1. Answers to Questions Page 345-346 #1 to 20

2. Exercise Answers Page 345 1.The conjugate acid of a base is the particle formed from it after it accepts a proton. The conjugate base of an acid is the negative ion formed after it loses a proton. For example, in the following reaction, H 3 O + is a conjugate acid, Cl - is a conjugate base: HCl + H 2 O  H 3 O + + Cl -

3. 2.a) H 2 Ob) HCO 3 - 3.The following are Arrhenius acids: (c) H 3 PO 4 phosphoric acid, and (d) HF Hydrogen Fluoride (if in aqueous solution) 4.The following are Arrhenius bases: (a) potassium hydroxide, (b) barium hydroxide 5.[H + ] = [H 2 SO 4 ]x2=0.004 so pH = 2.4 6.n=m/M =1.04/56.1 = 0.0185, so pOH = - log(0.0185) = 1.73, so pH = 14 – 1.73 or 12.27, or rounded to the correct significant digits, pH = 12.3

4. 7.

5. 8. 2NH 3  N 2 + 3H 2 213213 [NH 3 ][N 2 ][H 2 ] I2.0 mol/L0 mol/L C E0.60 mol/l +0.60 mol/L+0.20 mol/L-0.40 mol/L +1.6 mol/L+0.20 mol/L a) The [N 2 ] at equilibrium is 0.2 mol/l, the [NH 3 ] is 1.6 mol/L b) The equilibrium constant is : K c = [N 2 ] [H 2 ] 3 / [NH 3 ] 2 or 0.20(0.60) 3 / 1.6 2 so, K c = 0.016875 or, rounded to 2 S.D: 1.7 x 10 -2 0.3 mol in a 0.5 L flask =0.3/0.5 =0.6mol/L 1 mol in a 0.5 L flask =1.0/0.5 =2.0mol/L I am assuming that the author meant 1.0 moles of ammonia, and 0.30 moles of hydrogen. Otherwise there would only be one significant digit, and I never give chemistry problems with less than 2 significant digits.

6. 9. SO 2 + NO 2  NO + SO 3 K c =4.8 1 11 1 [SO 2 ][NO 2 ][NO][SO 3 ] I0.36 mol/L 0 mol/L C E X mol/L +X mol/L -X mol/L b) At equilibrium: K c = [NO] [SO 3 ] [SO 2 ][NO 2 ] 0.36-X X mol/L 4.8 = X 2 (0.36-X) 2 4.8 (0.1296 - 0.72X + X 2 ) = X 2 0.622 – 3.46X+4.8X 2 =X 2 3.8X 2 -3.46X+0.622 = 0 X=0.247 mol/L in a 5L container Total SO 3 = 1.2 mol

7. 10. (for this question, I am assuming the textbook intended to write 3.0 mol/L) a) 4NH 3(g) + 5O 2(g)  4NO (g) + 6H 2 O (g) b) if [NH 3 ]=[NO] then these two will cancel out in the equilibrium expression, so K c =0.26337 Answer: The equilibrium constant is 0.26

8. 11. 2NO 2(g)  N 2 O 4(g) K c =1.15 a) The equilibrium expression is: b) so x = 1.15 (0.05) 2 x = 2.875 × 10 -3 mol/L or 2.9 × 10 -3 mol/L c) decreasing the pressure (by giving the reactants a larger volume) or increasing the concentration of N 2 O 4 would shift the equilibrium towards the reactants.

9. 12. a) HF (aq) + SO 3 2- (aq)  F - (aq) + HSO 3 - (aq) b) CO 3 2- (aq) + CH 3 COOH (aq)  CH 3 COO - (aq) + HCO 3 - (aq) c) H 3 PO 4 - (aq) + OCl - (aq)  H 2 PO 4(aq) + HOCl (aq) d) HCO 3(aq) + HSO 4 - (aq)  SO 4 2- (aq) + H 2 CO 3(aq) ACIDBASE

10. 13. The answer is (a) explanation: hydronium (H 3 O + ) is a proton (H + ) that has been hydrated (ie. combined with water: H + +H 2 O  H 3 O + ) 14. The answer is (b) explanation: Calculate pOH: -log[OH - ] = -log(2.1) = -0.322 Calculate pH : 14-(-0.322) = 14.322 Calculate [H+] : 10-pH = 10 -14.322 = 4.8 x 10 -15 15. The answer is (b) explanation: pOH = 14 – pH, so 14 – 3.46 = 10.54 16. The answer is (b) explanation: calculate moles of NaOH 5g / 40.0 g/mol =0.125 mol calculate concentration: 0.125mol / 4L = 0.3125 mol/L calculate pOH:-log(.3125) = 1.505 calculate pH: 14 – 1.505= 12.495

