General Chemistry II CHM 1046 E Exam 2

General Chemistry II CHM 1046 E Exam 2 Dr. Shanbhag Name: 1. The formation of ammonia from elemental nitrogen and hydrogen is an exothermic process. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H= -92.2 kj Which of the following would drive the equilibrium system to the left? a. addition of hydrogen b. removal of ammonia c. increasing the pressure Le Chatelier s Principle d. decreasing the temperature e. removal of nitrogen 2. All of the following are weak acids EXCEPT a. HF b. CH 3 CO 2 H c. HBr HBr is one of the strong acids d. + NH 4 e. HCN 3. In the following reaction HCO - 3 (aq) + NH 3 (aq) CO 3 (aq) + NH + 4 (aq) a. HCO - 3 is an acid and NH 3 (aq) is its conjugate base. b. HCO - 3 is an acid and CO 3 is its conjugate base. c. NH 3 is an acid and HCO 3 is its conjugate base. d. NH 3 is an acid and NH + 4 is its conjugate base. e. NH + 4 is an acid and CO 3 is its conjugate base. HCO 3, and CO 3 are a conjugate pair NH 3, and NH 4 + are a conjugate pair 4. What is the conjugate acid of HPO 4 (aq)? a. H 3 PO 4 b. - H 2 PO 4 c. PO 4 d. H 3 O + e. OH - That means HPO 4 is produced by acid dissociation (by loss of a proton) 5. Molecules or ions that can behave as either a Brønsted-Lowry acid or base are called a. amphiprotic. b. hydronium. Definition c. polyprotic acids or bases. d. conjugate acids or bases. e. weak electrolytes. 6. A solution is prepared by diluting 0.25 mol HNO 3 to a volume of 750 ml. What is the ph of this solution? a. 0.33 b. 0.48 c. 0.60 = -log { (0.25/0.750)} d. 3.5 e. 4.2 7. Which one of the following solutions will have a ph of 3.0? a. 3.0 M HCl b. 3.0 M NH 3 c. 0.0010 M HBr ph = 3 means [H 3 O + ] = 0.001 M d. 0.0010 M CH 3 CO 2 H e. 0.0010 M NaOH

8. What is the ph of 1.3 10-5 M NaOH at 25ºC? a. 1.30 b. 4.89 c. 9.11 d. 11.58 9. What is the OH - concentration of a solution with a ph of 12.81? a. 1.6 10-13 M b. 2.7 10-8 M poh = 1.19, [OH] = 10-1.19 M c. 6.5 10-2 M d. 1.2 10 0 M e. 1.3 10 1 M 10. A solution with a ph of 2.00 is diluted from 1.0 L to 2.0 L. What is the ph of the diluted solution? a. 1.00 b. 1.70 c. 2.30 d. 3.00 e. 4.30 Since SB, poh = -log{1.3 10-5 } = 4.886 ph = 14-pOH = 9.11 11. Which of the following chemical reactions corresponds to the acid ionization constant, K a, for hydrogen carbonate ion (HCO 3 - )? a. HCO 3 - (aq) + OH - (aq) CO 3 (aq) + H 2 O( ) ph = 2 means [H 3 O + ] = 0.01 M when diluted new [H 3 O + ] = 0.005 M So new ph = 2.30 b. HCO - 3 (aq) + H 3 O + (aq) H 2 CO 3 (aq) + H 2 O( ) c. HCO - 3 (aq) + H 3 O + (aq) CO 2 (g) + 2 H 2 O( ) d. HCO - 3 (aq) CO 2 (g) + OH - (aq) e. HCO - 3 (aq) + H 2 O( ) CO 3 (aq) + H 3 O + (aq) definition 12. Which of the following weak acids has the strongest conjugate base? a. acetic acid, K a = 1.8 10-5 b. benzoic acid, K a = 6.3 10-5 c. dihydrogen