving a Strong Acid and a Strong Base What is the ph after the addition of 10.0 ml of M NaOH to 25.0 ml of M HCl? 10ml x 25ml x 0.1mmol NaOH = 1.0mmol OH - ml 0.1mmol HCl = 2.5mmol H ml H (aq) OH - (aq) start 2.5mmol H 1.0mmol 0 end 1.5mmol H 0 H 2 O (l ) 1.0mmol
8 Titrations Involving a Strong Acid and a Strong Base What is the ph after the addition of 10.0 ml of M NaOH to 25.0 ml of M HCl? H (aq) OH - (aq) H 2 O (l ) 1.5mmol H [H ] = = 4.28 x 10-2 M 35.0 ml (solution) ph = 1.37
9 Titrations Involving a Strong Acid and a Strong Base What is the ph after the addition of 25.0 ml of M NaOH to 25.0 ml of M HCl? 25ml x 25ml x 0.1mmol NaOH = 2.5mmol OH - ml 0.1mmol HCl = 2.5mmol H ml H (aq) OH - (aq) start 2.5 mmol H 2.5 mmol 0 end 0 0 H 2 O (l ) 2.5 mmol
10 Titrations Involving a Strong Acid and a Strong Base What is the ph after the addition of 25.0 ml of M NaOH to 25.0 ml of M HCl? H 2 O (l ) H 2 O (l ) H 3 O (aq) OH - (aq) [H ] = 1.0 x 10-7 M ph = 7.0
11 ph versus ml 0.10 M NaOH added volume (NaOH) volume (total) mmol (HCl) [H ] ph M ml ml 35.0 ml ml 45.0 ml ml 49.0 ml ml 49.5 ml ml 50.0 ml
12 15 ph at equivalence point Titration of ml of M HCl with M NaOH 0 0 volume (NaOH)
13 Titrations Involving a Strong Acid and a Strong Base What is the ph after the addition of 35.0 ml of M NaOH to 25.0 ml of M HCl? 35ml x 25ml x 0.1mmol NaOH = 3.5mmol OH - ml 0.1mmol HCl = 2.5mmol H ml H (aq) OH - (aq) H 2 O (l ) start 2.5 mmol 3.5 mmol 0 end mmol 2.5 mmol
14 Titrations Involving a Strong Acid and a Strong Base What is the ph after the addition of 35.0 ml of M NaOH to 25.0 ml of M HCl? H (aq) OH - (aq) H 2 O (l ) 1.0 mmol OH - [OH - ] = = 1.66 x 10-2 M 60.0 ml (solution) poh = 1.78 ph = 12.22
15 ph versus ml 0.10 M NaOH added volume (NaOH) volume (total) mmol (HCl) [H ] ph M 25.0 ml 50.0ml mmol [OH - ] (NaOH) M 25.5 ml 50.5 ml ml 51.5 ml ml 40.5 ml 50.0 ml 55.0 ml 65.0 ml 75.0 ml
16 15 ph at equivalence point Titration of ml of M HCl with M NaOH 0 0 volume (NaOH)
17 Titrations Involving a Strong Base and a Strong Acid just the opposite of the case we have just discussed
18 15 ph at equivalence point Titration of ml of M HCl with M NaOH 0 0 volume (HCl)
19 Titrations Involving a weak Acid and a Strong Base
20 Titrations Involving a weak Acid and a Strong Base Calculate the ph in the titration of 25.0 ml of M acetic acid with M NaOH after adding (a) ml of M NaOH, (b) 25.0 ml of M NaOH, and (c) 35.0mL of M NaOH. CH 3 COOH NaOH CH 3 COONa H 2 O CH 3 COOH H 2 O CH 3 COO - OH -
21 Titrations Involving a weak Acid and a Strong Base Calculate the ph in the titration of 25.0 ml of M acetic acid with M NaOH after adding (a) ml of M NaOH. CH 3 COOH OH - CH 3 COO - H 2 O 2.5mmol 1.0mmol mmol 0 1.0mmol 1.0mmol K a = 1.8 x 10-5 CH 3 COOH CH 3 COO - H 1.5mmol 1.0mmol
22 Titrations Involving a weak Acid and a Strong Base Calculate the ph in the titration of 25.0 ml of M acetic acid with M NaOH after adding (a) ml of M NaOH. CH 3 COOH CH 3 COO - H 1.5mmol 1.0mmol = 4.28 x 10-2 M = 2.8 x 10-2 M 35.0 ml 35.0 ml K a = [CH 3 COO - ] [H ] [CH 3 COOH] [H ] = [CH 3 COOH] [CH 3 COO - ] K a
23 Titrations Involving a weak Acid and a Strong Base Calculate the ph in the titration of 25.0 ml of M acetic acid with M NaOH after adding (a) ml of M NaOH. CH 3 COOH CH 3 COO - H [H ] = (4.2 x 10-2 M) (1.8 x 10-5 ) 2.8 x 10-2 M ph = 4.57 = 2.7 x 10-5
