[H3 O (aq)] = = = mol/l. = 0.13 mol mol [OH (aq)] = [NaOH(aq)] = = mol/l 4.00 L 14

Transcription

1 () Medications that are taken orally are often buffered due to the acidity of the stomach. Some medications are sensitive to acids and contain a buffer to keep the chemicals from reacting with stomach acid, as well as reducing their effectiveness. Other medications are acidic and are buffered to keep the ph level in the stomach from dropping too low and causing damage to the lining of the stomach or triggering acid reflux. Chapter 16 REVIEW Part 1 (Page 77) 1.. D.. 5. C 6. 5,, 1, 6 7.,, 6, C 10. D Solutions. 1 Kw [H O ].8 10 mol/l [OH ].110. poh 1.00 ph n 5.0 g NaOH 1 mol 0.00 g 0.1 mol 0.1 mol [OH ] [NaOH] 0.01 mol/l.00 L 1 Kw [H O ]. 10 mol/l [OH ] 0.01 ph log[h O ] log( ) 1.9 Part (Page77) 1. (a) rønstedlowry acid is a proton donor, in a specific reaction. (b) rønstedlowry base is a proton acceptor, in a specific reaction. (c) rønstedlowry acidbase reaction is a reaction involving the transfer of a proton from one entity (an acid) to another (a base). (d) n entity is one with the ability to either accept or donate a proton. (e) rønstedlowry strong acid is an entity that holds a proton very weakly and can easily donate it to almost any base. (f) rønstedlowry strong base is an entity that attracts a proton very strongly and can easily remove it from almost any acid. Copyright 007 Thomson Nelson Unit 8 Solutions Manual 661

2 1. (a) Solutions of sodium hydrogen sulfate and sodium carbonate are mixed. >50% HSO CO SO HCO (b) queous ammonia is added to a solution of potassium hydrogen sulfite. >50% HSO SO (c) Solutions of sodium hydrogen phosphate and acetic acid are mixed. >50% CHCOOH HPO HPO CHCOO (d) queous sodium hydroxide is added to a solution of sodium hydrogen phosphate. >50% HPO OH PO HO(l) 15. (a) >50% CHCOOH CN CHCOO HCN Conjugate pairs: CHCOOH / CHCOO and CN / HCN (b) >50% HSO HPO SO HPO Conjugate pairs: HSO / SO and HPO / HPO (c) >50% CO HCO Conjugate pairs: / and CO / HCO [HO ][F ] 16. (a) [HF] [ ][OH ] (b) Kb [ ] [HO ][HSO ] (c) [HSO ] (d) [Omit this question. The chemical equation as written is incorrect.] 17. (a) To predict [H O ] and ph for 0.10 mol/l hydroiodic acid: HI HO(l) I HO [HO ] [HI] 0.10 mol/l ph log[ho ] log(0.10) 1.00 ccording to the referenced information and the stoichiometry concept, the concentration of hydronium ions is 0.10 mol/l, and the ph is (b) (Optional: the following answer is provided for teachers who use this question for enrichment.) To predict [H O ] and ph for 0.10 mol/l methanoic acid: HCOOH HO(l) HCOO HO K a The quadratic formula is required for this prediction, as the acid is a weak acid and [HCOOH] which is less than K a 66 Unit 8 Solutions Manual Copyright 007 Thomson Nelson

3 (Note: students are not required to solve such problems requiring use of the quadratic equationin the current Chemistry curriculum. The following solution is provided for teacher reference only, in case this is done with a class as an enrichment exercise.) [HO ][HCOO ] mol/l [HCOOH] t equilibrium: Let x [HO ] [HCOO ] HCOOH (0.10 x) mol/l (Optional) ICE Table for HCOOH HO(l) HCOO HO mount [HCOOH] [H O ] {HCOO ] concentration Initial Change x x x Equilibrium 0.10 x x x x x x x b b ac Using the quadratic formula, x, a x 0.00 or x 0.00 (the negative root is discarded) [HO ]. 10 mol/l ph log [H O ] log. 10 mol/l.8 ccording to the referenced information and the stoichiometry concept, the concentration of hydronium ions is. 10 mol/l, and the ph is.8. (c) To predict [H O ] and ph for 0.10 mol/l hydrosulfuric acid: HS HO(l) 8 HS HO K a The quadratic formula is not required, as the acid is a weak acid and [HS] which is greater than K a [HO ][HS ] [HS] t equilibrium: Let x [HO ] [HS ] Then [HS] (0.10 x) 0.10 (using the assumption) (Optional) ICE Table for HS HO(l) HS H O mount [H S] [H O ] concentration [HS ] Initial Change x x x Equilibrium (0.10 x) 0.10 x x 8 x x [H O ] mol/l Copyright 007 Thomson Nelson Unit 8 Solutions Manual 66

