pH of (a)Weak Acid Solutions, (b) Bases, (c) Polyprotic Acids, and (d) Salts Chemistry 142 B Autumn 2004 J. B. Callis, Instructor Lecture 21

Calculation of the pH of Weak Acid Solutions - A Systematic Approach (1) Problem 21-1: What is the pH of a solution of 1.00 M nitrous acid, K a = 4.0 x Major species in solution: HNO 2 and H 2 O Which species can generate H + ions? –HNO 2 (aq) = H + (aq) + NO 2 - (aq) K a = 4.0 x –H 2 O(aq) = H + (aq) + OH - (aq) K w = 1.0 x –Ignore contribution from water, K a >> K w

Calculation of the pH of Weak Acid Solutions - A Systematic Approach (2) The equilibrium expression is The initial concentration are

Calculation of the pH of Weak Acid Solutions - A Systematic Approach (3) Let x be the change in concentration of HNO 2 that is required to achieve equilibrium. Then the equilibrium concentrations are:

Calculation of the pH of Weak Acid Solutions - A Systematic Approach (4) We rearrange this equation to yield a second order polynomial:

Calculation of the pH of Weak Acid Solutions - A Systematic Approach (5) For this example, the solutions are: x = Only the first solution is valid because it leads to all positive concentrations. [H + ] = [NO 2 - ] = [HNO 2 ] =

Calculation of the pH of Weak Acid Solutions - A Systematic Approach (6) Question 1(a): What is the pH of this solution? ans: Question 1(b): Were we correct to neglect H + from the water ans: Question 1(c): What % of the acid is ionized? ans:

Problem 21-2: Calculate of the pH of a Mixture of Weak Acids Calculate the pH of a mixture of 1.00 M of phenol (K a = 1.6 x and 5.00 M acetic acid (K a = 1.8 x ). Major Species in Solution: phenol (HPhe), acetic acid (HAc) and H 2 O Which Species Can Generate H + ions? –HAc(aq) = H + (aq) + Ac - (aq) K HAc = 1.8 x –HPhe(aq) = H + (aq) + Phe - (aq) K HPhc = 1.8 x –H 2 O(aq) = H + (aq) + OH - (aq) K w = 1.0 x –Ignore contribution from water and phenol, – K Ac >> K HPhe >> K w

Problem 21-2: Calculate of the pH of a Mixture of Weak Acids (2) Focusing on the Acetic Acid equilibrium: x = pH =

Problem 21-2: Calculate of the pH of a Mixture of Weak Acids (3) How much Phe - is generated?

Problem 21-3: Find the K a of a weak acid from % dissociation. If 0.10 M propanoic acid dissociates 1.1%, what is K a ?

Bases Definition (Bronsted-Lowry) – a proton acceptor. Strong bases dissociate completely. (e.g. metal hydroxides from Groups 1A and 1B. NaOH(s) - > Na + (aq) + OH - (aq)

Problem 21-4: Calculate the pH of a solution of 3.0 x M Ca(OH) 2 (aq)

Note: A base doesn’t have to contain OH -, it just needs to be able to accept H +, e.g. aqueous ammonium. NH 3 (aq) + H 2 O(l) = NH 4 + (aq) + OH - (aq) Many nitrogen containing compounds are bases. General reaction: B(aq) + H 2 O(l) = BH + (aq) + OH - (aq) base acid conj. Acid conj. base

Problem 21-5: What is the pH of 1.5 M Dimethylamine (CH 3 ) 2 NH (K b = 5.9 x ). Let x be the amount of dimethylamine that has dissociated. Conc. (M) DMA =DMA + OH - Initial Change Equil.

Problem 21-5 (cont.)

Polyprotic Acids Can furnish more than one proton per molecule of acid. They do this in a step-wise manner. Example: oxalic acid: H 2 C 2 O 4 (aq) = H + (aq) + HC 2 O 4 - (aq) K a1 = 5.6 x HC 2 O 4 - (aq) = H + (aq) + C 2 O 4 2- (aq) K a2 = 5.4 x 10 -5

Problem 21-6 – Calculate the pH of M Ascorbic Acid (K a1 = 1.0 x ; K a2 = 5.0 x ). Let x be the amount of [H + ] that is produced. Assume that it all comes from the first ionization. Then let the [H + ] determine the amount of the doubly ionized base, Asc 2-.

Acid-Base Properties of Salts Salt – an ionic compound that dissolves in H 2 O to give ions. Sometimes the ions can behave as acids or bases. (a)Anions that correspond to strong acids, e.g. Cl - and NO 3 - are weak, weak bases. Also, cations from strong bases, e.g. Na +, K + are weak, weak acids. (b)Salts that consist of cations from strong bases and anions from strong acids produce neutral solutions (pH= 7). (c)Salts of weak acids produce basic solutions. (d)Salts of weak bases produce acid solutions.

Problem 21-7: What is the pH of 0.25 M Sodium Acetate (NaAc)? Let x be the amount of acetic acid that is formed by the following reaction: Ac - (aq) + H 2 O(l) = HAc(aq) + OH - (aq). Conc. (M) Ac - =HAc +OH - Initial Change Equil.

Problem 21-7 (cont.)

Answers to Problems in Lecture [H + ] = [NO 2 - ] = M; [HNO 2 ] = 0.98 M (a) pH = 1.70 (b) Yes (c) 2.0 % 2. pH = 2.023; Very little dissociation of HPhe ;Very little H + from HPhe pH = x M 7.pH = 9.07