DEPARTMENT OF CHEMISTRY AND CHEMICAL TECHNOLOGY CHEMISTRY OF SOLUTIONS 202-NYB-05 05 TEST 2 7 NOVEMBER 2011 INSTRUCTOR: I. DIONNE Print your name: Answers INSTRUCTIONS: Answer all questions in the space provided. 1. Duration of this test is 75 minutes. 2. No books or extra paper are permitted. 3. Answer the questions in ink in order to preserve the right to grieve. 4. In order to obtain full credit for your answers, you must clearly show your work. Answers to problems involving calculations must be expressed to the correct number of significant figures and proper units. 5. Calculators may not be shared. Programmable calculators are not permitted. 6. Your attention is drawn to the College policy on cheating. This policy will be enforced. 7. A Periodic Table with constants is provided. Problem 1: / 3 Problem 5: / 4 Problem 2: Problem 3: / 11 / 9 Problem 6 Sig. Fig.: / 4 / 1 Total: / 36 Problem 4: / 3 Units: / 1
PROBLEM 1 (3 marks) In which direction will the position of the following equilibrium be shifted for each of the following changes? 6 CO 2 (g) 6 H 2 O (l) C 6 H 12 O 6 (s) 6 O 2 (g) ΔH = 2801.69 kj (a) (b) (c) (d) (d) (e) CO 2 (g) is added. O 2 (g) is removed. C 6 H 12 O 6 (s) is removed. In a rigid container, some Ar (g) is added. The volume of the container is doubled. The temperature is decreased. to the left equilibrium position no change to the right PROBLEM 2 (11 marks) Short-answer questions. (a) Write a balanced equation that describe the dissociation of perchloric acid in water. HClO 4 H ClO 4 or HClO 4 H 2 O (l) H 3 O ClO 4 (b) For the following aqueous reaction, identify the acid, the base, the conjugate base, and the conjugate acid. HCO 3 C 5 H 5 NH H 2 CO 3 C 5 H 5 N base acid conjugate acid conjugate base (c) Provide the chemical formula for the following species: (i) the conjugate base of HSO 4 (ii) the conjugate acid of HCO 3 SO 4 2 H 2 CO 3 Page 2
(d) Classify each of the following as a strong acid, a weak acid, a strong base or a weak base. (i) NH 4 (ii) HF (iii) CH 3 NH 2 (iv) Sr(OH) 2 (v) HNO 3 (vi) HClO 3 strong acid weak acid strong base weak base (e) Order the following from the strongest to the weakest acid. KOH, KNO 3, KCN, NH 4 Cl, HCl HCl > NH 4 Cl > KNO 3 > KCN > KOH strongest weakest (f) Are solutions of the following species acidic, basic, or neutral? acidic basic neutral (i) NaNO 2 (ii) CaO (iii) NH 4 OCl (iv) KF (g) Which of the following solutions can act as a buffer? (i) HF/KF (ii) HBr/NaBr (iii) NH 3 /KNO 3 (iv) HNO 2 /KNO 2 yes no Page 3
PROBLEM 3 (9 marks) Calculate the ph of each of these solutions. (a) 0.0040 M Ca(OH) 2 [OH ] = 2 x 0.0040 = 0.0080 M log(0.0080) = 2.09691 = poh ph = 14 2.09691 = 11.90309 11.90 (b) 0.45 M NH 4 Cl NH 4 Cl is a salt. NH 4 will make the solution acidic, Cl will not affect the ph. NH 4 NH 3 H I 0.45 M 0 0 C x x x E 0.45 x x x K a = 1.0 x 10 14 /1.8 x 10 5 = 5.56 x 10 10 x 2 /0.45 x = 1.5811 x 10 5 and 5% rule works ph = log(1.5811 x 10 5 ) = 4.801 4.80 Page 4
(c) 0.20 M C 2 H 5 NH 2 C 2 H 5 NH 2 H 2 O (l) C 2 H 5 NH 3 OH I 0.20 M 0 0 C x x x E 0.20 x x x K b = 5.6 x 10 4 x 2 /0.20 x = 1.0853 x 10 2 and 5% rule does not work Solve x using quadratic equation: x = 1.0307 x 10 2 poh = log(1.0307 x 10 2 ) = 1.9869 ph = 12.01 12.01 PROBLEM 4 (3 marks) A 0.15 M solution of a weak acid is 3.0% dissociated. Calculate K a. 3.0% = (x/0.15) x 100 X = 0.0045 K a = (0.0045) 2 /(0.15 0.0045) = 1.392 x 10 4 1.4 x 10 4 Page 5
PROBLEM 5 (4 marks) Calculate the mass of sodium acetate (NaC 2 H 3 O 2, MM = 82.034 g/mol) that must be added to 500.0 ml of 0.200 M acetic acid (HC 2 H 3 O 2 ) to form a ph = 5.00 buffer solution. HC 2 H 3 O 2 H C 2 H 3 O 2 I 0.200 M 0 y C x x x E 0.200 x x y x Since ph = 5.00, [H ] = 1.0 x 10 5 K a = 1.8 x 10 5 = (x)(y x)/(0.200 x) = (1.0 x 10 5 )(y 1.0 x 10 5 )/(0.200 1.0 x 10 5 ) Y = 0.359972 M For 0.5000L, we have 0.179986 mol 0.179986 x 82.034 g/mol = 14.764972 g Or ph = pka log ([base]/[acid]) 5.00 = 4.7447 log (x/0.200) x = 0.3575 M For 0.5000L, we have 0.179986 mol 0.179986 x 82.034 g/mol = 14.764972 g 15 g Page 6
PROBLEM 6 (4 marks) Calculate the ph after 0.010 mol gaseous HCl is added to 250.0 ml of the a buffered solution made of 0.050 M NH 3 and 0.15 M NH 4 Cl. HCl NH 3 NH 4 0.050 M 0.15 M x 0.2500 L x 0.2500 L start with: 0.010 mol 0.0125 mol 0.0375 mol react: 0.010 0.010 0.010 left with: 0 0.0025 mol 0.0475 mol These moles (left over) are in a total volume of 250.0 ml [NH 3 ] = 0.010 M and [NH 4 Cl] = 0.19 M NH 3 H 2 O (l) OH NH 4 I 0.010 M 0 0.19 C x x x E 0.010 x x 0.19 x K b = 1.8 x 10 5 (x)(0.19)/(0.010) x = 9.4737 x 10 7 poh = 6.0235 and ph = 7.9765 7.98 Page 7