Types of Ka Calculation problems

Transcription

1 Chemistry 12 UNIT 4 ACIDS AND BASES PACKAGE #3 Ka HB (aq) H + (aq) + B - (aq) Ka = [ H + ] [ B - ] [HB] STRONG ACID: 100% dissociation (ionization) HCl à H + + Cl - Ka ~ infinity since [large] Equil: ~0 large large ~0 WEAK ACID: small % dissociation (ionization) Ka usually << 1 since [small] large CH3COOH H + + CH3COO - Equil: large small small Types of Ka Calculation problems UNIDENTIFIED ACID: Ka not known Arithmetic problem: start with [ H + ] E or [OH - ] E known start with ph, poh, % dissociation IDENTIFIED ACID: Ka known Algebra Problem with weak acid: Assume for now: monoprotic acids: HB Simple Arithmetic problem with strong acid % dissociation = [ H + ] E x 100 [HB] I

2 oxy acids of chlorine and their Ka HClO hypochlorous acid 5.6 x 10-8 HClO2 chlorous acid 1 x 10-2 HClO3 chloric acid large HClO4 perchloric acid very large -As the central chlorine atom becomes more positive, then the electrons shared between the H-O bond are pulled closer toward the Cl atom. Therefore the H-O bond becomes more polar. SEE P. 409 The strength of oxyacids is determined by the number of O's attached to the oxyacid. (more O's = stronger) H2SO5 persulphuric acid very very strong H2SO4 sulphuric acid very strong H2SO3 sulphurous acid 1.7 x 10-2 H2SO2 hyposulphurous acid very weak The more positive the S becomes, the more easily the H is removed.

3 Sample Ka calcs: UNIDENTIFIED ACID PROBLEMS M HB (aq) has a ph of Find Ka, % dissociation, and possible identity M HB (aq) has poh = Find [ H + ], [OH - ], ph, Ka, identity, % dissocation M HX (aq) is 5.0 % dissociated. Find ph, Ka, identity. 4. Find identity and % dissociation if M HB (aq) has ph = M HB (aq) has poh = Find ph, [ H + ], [OH - ], Ka, identity, % dissocation. Sample Ka calcs: IDENTIFIED ACID PROBLEMS 1. Find ph of M H2O2 (aq). Ka = 2.4 x Find ph of M HBr (aq). 3. Find ph, poh, [ H + ], [OH - ] and % dissocation for M HCO3 - (aq). 4. Find % dissocation for 0.96 M HNO2 (aq). NOW YOU CAN COMPLETE WORKSHEET 4.3

4 Wkst 4.3: K a Problems 1- A M solution of HB (aq) has [H + ] e = 8.0 x 10-4 M. Determine the ph, % dissociation, poh, [OH - ], K a and the possible identity of the acid from your K a table. 2- Determine the ph, poh, [H + ], [OH - ], and % dissociation of 0.50 M H 2 O 2(aq). 3- Find the ph, poh of M HNO 3(aq). 4- If a M solution of an unidentified acid HB (aq) has a ph of 5.00, find its K a, % dissociation and possible identity. 5- A M solution of acid HX (aq) is 2.0% dissociated. Find the ph, poh, K a and possible identity of the acid. 6- Find the ph and the % dissociation of a M solution of HBrO (aq). K a = 2.5 x Find the ph and poh for M HI (aq). 8- An unknown acid HB(aq), in a M solution, has a poh of Find the K a, % dissociation and possible identity of the acid. 9- A 0.20 M solution of HAsO 2(aq) is 5.5 x 10-3 % dissociated. Find its K a.

5 Wkst 4.3: K a Problems 1- ph = 3.10, % dissociation = 8.0 %, [OH - ] = 1.3 x M, poh = 10.90, K a = 7.0 x 10-5 (similar to benzoic acid) 2- [H + ] = 1.1 x 10-6 M, ph = 5.96, % dissociation = 2.2 x 10-4 %, [OH - ] = 9.1 x 10-9 M, poh = ph = 1.120, poh = % dissociation = 1.0 %, K a = 1.0 x 10-7 (similar to hydrogen sulphite ion) 5- ph = 2.70, poh = 11.30, K a = 4.1 x 10-5 (similar to hydrogen oxalate) 6- ph = 5.00, % dissociation = 2.3 x 10-2 % 7- ph = 2.19, poh = % dissociation = 0.71 %, K a = 2.3 x 10-7 (similar to hydrogen sulphite or monohydrogen citrate ions) 9- K a = 6.05 x 10-10

6 Kn Kn = Ka (forward reaction) Ka (reverse reaction) (this is a division or really, just a RATIO) Example: HC2O4 - + CN- HCN + C2O4 2- a) identify the acids: HC2O4 - and HCN b) identify the acid strength by looking up the Ka values on page 6 of the data booklet HC2O4 - : 6.4 x 10-5 and HCN : 4.9 x HC2O4 - is the stronger acid (higher Ka value) c) predict whether the forward or reverse reaction is favoured: HC2O4 - dissociating is the forward reaction and HCN dissociating is the reverse reaction - since the Ka of the hydrogen oxalate ion is higher, I predict that the forward reaction is favoured. d) back up your answer by calculating Kn (which is just a ratio of the Ka values of the two acids) Kn = Ka (forward reaction) OR the acid on the reactants side Ka (reverse reaction) the acid on the products side = 6.4 x x = 1.3 x 10 5 Since Kn >>1, my previous prediction is correct. The forward reaction is favoured. That is, the products are favoured. NOW WORKSHEET 4.4 CAN BE COMPLETED

7 Wkst 4.4: vs. Competition For each of the following neutralisation reactions, identify: the acids, their relative strengths and whether reactants or products are favoured. After predicting the winner, offer numerical proof by calculating K n. a) HSO 4 -(aq) + C 6 H 5 COO - (aq) ó C 6 H 5 COOH (a) + SO 4-2 (aq) b) HCO 3 - (aq) + NO 2 -(aq) ó HNO 2(aq) + CO 3-2 (aq) c) H 2 O 2(aq) + F - (aq) ó HF (aq) + HO 2 -(aq) d) HSO 4 - (aq) + NH 4 +(aq) ó NH 3(aq) + H 2 SO 4(aq) e) S -2 (aq) + H 2 PO 4 -(aq) ó HPO 4-2 (aq) + HS - (aq) f) CH 3 COO - (aq) + HBr (aq) ó Br - (aq) + CH 3 COOH (aq)

8 Wkst 4.4: vs. Competition For each of the following neutralisation reactions, identify: the acids, their relative strengths and whether reactants or products are favoured. After predicting the winner, offer numerical proof by calculating K n. a) HSO 4 -(aq) + C 6 H 5 COO - -2 (aq) ó C 6 H 5 COOH (a) + SO 4 (aq) K n = 1.8 x 10 2 ; products are favoured. - b) HCO 3 (aq) + NO (aq) ó HNO 2(aq) + CO 3 (aq) K n = 1.2 x 10-7 ; reactants are favoured. c) H 2 O 2(aq) + F - (aq) ó HF (aq) + HO 2 -(aq) K n = 6.9 x 10-9 ; reactants are favoured. - d) HSO 4 (aq) + NH 4 +(aq) ó NH 3(aq) + H 2 SO 4(aq) K n = 0; reactants are favoured. e) S -2 (aq) + H 2 PO 4 -(aq) ó -2 HPO 4 (aq) + HS - (aq) K n = 4.8 x 10 5 ; products are favoured. f) CH 3 COO - (aq) + HBr (aq) ó Br - (aq) + CH 3 COOH (aq) K n = ; products are favoured.