phosphate ion, K a = 6.2 10-8 d. formic acid, K a = 1.8 10-4 e. hydrocyanic acid, K a = 4.0 10-10 13. The ph of 0.010 M trimethylamine is 10.88. What is the value of K b for this base? a. 1.3 10-11 b. 9.8 10-8 c. 4.8 10-7 d. 6.2 10-5 e. 7.6 10-4 14. The ph of 1.5 M formic acid, HCO 2 H, is 1.78. What is the pk a for this acid? a. 1.86 b. 2.43 [H c. 3.73 3 O + ] = So approximate K a = [10-1.78 x 10-1.78 ]/1.5 = 0.0001836 d. 5.14 So pka = 3.736 e. 6.92 15. Of the following salts, which one forms a 0.1 M solution with the highest ph? a. KCl b. NH 4 Cl c. FeCl 3 d. KNO 2 e. Ca(NO 3 ) 2 As Ka decrease, the acid strength decreases and the strength of the conjugate base increases poh = 3.14, [OH] = 10-3.14 M Kb = [(10-3.14 )( 10-3.14 )] / 0.010 approximately!!! Salts produced from weak acids produce basic solution because of the reactivity of their conjugate base anion, in this bunch KNO 2 produced from HNO 2 which is a WA

16. Which of the following chemical reactions corresponds to K a2 for phosphoric acid? a. HPO 4 (aq) + H 2 O( ) PO 4 3- (aq) + H 3 O + (aq) b. PO 3-4 (aq) + H 2 O( ) HPO 4 (aq) + OH - (aq) c. H 3 PO 4 (aq) + H 2 O( ) H 2 PO - 4 (aq) + H 3 O + (aq) d. H 3 PO 4 (aq) + 2 H 2 O( ) HPO 4 (aq) + 2 H 3 O + (aq) e. H 2 PO - 4 (aq) + H 2 O( ) HPO 4 (aq) + H 3 O + (aq) definition 17. Which one of the following equations represents the reaction of a weak acid with a strong base? a. H+(aq) + OH (aq) H2O(aq) b. H+(aq) + CH3NH2(aq) CH3NH3 + (aq) definition c. OH (aq) + HCN(aq) H2O(aq) + CN (aq) d. HCN(aq) + CH3NH2(aq) CH3NH3 + (aq) + CN (aq) 18. Calculate the hydrogen ion concentration in a solution of fruit juice whose ph is 4.25. a. 1.0 10 14 M b. 5.6 10 5 M c. 4.0 10 25 M [H 3 O + ] = 10-4.25 M d. 2.5 10 4 M e. 5.6 10 4 M 19. The ph of a certain solution is 2.0. How many H+(aq) ions are there in 1.0 L of the solution? a. 0.010 ions b. 100 ions The solution has 0.01 moles of H + per liter c. 2 ions That means it has 0.01 x 6.022 x 10 23 H + ions per liter of solution d. 6.02 1021 ions e. 6.02 1023 ions 20. A 0.14 M HNO2 solution is 5.7% ionized. Calculate the H+ ion concentration. a. 8.0 1 0 3 M b. 0.057 M If 5.7 % ionized, then [NO 2 ] = (5.7/100)*0.14 M = [H 3 O + ] = 0.00798 M c. 0.13 M d. 0.14 M e. 0.80 M 21. Which solution will have the lowest ph? a. 0.10 M HCN b. 0.10 M HNO3 Strong acid solution of a given molarity will have lower ph than a c. 0.10 M NaCl weak acid of same molarity (even if it is a diprotic acid)! d. 0.10 M H2CO3 e. 0.10 M NaOH 22. In which of the following gas-phase equilibria is the yield of products increased by increasing the total pressure on the reaction mixture? a. b. c. d. CO(g) + H 2 O(g) 2NO(g) + Cl 2 (g) 2SO 3 (g) PCl 5 (g) CO 2 (g) + H 2 (g) 2NOCl(g) 2SO 2 (g) + O 2 (g) PCl 3 (g) + Cl 2 (g) Whichever has fewer gaseous moles on right side!

