s for Ch. 14 Acids and Bases Practice Questions 1. In the following reactions, label the Acid, Base, Conjugate Acid, and Conjugate Base. Also indicate the two conjugate acid-base pairs Plan Your Strategy Identify the proton donor on the left side of the equation as the acid and the proton receiver on the left side as the base. Identify the conjugate acid and base on the right side of the equation by the difference of a single proton from the acid and base on the left side a. HClO 4(aq) H 2 O (l) H 3 O (aq) ClO 4 (aq) Acid Base Conjugate Acid Conjugate Base b. H 3 PO 4 (aq) H 2 O (l) H 3 O (aq) H 2 PO 4 (aq) Acid Base Conjugate Acid Conjugate Base c. CN (aq) HCOOH (aq) HCN (aq) HCOO (aq) Base Acid Conjugate Acid Conjugate Base d. NH 3 (aq) H 2 O (l) NH 4 (aq) OH (aq) Base Acid Conjugate Acid Conjugate Base e. CH 3 COOH (aq) H 2 O (l) CH 3 COO (aq) H 3 O (aq) Acid Base Conjugate Base Conjugate Acid 2. Write the formula for the conjugate base of each molecule or ion. a. HCl b. HCO 3 c. H 2 SO 4 d. N 2 H 5 Cl - 2 - CO 3 HSO 4 A conjugate base differs from the molecule or ion by having one less proton. N 2 H 4 3. Write the formula of the conjugate acid of each molecule or ion. a. NO 3 b. OH c. H 2 O d. HCO 3 HNO 3 HOH, H 2 O H 3 O H 2 CO 3 A conjugate acid differs from the molecule or ion by having one more proton.
4. An amphoteric species is a substance that can act as an acid in one reaction and as a base in a different reaction. Examine the reactions in question 1 d and e. Identify the amphoteric chemical species, and identify its role as either an acid or a base. In 1d, H 2 O (l) is acting as a proton donor and is therefore an acid In 1e, H 2 O (l) is acting as a proton receiver and is a base. 5. HS, can be classified as amphoteric. Write equations to show the hydrogen sulfide ion acting as a) an acid b) a base HS (aq) H 2 O (l) S 2- (aq) H 3 O (aq) HS (aq) H 2 O (l) H 2 S (aq) OH (aq) 10. Predict the direction for the following equations. State whether reactants or products are favoured, and give reasons to support your decision. Using Figure 14.12, identify the weaker acid and the weaker base. The reaction will proceed towards the formation of these two weaker species. a. NH 4 (aq) H 2 PO 4 (aq) NH 3 (aq) H 3 PO 4 (aq) Reactants will be favoured in the above reaction (location of weaker substances) b. H 2 O (l) HS (aq) OH (aq) H 2 S (aq) Reactants will be favoured in the above reaction. (location of weaker substances) c. HF (aq) SO 4 2 (aq) F (aq) HSO 4 (aq) Reactants will be favoured in the above reaction. (location of weaker substances) 11. In which direction will the following reactions proceed? Explain why in each case. a. HPO 4 2 (aq) NH 4 (aq) H 2 PO 4 (aq) NH 3 (aq) SHIFT LEFT (location of weaker substances) H 2 O (l) HSO 4 (aq) H 3 O (aq) b. H 2 SO 4 (aq) Strong Acid Strong Base Weak Base Weak Acid SHIFT RIGHT (location of weaker substances) c. H 2 S (aq) NH 3(aq) HS (aq) NH 4 (aq) Strong Acid Strong Base Weak Base Weak Acid SHIFT RIGHT (location of weaker substances)
12. Determine [H 3 O ] and [OH - ] in each solution. (a) 0.45 mol/l hydrochloric acid (b) 1.1 mol/l sodium hydroxide i) Verify whether you have a strong or a weak solution. The above substances are both strong so write a balanced dissociation equation using a single arrow. a) HCl H (aq) Cl - (aq) 1:1 mole ratio so [H ] or [H 3 O ] =0.45 mol/l Kw = [H 3 O ] [OH ] 1.0 10 14 = 0.45 mol/l [OH ] 1.0 10 14 = [OH ] 0.45 mol/l 2.2 10 14 mol/l = [OH ] b) NaOH (aq) Na (aq) OH (aq) 1:1 mole ratio so [OH ] = 1.1 mol/l Kw = [H 3 O ] [OH ] 1.0 10 14 = [H 3 O ] (1.1 mol/l) 1.0 10 14 = [H 3 O ] 1.1 mol/l 9.1 10 15 mol/l = [H 3 O ] Check Your (a) is a strong acid. Therefore, [H 3 O ] should be greater than 1.0 10 7, and [OH ] should be less than 1.0 10 7. For a solution of a strong base, as in (b), [OH ] should be greater than 1.0 10 7, and [H 3 O ] should be less than 1.0 10 7. 