Acid-Base Equilibria (Titrations & Indicators) Green & Damji Chapter 8, Sections 18.3 & 18.4 Chang Chapter 16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Section Acid-Base Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4.7 End point – point at which the indicator changes color

Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l) 16.4

Strong Acid-Strong Base Titrations 16.4 What do you notice about… the pH at equivalence point? the shape of the line around equivalence point?

Strong Acid-Strong Base Titrations Ex 18.4, # 1 1 M nitric acid is being titrated with aqueous sodium hydroxide. When 99.9% of the acid has been neutralised, the pH of the solution, ignoring changes in the total volume, will be… A 3B 6 C 6.900D 6.999

Strong Acid-Strong Base Titrations Ex 18.4, # 1 1 M nitric acid is being titrated with aqueous sodium hydroxide. When 99.9% of the acid has been neutralised, the pH of the solution, ignoring changes in the total volume, will be… A 3B 6 C 6.900D Why? If only 0.1% of the acid is left, the [H + ] = 0.001M which is 1 x … pH = 3

Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) 16.4

Weak Acid-Strong Base Titrations CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) 16.4 What do you notice about… the pH at equivalence point? the shape of the line around equivalence point? the shape of the line near the half-equivalence point note – at the half-equivalence point pH = pKa of WA

Weak Acid-Strong Base Titrations Ex 18.4, # 2 During the titration of a weak acid with a strong base, the pH of the solution will equal the pKa of the weak acid A at the start of the titration B when half the volume required to reach the end point has been added C at the end point D when twice the volume required to reach the end point has been added ANSWER:

Weak Acid-Strong Base Titrations Ex 18.4, # 3 During the titration of a weak acid using a strong base, at the end point there will be a rapid change in pH between A 4 and 10 B 3 and 7 C 7 and 11 D 6 and 8 ANSWER:

Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) 16.4 H + (aq) + NH 3 (aq) NH 4 Cl (aq)

Strong Acid-Weak Base Titrations 16.4 What do you notice about… the pH at equivalence point? the shape of the line around equivalence point? the shape of the line near the half-equivalence point note – at the half-equivalence point pOH = pKb of WB

Weak Acid-Weak Base Titrations 16.4 What do you notice about… the pH at equivalence point? the general shape of the titration curve? Why is this type of titration difficult to work with?

Calculations involving Titrations Ex 18.4, # 4 When 20 mL of a solution of aqueous ammonia is titrated with 0.20 M HCl, 15 mL of the acid were needed to reach the equivalence point. a)What is the concentration (initial) of the ammonia?

Calculations involving Titrations Ex 18.4, # 4 When 20 mL of a solution of aqueous ammonia is titrated with 0.20 M HCl, 15 mL of the acid were needed to reach the equivalence point. a)What is the concentration (initial) of the ammonia? Chem equivalence point mol acid = mol base VaCa = VbCb M = # mol / vol (in L)

Calculations involving Titrations Ex 18.4, # 4 When 20 mL of a solution of aqueous ammonia is titrated with 0.20 M HCl, 15 mL of the acid were needed to reach the equivalence point. b)Given the pKa for the mmonium ion is 9.3, calculate the pH of the solution i)at the start ii)when 7.5 cm3 of the acid has been added. iii)at the equivalence point

Calculations involving Titrations Ex 18.4, # 4 When 20 mL of a solution of aqueous ammonia is titrated with 0.20 M HCl, 15 mL of the acid were needed to reach the equivalence point. b)Given the pKa for the ammonium ion is 9.3, calculate the pH of the solution i)at the start

Calculations involving Titrations Ex 18.4, # 4 When 20 mL of a solution of aqueous ammonia is titrated with 0.20 M HCl, 15 mL of the acid were needed to reach the equivalence point. b)Given the pKa for the mmonium ion is 9.3, calculate the pH of the solution ii) when 7.5 cm3 of the acid has been added.

Calculations involving Titrations Ex 18.4, # 4 When 20 mL of a solution of aqueous ammonia is titrated with 0.20 M HCl, 15 mL of the acid were needed to reach the equivalence point. b)Given the pKa for the mmonium ion is 9.3, calculate the pH of the solution iii)at the equivalence point

Calculations involving Titrations Ex 18.4, # 4 When 20 mL of a solution of aqueous ammonia is titrated with 0.20 M HCl, 15 mL of the acid were needed to reach the equivalence point. c) Bearing these values in mind, sketch the shape of the graph of pH against titre you would expect for this titration

Calculations involving Titrations Ex 18.4, # 4 When 20 mL of a solution of aqueous ammonia is titrated with 0.20 M HCl, 15 mL of the acid were needed to reach the equivalence point. d) Which section of the curve is known as the ‘buffering region’ and why is it so called?

Calculations involving Titrations Ex 18.4, # 4 When 20 mL of a solution of aqueous ammonia is titrated with 0.20 M HCl, 15 mL of the acid were needed to reach the equivalence point. e) Identify two important ways in which the curve would differ if the titration where carried out with aqueous barium hydroxide of the same concentration as the ammonia

TEXTWORK: Section 18.4 Read Section 18.4 pp Do Ex 18.4 pp # 1,2,3,4 Surprise – You’ve already DONE this.