23. The ph of a Ba(OH)2 solution is 10.0. What is the H+ ion concentration? a. 4.0 10 11 M b. 1.6 10 10 M c. 1.3 10 5 M d. 1.0 10 10 M e. 10.1 M 24. For the following reaction at equilibrium in a reaction vessel, which one of the changes below would cause the Br2 concentration to decrease? 2NOBr(g) 2NO(g) + Br 2 (g) DH rxn = 30 kj a. Increase the temperature. b. Remove some NO. c. Add more NOBr. d. Compress the gas mixture into a smaller volume. 25. Sulfurous acid is a diprotic acid. Its two stages of ionization are H 2 SO 3 (aq) H+ + HSO 3 - K a1 = 1.3 10-2 HSO 3 - H+ + SO 3 K a2 = 6.3 10-8 If ph s 10, then regardless of what is responsible for it, [H 3 O + ] has to be 1.0 x 10-10 M!!! Increasing pressure by decreasing volume will shift the equilibrium towards fewer gaseous moles Which of the following species would have the highest concentration in a 0.10 M H2SO3 solution with an adjusted ph of 3? a. H+ Since only small part of starting acid is dissociated and this conjugate species further dissociates with a very low Ka, the conc of all species is as follows: b. H2SO3 c. HSO3 d. SO3 2 H2SO3 >>> H+ >. HSO3 >>>> SO3 2 26. Determine the ph of a KOH solution made from 0.251 g KOH and enough water to make 1.00 102 ml of solution. a. 1.35 b. 2.35 [OH] = (0.252/56.1)/0.100, mw of KOH = 56.1, and it is a SB c. 7.00 poh = 1.349 d. 11.65 e. 12.65 27. Acid strength increases in the series: HCN < HF < HSO4. Which of the following species is the strongest base? a. H2SO4 b. SO4 2 The weaker the acid, the stronger is its conjugate base! c. F d. CN e. HSO4 28. The ph of coffee is approximately 5.0. How many times greater is the [H3O+] in coffee than in tap water with a ph of 8.0? a. 0.62 b. 1.6 They differ in 3 ph units that means they c. 30 are 10 3 apart in [H 3 O + ] d. 1,000 e. 1.0 104 29. Which of the following has an effect on the magnitude of the equilibrium constant? a. activation energy of the forward reaction b. concentrations of the reactants and products K eq can be changed by temp. only c. presence of a catalyst d. change in temperature 30. For the system, NH 2 OH + CH 3 NH 3 +!"CH 3 NH 2 + NH 3 OH + 95% 95% 5% 5% the state of equilibrium in the system is described by the numbers given. Which is the weakest base in the system? a. NH 2 OH b. CH 3 NH 3 + Weakest base will be least reactive, that means it will have larger concentration, c. CH 3 NH 2 in this NH 2 OH and CH 3 NH 2 are functioning as bases, NH 2 OH is 95% d. NH 3 OH +

Name: 1. Write a balanced dissociation reaction for the Bronsted base- HSO 4 HSO 4 + H 2 O #$ H 2 SO 4 + OH 2. Calculate the degree of dissociation (α) for a 0.240 solution of a weak acid that has a Ka = 4.9 x 10-7 HA + H 2 O #$ H 3 O + + A x =square root ( 4.9 x 10-7 *0.24) = 3.429x10-4 0.24 0 0 α = 3.429x10-4 /0.240 = 0.00143 -x x x 3. Give an example each of a strong acid, a weak acid, a strong base and a weak base SA = HClO 4 SB = KOH WA = HCN WB ==NH 3 4. Give an example to demonstrate the conjugate acid-base pairs in equilibrium by selecting appropriate compounds. Comment on the strength or reactivity of each of the species involved in the equilibrium Consider HCl and NH 3 as the acid and the base, HCl is SA, NH 3 is a WB HCl + NH 3 #$ NH 4 + + Cl Since HCl is a SA, its conjugate base is weak, NH 3 being a weak base, NH 4 + is a strong CA HCl and NH 4 + are highly reactive while NH 3 and Cl are quite inert 5. Consider the equilibrium: A(s)!"B(s) + C(g) H o > 0 Predict and explain how or whether the following actions would affect this equilibrium. 