13. Determine [H 3 O ] and [OH ] in each solution. (a) 0.95 mol/l hydroiodic acid (b) 0.012 mol/l calcium hydroxide i) Verify whether you have a strong or a weak solution. The above substances are both strong so write a balanced dissociation equation using a single arrow. a) HI H (aq) I - (aq) 1:1 mol ratio so [H ] or [H 3 O ] =0.45 mol/l Kw = [H 3 O ] [OH ] 1.0 10 14 = 0.95 mol/l [OH ] 1.0 10 14 = [OH ] 0.95 mol/l 1.1 10 14 mol/l = [OH ] b) Ca(OH) 2(aq) Ca 2 (aq) 2OH (aq) 0.012 mol/l Ba(OH) 2 2 mol OH = 0.024 mol/l OH 1 mol Ba(OH) 2 ii) Kw = [H 3 O ] [OH ] 1.0 10 14 = [H 3 O ] (0.024 mol/l) 1.0 10 14 = [H 3 O ] 0.024 mol/l 4.2 10 13 mol/l = [H 3 O ] 14. [OH ] is 5.6 10 14 M in a solution of hydrochloric acid. What is the molar concentration of the HCl (aq)? i) Kw = [H 3 O ] [OH ] 1.0 10 14 = [H 3 O ] (5.6 10 14 mol/l) 1.0 10 14 = [H 3 O ] 5.6 10 14 mol/l 0.18 mol/l = [H 3 O ] ii) HCl H (aq) Cl - (aq) Hydrochloric acid is a strong acid. The dissociation equation shows 1:1 mol ratio so [HCl] = [H ] =0.18 mol/l
15. [H 3 O ] is 1.7 10 14 in a solution of calcium hydroxide. What is the molar concentration of the Ca(OH) 2(aq)? i) Kw = [H 3 O ] [OH ] 1.0 10 14 = (1.7 10 14 ) [OH ] 1.0 10 14 = [OH ] 1.7 10 14 0.588 mol/l = [OH ] ii) Ca(OH) 2(aq) Ca 2 (aq) 2OH (aq) 0.588 mol/l OH 1 mol Ca(OH) 2 = 0.294 mol/l Ca(OH) 2(aq) 2 mol OH 16. Calculate the ph of each solution, given the hydronium ion concentration. Use the formula ph = - log [H 3 O ] to calculate each answer [H 3 O ] in mol/l ph value Acidic/Basic A 0.0027 2.57 Acidic B 7.28 x 10-8 7.14 Neutral (slightly basic) C 9.7 x 10-5 4.01 Acidic D 8.27 x 10-12 11.08 Basic Substance [H 3 O ] in mol/l ph value Acidic/Basic 17 Cola 5.0 x 10-3 2.30 Acidic 18 Orange Juice 2.9 x 10-4 3.54 Acidic 19a nitric acid 6.3 10 3 2.2 19b sodium hydroxide 6.59 10 10 9.18 20. [H 3 O ] of a sample of milk is found to be 3.98 10 7 mol/l. Is the milk acidic, neutral, or basic? Calculate the ph and [OH ] of the sample. i) ph = - log [H 3 O ] = - log 3.98 10 7 = 6.400 ii) 1.00 x 10-14 = [H 3 O ] [OH - ] 1.00 x 10-14 = (3.98 10 7 ) [OH - ] 3.98 10 7 2.51 x 10-8 mol/l = [OH - ] 21. A sample of household ammonia has a ph of 11.9. What is the poh and [OH ] of the sample? i) ph poh 14.0 11.9 poh 14.0 poh 14.0 11.9 poh = 2.1 ii) [OH ] = 10 -poh [OH ] = 10-2.1 [OH ] = 8 x 10-3 mol/l
22. Phenol, C 6 H 5 OH, is used as a disinfectant. An aqueous solution of phenol was found to have a ph of 4.72. Is phenol acidic, neutral, or basic? Calculate [H 3 O ], [OH ], and poh of the solution. i) As the ph is less than 7, the solution is acidic ii) [H 3 O ] = 10 -ph [H 3 O ] = 10-4.72 [H 3 O ] = 1.9 x 10-5 mol/l iii) 1.00 x 10-14 = [H 3 O ] [OH - ] 1.00 x 10-14 = (1.9 x 10-5 ) [OH - ] 1.9 x 10-5 5.3 x 10-10 mol/l = [OH - ] iv) ph poh 14.0 4.72 poh 14.0 poh 14.0 4.72 poh = 9.28 23. At normal body temperature, 37 C, the value of Kw for water is 2.5 10 14. Calculate [H 3 O ] and [OH ] at this temperature. Is pure water at 37 C acidic, neutral, or basic? Pure water is neutral since the concentration of [H 3 O ] = [OH - ] Note the following equation: H 2 O (l) H 2 O (l) H 3 O (aq) OH - (aq) 2.5 10 14 = [H 3 O ] [OH - ] 2.5 10 14 = x 2 /- 1.6 10 7 = x It is not possible to have a negative concentration, therefore [H 3 O ] and [OH - ] are both 1.6 10 7 mol/l 24. A sample of baking soda was dissolved in water and the poh of the solution was found to be 5.81 at 25 C. Is the solution acidic, basic, or neutral? Calculate the ph, [H 3 O ], and [OH ] of the solution. i) ph poh 14.0 ph 5.81 14.0 ph 14.0 5.81 ph = 8.19 As the ph is larger than 7, baking soda is a base ii) [OH - ] = 10 -poh [H 3 O ] = 10-5.81 [H 3 O ] = 1.5 x 10-6 mol/l iv) 1.00 x 10-14 = [H 3 O ] [OH - ] 1.00 x 10-14 = [H 3 O ] (1.5 x 10-6 M) 1.5 x 10-6 7. x 10-9 mol/l = [OH - ] 25. A chemist dissolved some Aspirin in water. The chemist then measured the ph of the solution and found it to be 2.73 at 25 C. What are [H 3 O ] and [OH ] of the solution? i) [H 3 O ] = 10 -ph [H 3 O ] = 10-2.73 [H 3 O ] = 1.9 x 10-3 mol/l ii) 1.00 x 10-14 = [H 3 O ] [OH - ] 1.00 x 10-14 = [H 3 O ] (1.9 x 10-3 M) 1.9 x 10-3 5.3 x 10-12 mol/l = [OH - ]