24 Titrations Involving a weak Acid and a Strong Base Calculate the ph in the titration of 25.0 ml of M acetic acid with M NaOH after adding (b) 25.0 ml of M NaOH. CH 3 COOH OH - CH 3 COO - H 2 O 2.5mmol 2.5mmol mmol 2.5mmol 2.5mmol 50.0 ml K b = CH 3 COO - H 2 O CH 3 COOH OH - = 5.0 x 10-2 M 5.6 x 10-10
25 Titrations Involving a weak Acid and a Strong Base Calculate the ph in the titration of 25.0 ml of M acetic acid with M NaOH after adding (b) 25.0 ml of M NaOH. CH 3 COO - H 2 O CH 3 COOH OH x 10-2 M - x x x 5.6 x = [OH - ] = x 2 (5.0 x x) x = 5.29 x 10-6 poh = 5.27 ph = 8.72
26 Titrations Involving a weak Acid and a Strong Base Calculate the ph in the titration of 25.0 ml of M acetic acid with M NaOH after adding (c) 35.0mL of M NaOH. CH 3 COOH OH - CH 3 COO - H 2 O 2.5mmol 3.5mmol mmol 2.5mmol 2.5mmol 1.0mmol OH ml (solution) = 1.67 x 10-2 M poh = 1.78 ph = 12.22
27 ph versus ml 0.10 M NaOH added volume volume mmol mmol [H ] ph (NaOH) (total) (HOAc) (NaOAc) 0 ml 25.0ml x ml 35.5 ml x ml 50.0 ml x mmol [OH - ] (NaOH) 35.0 ml 60.0 ml x
28 15 ph Maximum buffer Titration of 25.0 ml of 0.10 M acetic acid with 0.10 M NaOH 0 volume (0.10 M NaOH)
29 Titrations Involving a Strong Acid and a Weak Base
30 Titrations Involving a Strong Acid and a Weak Base Calculate the ph in the titration of 25.0 ml of M HCl with M NH 3 after adding (a) ml of M NH 3, (b) 25.0 ml of M NH 3, and (c) 35.0mL of M NH 3. HCl NH NH Cl H NH 4 NH 3
31 Titrations Involving a Strong Acid and a Weak Base Calculate the ph in the titration of 25.0 ml of M HCl with M NH 3 after adding (a) ml of M NH 3. H NH NH mmol 1.0mmol 0 1.5mmol 0 1.0mmol 1.5 mmol H [H ] = = 35.0 ml 4.29 x 10-2 ph = 1.37
32 Titrations Involving a Strong Acid and a Weak Base Calculate the ph in the titration of 25.0 ml of M HCl with M NH 3 after adding (b) 25.0 ml of M NH 3. H NH NH mmol 2.5 mmol mmol K a = 5.6 x NH NH 3 H 4 2.5mmol 50.0 ml = 5.0 x 10-2 M
33 Titrations Involving a Strong Acid and a Weak Base Calculate the ph in the titration of 25.0 ml of M HCl with M NH 3 after adding (b) 25.0 ml of M NH 3. K a = 5.6 x NH NH 3 H x x x x 5.6 x = [H ] = x 2 (5.0 x x) x = 5.29 x 10-6 ph = 5.28
34 Titrations Involving a Strong Acid and a Weak Base Calculate the ph in the titration of 25.0 ml of M HCl with M NH 3 after adding (c) 35.0mL of M NH 3. H NH 3 NH 4 2.5mmol 3.5 mmol mmol 2.5 mmol K a = 5.6 x NH NH 3 H 4 2.5mmol 60.0 ml = 4.1 x 10-2 M 1.0mmol 60.0 ml = 1.66 x 10-2 M
35 Titrations Involving a Strong Acid and a Weak Base Calculate the ph in the titration of 25.0 ml of M HCl with M NH 3 after adding (c) 35.0mL of M NH 3. K a = 5.6 x NH NH 3 H x x 10-2 [H ] = [NH 4 ] [NH 3 ] K a = (4.16 x 10-2 )(5.6 x ) = 1.4 x x 10-2 ph = 8.85
36 ph versus ml 0.10 M NH 3 added volume (NH 3 ) volume mmol [H ] ph (total) (HCl) 0 ml 25.0ml ml 35.0 ml mmol (NH 4 ) mmol (NH 3 ) 25.0 ml 50.0 ml ml 60.0 ml
37 12.5 ph at equivalence point Titration of 25.0 ml of 0.10 M HCl acid with 0.10 M NH volume (0.10 M NH 3 )
38 ph when [NH 3 ] = [NH 4 ] Titration of 25.0 ml of 0.10 M HCl acid with 0.10 M NH volume (0.10 M NH 3 )
A 95 g/mol B 102 /mol C 117 g/mol D 126 g/mol E 152 g/mol
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