version 496 Exam 2 Sparks 51815 cid 74 This MC portion of the exam should have 17 questions The point values are given with each question Bub ble in your answer choices on the bubblehseet provided Your score is based on what you bubble on the bub blesheet and not what is circled on the exam Below are some ionization constants you will nd useful Acid acetic acid benzoic acid chlorous acid formic acid hypochlorous acid malonic acid oxalic acid Ka 1 8 10 5 6 4 10 5 1 2 10 2 1 8 10 4 3 5 10 8 1 1 4 10 3 2 2 0 10 6 1 5 6 10 2 2 5 4 10 5 Base ammonia hydrazine hydroxylamine pyridine ethylenediamine Kb 1 8 10 5 1 7 10 6 9 1 10 9 1 7 10 9 1 8 5 10 5 2 7 0 10 8 1 The following reaction is at equilibrium at 250 K I2 g C5H8 g cid 42 cid 41 C5H6 g 2HI g You increase the temperature to 500 K How will the equilibrium concentration of I2 g change 4 pts A It will decrease B It will increase H 92 5 kJ C It will not change Explanation For this endothermic reaction you can consider the reaction to be heat I2 g C5H8 g cid 42 cid 41 C5H6 g 2HI g At increased tempertures the equilibrium constant increases and our reaction will shift to make more products 2 75 mL of 0 015 M HNO3 is added to 100 mL of 0 012 M KOH What is the nal concentration of H or H3O if you prefer 4 pts A 1 71 10 5 M B 7 5 10 11 M C 2 33 10 11 M D 1 33 10 11 M E 7 5 10 4 M Explanation 75 0 015 1 125 mmol H 100 012 1 200 mmol OH Limiting reactant is the acid and 1 2 1 125 0 075 mmol of OH left over in 175 mL of solution which means the concentration of OH is 4 29 10 4 M This makes the H concentration equal to 2 33 10 11 M version 496 page 2 Exam 2 Sparks 51815 cid 74 3 At equilibrium at a certain temperature you nd concentrations of 0 600 M SO2 g 0 300 M O2 g and 3 28 M SO3 g Given the reaction 2 SO2 g O2 g cid 42 cid 41 2 SO3 g Calculate Kc at this temperature 4 pts A 1 00 x 10 2 B 2 36 x 10 4 C 99 6 D 18 2 E 54 6 Explanation Kc SO3 2 SO2 2 O2 3 28 2 6 2 3 4 A 0 019 M solution of 2 chloroproprionic acid is 21 3 ionized at 25 C What is the ionization constant Ka for this organic weak acid 4 pts A 4 2 10 3 B 1 1 10 3 C 6 3 10 2 D 8 6 10 4 E 3 5 10 2 F 2 1 10 4 G 2 1 10 1 Explanation 21 3 of 019 is 0 00405 M which is the amount ionized the H and the A This leaves 0 019 00405 0 0150 M of the unionized acid HA in solution Now plug in those numbers in the mass action expression for Ka Ka H A HA 0 00405 2 0 015 1 1 10 3 5 Determine the pH of a 0 42 M solution of hypochlorous acid HClO 4 pts Explanation Ka 3 5 10 8 H cid 112 Ka 42 This gives you H 1 21 10 4 M The pH log H 3 92 6 Which of the following is not a conjugate acid base pair 4 pts A 2 88 B 3 92 C 5 61 D 0 38 E 4 95 A CN HCN B CH3COOH CH3COO C H3O OH D HF F E NH3 NH 4 Explanation Conjugated acid base pairs di er in formula by a single proton H3O is the conjugate acid of H2O OH is the conjugate base of H2O These are not a conjugate pair with respect to each other version 496 page 3 Exam 2 Sparks 51815 cid 74 7 Which of the following weak bases will have the strongest conjugate acid 4 pts A CH3NH2 Kb 3 6 10 4 B CH3 3N Kb 6 5 10 5 C C2H5NH2 Kb 6 5 10 4 D CH3 2NH Kb 5 4 10 4 Explanation For every conjugate acid base pair Ka Kb Kw Therefore the weakest base will have the strongest conjugate acid partner The larger the value of the equilibrium constant the greater the extent of ionization 8 Determine the pH of a 0 0146 M solution of HBr 4 pts A 3 67 B 1 84 C 4 23 D E 12 2 impossible Ka was not given Explanation HBr is a strong acid Strong acids ionize 100 in solution Therefore the concentration of the hydronium ion is the same as the stated concentration of the strong acid pH log 0 0146 9 A sample of 25 0 mL of a weak acid HZ was Titration of HZ titrated with a 0 038 M solution of KOH The re sulting pH curve for this titration is shown in the gure to the right Using this data determine the original concentration of the weak acid HZ 4 pts A 0 085 M B 0 060 M C 0 070 M D 0 055 M E 0 075 M F 0 080 M G 0 065 M Explanation Analysis of the pH curve reveals that the endpoint of the titration corresponds to a volume of approximately 46 mL of the KOH solution 46mL 0 038M 1 748 mmol of OH There is the same number of mmol of the weak acid HZ Divide mmol by volume to get concentration of the HZ 1 748 25 0 70 M HZ 10 Please refer to the titration curve shown in the previous problem Analyze the titration curve and determine the value of Ka for the weak acid HZ 4 pts A 3 1 10 9 B 1 2 10 8 C 1 0 10 6 D 8 5 10 6 E 2 5 10 7 F 5 0 10 7 G 4 5 10 10 Explanation The endpoint of the titration is about 46 mL Half of that value is 23 mL The pH from the curve at 23 mL is about 6 3 which will be the same as the pKa for the weak acid So that Ka 10 6 3 5 0 10 7 13119753110203040506070pHmL of KOH version 496 page 4 Exam 2 Sparks 51815 cid 74 11 What is the pOH of a 0 0048 M solution of chloric acid HClO3 4 pts A 12 74 B 3 23 C 1 36 D 11 68 E 10 77 F 2 32 Explanation OH Kw H 1 0 10 14 0 0048 2 0833 10 12 Take the log for pOH and …
View Full Document
Unlocking...