1. General Chemistry II 2302102 Acid and Base Equilibria Lecture 2 i.fraser@rmit.edu.au Ian.Fraser@sci.monash.edu.au

2. Acids and Bases - 3 Lectures 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 Outline - 5 Subtopics

3. 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 (K b ) & relate to K a 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. Objectives - Lecture 2 Acids and Bases

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

5. Acids & Bases - Lecture 2 By the end of this lecture AND completion of the set problems, you should be able to: Objectives Distinguish between strong & weak bases Calculate equilibrium concentrations of acids & bases using basicity constants (K b ) & relate to K a 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 pK a from titration of a weak acid

6. BASES

7. BASES Bases accept Protons and form acids. B(aq) + H 2 O HB + (aq) + OH - (aq) Equilibrium (‘Basicity’) constant, K b.

8. BASES Bases accept Protons and form acids. B(aq) + H 2 O HB + (aq) + OH - (aq) pK b = -log 10 K b Strong Bases Large K b equilibrium lies to the right e.g. O 2- (aq) Weak Bases Small K b, large pK b equilibrium lies to left e.g. NH 3 (aq)

9. BASES Relation between K b and K a of the conjugate acid. NH 4 + (aq) + H 2 O NH 3 (aq) + H 3 O + (aq) K a = 5.5 x 10 -10 NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) K b = 1.8 x 10 -5 add equations: 2 H 2 O H 3 O + (aq) + OH - (aq) multiply equilibrium constants K w = K a x K b

10. WEAK BASES Calculate [OH - ], [H + ] and pH when 0.1 mol of NaNO 2 is dissolved in sufficient water to produce 1.0 L of solution. K a for HNO 2 = 4.5 x 10 -4. Assume NaNO 2 (a salt) is completely ionised when dissolved, hence have 0.1 M NO 2 -. initial 0.10 M 0 0 equilibrium (0.10 - x) M x M x M H 2 O( ) + NO 2 - (aq) HNO 2 (aq) + OH - (aq)

11. WEAK BASES initial 0.10 M 0 0 equilibrium (0.10 - x) M x M x M H 2 O( ) + NO 2 - (aq) HNO 2 (aq) + OH - (aq) Assuming x << 0.10, then 2.2 x 10 -12 = x 2 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 = -log 10 (6.7 x 10 -9 ) = 8.17

12. 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: NH 4 + is a stronger acid than H 2 O, and OH - is a stronger base than NH 3 So reaction proceeds spontaneously to the left H 2 O (l) + NH 3 (aq) NH 4 + (aq) + OH - (aq)

13. Relative Strengths of Acids (Bases) Pick the strongest acid (smallest pK a ) NH 4 + is a stronger acid than H 2 O, hence OH - is a stronger base than NH 3 H 2 O( ) + NH 4 + (aq) H 3 O + (aq) + NH 3 (aq) For NH 3, K b = 1.8 x 10 -5, pK b = 4.74 So pK a = 9.26 H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) For H 2 O, K a = K b = K w = 1.0 x 10 -14, So pK a = 14.00

14. Relative Strengths of Acids (Bases) HNO 2 is a stronger acid than HCN, hence CN - is a stronger base than NO 2 -. Reaction strongly favours the right For HNO 2, K a = 4.5 x 10 -4 so pK a = 3.35 For HCN, K a = 7.2 x 10 -10 so pK a = 9.14 Which direction is favoured here? HNO 2 (aq) + CN - (aq) NO 2 - (aq) + HCN (aq)

15. BUFFERSOLUTIONS(“BUFFERS”)

16. 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 - (K a = 3.7 x 10 -8 ) [HOCl] = 0.05 [OCl - ] = 0.1 [H 3 O + ] = 1.9 x 10 -8 pH =7.72 [OH - ] = 5.3 x 10 -7 HOCl(aq) + H 2 O(l) H 3 O + (aq) + OCl - (aq) initial 0.05 0 0.1 equilibrium (0.05 - z) z 0.1 + z

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 [H 3 O + ] 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. 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 H 3 O + to a solution containing of 0.05 M HOCl in water and 0.1M OCl - ion with the effect of adding 10 -3 M H 3 O + to pure water. OCl - (aq) + H 3 O + (aq) HOCl(aq) + H 2 O(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

19. Acids and Bases - End of 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 (K b ) & relate to K a 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. Objectives Covered in Lecture 2