4 5 ph log [HO ] log mol/l.0 ccording to the referenced information and the stoichiometry concept, the concentration of hydronium ions is mol/l, and the ph is (a) HC6H5O 7 HO(l) C6H5O 7 HO [HO ][C6H5O 7 ] HC6H5O 7 ph. [HO ] [C6H5O 7 ] mol/l [HC6H5O 7 ] ( ) mol/l 0.5 mol/l (.8 10 mol/l) mol/l ccording to the information provided and the equilibrium equation expression, the K a of this solution of potassium hydrogen citrate is (b) solution is tested with three indicators to establish its ph. The indicators chosen are methyl orange, bromothymol blue, and phenolphthalein. Each is placed in a sample of the test solution, and the colour is recorded. Critique: This design will provide only a very approximate ph value, of very low precision. t best, it could only determine that the solution was one of: ph>9, 9>pH>7, 7>pH>, or ph<. To be useful and practical, such a design must involve many more indicators. 19. (a) litmus: red (b) methyl red: redorange (intermediate colour, closer to red) (c) methyl orange: yellow (d) phenolphthalein: colourless 0. oth sulfate ions and benzoate ions are rønstedlowry bases, and both will react with H O(l), by removing protons from water molecules to create hydroxide ions. This increase in [OH ] will decrease the [H O ] according to the K w equilibrium, thus increasing the ph. enzoate ion is a stronger base than sulfate ion, and so the benzoate ion will react to a greater extent, and the benzoate solution will have the lower ph value. 1. Since indigo carmine turns blue, the ph of the solution must be less than or equal to 11.. Since phenolphthalein turns red, the ph of the solution must be greater than or equal to Therefore, the household cleaning solution ph is between 10.0 and 11., or about poh [OH ] mol/l. NaCHCOO Na CHCOO CHCOO HO(l) CHCOOH OH t equilibrium: Let x [OH ] [CHCOOH] ICE Table for CHCOO HO(l) CHCOOH OH mount [CH COO ] [OH ] [CH concentration COOH ] Initial Change x x x Equilibrium x x x 66 Unit 8 Solutions Manual Copyright 007 Thomson Nelson

5 K K w b The quadratic formula is not required, as the base is a weak base and [CHCOO ] , which is greater than Kb [OH ][CHCOOH] Kb [CHCOO ] [CH COO ] (0.015 x ) mol/l mol/l (using the assumption) x x [OH ].9 10 mol/l. 6 poh log [OH ] log(.9 10 ) 5.5 ph 1.00 poh [OH ].9 10 mol/l p 100% [CH COO ] mol/l 100% 0.019% ccording to the information provided and the stoichiometric method, the percent reaction is 0.019%, the ph is 8.6, the poh is 5.5, and concentration of hydroxide ions is mol/l.. (a) Copyright 007 Thomson Nelson Unit 8 Solutions Manual 665

6 (b). Each solution is tested with a ph meter. Once tested, the acids are ranked by ph. The strongest acid will have the lowest ph, while the weakest acid will have the highest ph. (Indicators may be used instead of the ph meter, although this method involves more trial and error by adding different indicators to samples of the solutions to obtain a fairly precise ph of each.) manipulated variable: acid solution responding variable: ph controlled variables: concentration of acid, temperature 5. (a) S Na SO K C6H5COO HO(l) S 50% C6H5COO HO(l) HO C6H5COOH (b) S NO Na PO HO(l) S 50% PO HPO (c) S HC6H5O 7 Na HCO HO(l) S 50% HC6H5O 7 HCO HC6H5O 7 HCO 666 Unit 8 Solutions Manual Copyright 007 Thomson Nelson