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General Chemistry II 2302102

General Chemistry II 2302102. Acid and Base Equilibria. Lecture 2. i.fraser@rmit.edu.au Ian.Fraser@sci.monash.edu.au. Acids and Bases - 3 Lectures. Outline - 5 Subtopics. Autoionization of Water and pH (completed) Defining Acids and Bases (completed)

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General Chemistry II 2302102

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  1. General Chemistry II2302102 Acid and Base Equilibria Lecture 2 i.fraser@rmit.edu.au Ian.Fraser@sci.monash.edu.au

  2. Acids and Bases- 3 Lectures Outline - 5 Subtopics • Autoionization of Water and pH (completed) • Defining Acids and Bases (completed) • Interaction of Acids and Bases with Water • (part-completed) • Buffer Solutions • Acid-Base Titrations

  3. Acids and Bases Objectives - Lecture 2 • By the end of this lecture AND completion of the set problems, you should : • Distinguish between strong & weak bases • Calculate equilibrium concentrations of acids & bases using basicity constants (Kb) & relate to Ka • Determine the direction of acid-base reactions • Understand the concept of buffer solutions, buffer action following dilution, following the addition of strong acid and following the addition of strong base.

  4. Acids & Bases - 2 Lectures Outline Introduction to acids and bases Strong & Weak Acids Conjugate acid-base pairs Common Ion Effect Lecture 1 • Bases • Buffers • Indicators • Titrations • Strong Acids • Weak Acids Lecture 2

  5. Acids & Bases - Lecture 2 Objectives By the end of this lecture AND completion of the set problems, you should be able to: • Distinguish between strong & weak bases • Calculate equilibrium concentrations of acids & bases using basicity constants (Kb) & relate to Ka • Determine the direction of acid-base reactions • Describe and calculate buffering of weak acids • Calculate concentrations from titration of a strong acid • Calculate concentrations and pKa from titration of a weak acid

  6. BASES

  7. B(aq) + H2O HB+(aq)+ OH-(aq) Equilibrium (‘Basicity’) constant, Kb. BASES Bases accept Protons and form acids.

  8. Bases accept Protons and form acids. B(aq) + H2O HB+(aq)+ OH-(aq) BASES pKb = -log10Kb Strong Bases Large Kb equilibrium lies to the right e.g. O2-(aq) Weak Bases Small Kb , large pKb equilibrium lies to left e.g.NH3(aq)

  9. NH4+(aq) + H2O NH3(aq) + H3O+(aq) Ka = 5.5 x 10-10 NH3(aq) + H2O NH4+(aq) + OH-(aq) Kb = 1.8 x 10-5 add equations: 2 H2O H3O+(aq) + OH-(aq) multiply equilibrium constants Kw = Ka x Kb BASES Relation between Kb and Kaof the conjugate acid.

  10. Assume NaNO2 (a salt) is completely ionised when dissolved, hence have 0.1 M NO2-. H2O(l) + NO2-(aq) HNO2(aq) + OH-(aq) initial 0.10 M 0 0 equilibrium (0.10 - x) M x M x M WEAK BASES Calculate [OH-], [H+] and pH when 0.1 mol of NaNO2 is dissolved in sufficient water to produce 1.0 L of solution. Ka for HNO2 = 4.5 x 10-4.

  11. H2O(l) + NO2-(aq) HNO2(aq) + OH-(aq) initial 0.10 M 0 0 equilibrium (0.10 - x) M x M x M WEAK BASES Assuming x << 0.10, then 2.2 x 10-12 = x2 thus, x = 1.5 x 10-6 M [OH-] = 1.5 x 10-6, [H+] = 1.0 x 10-14 / 1.5 x 10-6 so, [H+] = 6.7 x 10-9, pH = -log10(6.7 x 10-9) = 8.17

  12. H2O (l) + NH3 (aq) NH4+ (aq) + OH- (aq) Direction of Acid-Base Reactions • Acid-Base Reactions proceed spontaneously with the strongest acid and strongest base forming the weakest acid and the weakest base • Returning to: NH4+ is a stronger acid than H2O, and OH- is a stronger base than NH3 So reaction proceeds spontaneously to the left

  13. H2O(l) + NH4+(aq) H3O+(aq) + NH3(aq) For NH3, Kb = 1.8 x 10-5, pKb = 4.74 So pKa = 9.26 H2O (l) + H2O (l) H3O+(aq) + OH- (aq) For H2O, Ka = Kb = Kw = 1.0 x 10-14, So pKa = 14.00 Relative Strengths of Acids (Bases) • Pick the strongest acid (smallest pKa) NH4+ is a stronger acid than H2O, hence OH- is a stronger base than NH3

  14. Which direction is favoured here? HNO2 (aq) + CN- (aq) NO2-(aq) + HCN(aq) Relative Strengths of Acids (Bases) For HNO2, Ka = 4.5 x 10-4 so pKa = 3.35 For HCN, Ka = 7.2 x 10-10 so pKa = 9.14 HNO2 is a stronger acid than HCN, hence CN- is a stronger base than NO2-. Reaction strongly favours the right

  15. BUFFER SOLUTIONS (“BUFFERS”)

  16. HOCl(aq) + H2O(l) H3O+(aq) + OCl-(aq) initial 0.05 0 0.1 equilibrium (0.05 - z) z 0.1 + z BUFFER SOLUTIONS The presence of the conjugate base of an acid inhibits the ionization of the acid. e.g. a solution that contains 0.05M HOCl and 0.1M OCl- (Ka = 3.7 x 10-8) [HOCl] = 0.05 [OCl-] = 0.1 [H3O+] = 1.9 x 10-8 pH =7.72 [OH-] = 5.3 x 10-7

  17. BUFFER SOLUTIONS The presence of the conjugate base of an acid inhibits the ionisation of the acid. Compare the dissociation of 0.05 M HOCl in water with its dissociation in a solution containing 0.1 M OCl- ion. in water in 0.1 M OCl- [HOCl] 0.05 0.05 [OCl-] 4.3 x 10-5 0.1 [H3O+] 4.3 x 10-5 1.9 x 10-8 [OH-] 2.3 x 10-10 5.3 x 10-7 pH 4.37 7.73

  18. OCl-(aq) + H3O+(aq) HOCl(aq) + H2O(l) initial 0.1 M 1.85 x 10-8 0.05 M final 0.099 M 1.91 x10-8 0.051 M pH change 7.73 to 7.72 acid to water pH change 7.0 to 3.0 BUFFER SOLUTIONS BUFFER SOLUTION A mixture that contains a weak acid and its conjugate base. Addition of small amounts of acid or base result in only small changes of pH. Compare the effect of adding 10-3 M H3O+ to a solution containing of 0.05 M HOCl in water and 0.1M OCl- ion with the effect of adding 10-3 M H3O+ to pure water.

  19. Acids and Bases - End of Lecture 2 Objectives Covered in Lecture 2 • After studying this lecture and the set problems, • you should be able to: • Distinguish between strong & weak bases • Calculate equilibrium concentrations of acids & bases using basicity constants (Kb) & relate to Ka • Determine the direction of acid-base reactions • Understand the concept of buffer solutions, buffer action following dilution, following the addition of strong acid and following the addition of strong base.

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