11. 17. Use the general formula: HA + H 2 O  H 3 O + + A - or simplified: HA  H + + A - This is a weak acid, only five percent ionizes, so 5% of a 0.10 mol/L = 0.05 x 0.10 mol/L =0.005 mol/L of H + (or H 3 O + ) ions and A - ions The acid is 95% undissociated, so its concentration at equilibrium is 0.95 x 0.10 = 0.095 mol/L K a = [H+][A-] =(0.005)(0.005) = 2.6x10 -4 [HA] 0.095 The K a of the weak acid is 2.6x10 -4 Note: for very weak acids (less than 1% ionization) we don’t usually bother changing the denominator. Just leave the original concentration there, and your answer will still work out so close to correct that it will be the same after rounding Sig. Digits. 95%5%

12. 18: a – d – c – b Explanation: The perchloric acid and hydrochloric acids are both “strong” acids, but the perchloric acid is more concentrated so it will be the most acidic and have the lowest pH. Acetic acid is a “weak” acid. Even though its concentration is slightly higher than the hydrochloric, less than 10% of it ionizes, so it will have be less acidic and have a higher pH. Sodium chloride does not have any H+ ions, so its pH will be close to that of pure water (about pH=7), the highest of the four.

13. 19. K c =[NO 2 ] 2 K c = 4.8x10 -3 [N 2 O 4 ] So, we just have to substitute the five sets of values to see which work out to 4.8x10 -3 a) (1.0x10 -4 ) 2 ÷ 4.8x10 -1 = 2.08x10 -8 b) (4.8x10 -4 ) 2 ÷ 1.0x10 -1 = 2.30x10 -6 c) (2.2x10 -2 ) 2 ÷ 1.0x10 -1 = 4.84x10 -3 d) (1.0x10 -1 ) 2 ÷ 2.2x10 -2 = 4.55x10 -1 e) (1.1x10 -2 ) 2 ÷ 5.0x10 -2 = 2.42x10 -3 So the best answer is (c).

14. 20. The best answer is (a) explanation: (a) was the only case where water gave away its H + (donated a proton). It gave the proton to the NH 3 molecule, thus changing it into an NH 4 + ion. According to the Bronsted-Lowry theory, any substance that donates a proton must be an acid. In (b) it absorbed a hydrogen it got from the phosporic acid (accepted a proton), so it was acting like a base. In (c) it decomposed into two covalent compounds, In (d) the water did not change, In (e) there was a chemical reaction, but there is no clear indication of any acid or base behavior.

15. 21. Answer (b) 22. Answer: pH=11.08 or pH=11.1 (rounded to reasonable sig. Digits) 23. The soft drink is acidic because its pH is 2.3 24. Answer: c – b – a – e – d 25. Answer: (a) the pH of tears is 7.4 (b) the pH of gastric acid is 1.4 26. At equilibrium, the concentrations of [H 2 ] and [I 2 ] are both 2.5x10 -4 mol/L 27. The system is not at equilibrium. Since no solid solute exists, the reverse reaction is not occurring. 28. The pH of codeine solution is 10.3 29. The calculation of the equilibrium constant uses stoichiometric coefficients. If the equation is not balanced, the coefficients will be incorrect.

16. 30.Raising the temperature increases the equilibrium constant. Since the constant is products over reactants, the products have been favoured. If products were favoured by raising the temperature, then it must have been endothermic (Le Chatelier’s principle) 31.The pOH of the sodium hydroxide solution is 0.372 32.The pH of this acid is 2.42, and the pOH is 11.58 33.You must dissolve 0.018g of sodium hydroxide (NaOH) 34.The equilibrium constant is 1.7x10 -2.

17. 35. Sulphur trioxide and hydrogen fluoride a.SO 3(g) + 6 HF (g)  SF 6(g) + 3H 2 O(g) b.The equation is: The initial concentrations are found using C=n/V for SO 3 and HF. They are [SO 3 ]=0.617 mol/L and [HF]=1.936 mol/L Using an ICE Table: We could substitute the expressions in the bottom row into the equation, but that is about as far as we can go with this. c. It is difficult to solve this because it contains terms that are to the sixth power, which cannot be solved by simple mathematical equations. SO 3 6 HFSF 6 3 H 2 O 0.6171.93600 -x-6x+x+3x 0.617-x1.936-6xx3x

18. 36.Bronsted-Lowry vs. Arrhenius a.According the Bronsted-Lowry theory, an acid is a substance that donates a proton (H+) and a base is a substance that accepts a proton. b.In both theories, an acid gives off H+ ions, but in the Bronsted-Lowry theory, a base can be any negative ion or radical, not just a compound containing OH 37.Thirty seven a.The conjugate acid of HCO 3 - is H 2 CO 3, the conjugate base is CO 3 2- b.HCO 3 - + H 3 O  H 2 CO 3 + H 2 O B/L base c.HCO 3 - + OH-  CO 3 2- + H 2 O B/L acid

19. 38.Tables don’t usually list the K sp of soluble salts, because they would be extremely large numbers. 39.Most salts would have a higher solubility product at higher temperatures 40.If there is no solid left, then there is no reverse change taking place, therefore no equilibrium ( this is just a different way of asking question 27 ).