1. increasing the pressure on the system by reducing its volume Increase in pressure by reducing volume will shift a gaseous equilibrium towards fewer gaseous moles. In this case there are no reactant gases. So when the volume is reduced and pressure is increases, equilibrium will shift left, no change in K eq 2. adding helium gas to increase the total pressure Addition of unreactive gas will will have no effect on neither the position of the equilibrium nor the K eq 3. adding more solid A Since A is not taking part in the equilibriumconstant or quotient, any increase in the amount of solids will have no effect on the position of equilibrium or K eq 4. lowering the temperature The reaction being endothermic lowering temperature will shift the equilibrium to left. Moreover, the K eq will be lowered as well, as a result, the equilibrium will further shift towards left. 6. I want to prepare a solution of weak monoprotic acid that has a ph of 3.85. How many moles of acid should I dissolve in one liter to prepare this solution? (Ka for the acid is 3.18x 10-4 ) HA + H 2 O #$ H 3 O + + A ph = 3.85 means [H 3 O + ] = [A ] = 10-3.85 M 3.18 x 10-4 = (10-3.85 ) x(10-3.85 ) [HA] eq [HA] eq = 6.2744 x10-5 M [HA] initial = [HA] eq +[A ] = 6.2744 x10-5 M + 1.4125 x 10-4 M = 2.04x10-4 M So you need 2.04 x 10-4 moles of the weak acid to dissolve in 1 L solution to produce a ph of 3.85 7. A 0.248 M solution of a weak base has a ph of 10.96. Calculate the Kb for this base. ph = 10.96 means poh = 3.04 that means [OH ] = 10-3.04 M K b = (10-3.04 ) x (10-3.04 ) = 3.4 x 10-6 0.248

8. Consider the reaction CaCrO 4(s) #$ Ca 2+ (aq) + CrO 4 (aq) The equilibrium constant for this reaction is 7.1 x 10-4 @ 298 K. If I place 20 g of CaCrO 4 in 1L water, what are the equilibrium concentrations of Ca 2+ (aq) and CrO 4 (aq)? (Hint, the mass and the volume may be distractors!) CaCrO 4(s) #$ Ca 2+ (aq) + CrO 4 (aq) K eq = [Ca 2+ (aq)][ CrO 4 (aq) ]???? 0 0 since Q<K shift right -x +x +x @eq.??? x x so 7.1 x 10-4 = x 2 x = 2.665 x 10-2 M = [Ca 2+ ] eq = [CrO 4 ] eq 9. Consider a 0.0200 M solution of formic acid (HFo) This solution is found to have 2.13 % dissociated, determine 1. the [H 3 O + ] at equilibrium, then determine the ph HFo + H 2 O #$ H 3 O + + Fo If 2.13% dissociated, then [Fo ] =(2.13/100)*0.0200 M = 0.000426 M [Fo ] = [ H 3 O + ] = 0.000426 M ph = 3.37 2. the value of Ka and then the value of pka Ka = (0.000426 x 0.000426) / (0.0200-0.000426) = 9.3 x 10-6 pka = 5.033 3. Give the conjugate pairs involved in the dissociation equilibria of HFo {HFo and Fo } are conjugate pairs and so are {H 2 O, H 3 O + } 4. Calculate the ph of 0.0200 M HNO 3, what is the difference between 0.0200 M HFo and 0.0200 M HNO 3? Since HNO 3 is a SA, ph = -log (0.02) = 1.699 HFo being the weak base, the ph is 3.37 That means HNO 3 will produce [10 (3.37-1.699) ] = 46.88 times as strong acidic solution as compared to HFo 10. H 3 PO 4 has 3 pka values, 2.12, 7.13 and 12.3. Determine the ph of 0.2 M H 3 PO 4 H 3 PO 4 + H 2 O #$ H 3 O + + H 2 PO 4 H 2 PO 4 + H 2 O #$ H 3 O + + HPO 4 HPO 4 + H 2 O #$ H 3 O + + PO 4 3- K 1 = 10-2.13 K 2 = 10-7.13 K 3 = 10-12.3 Considering the first equilibrium H 3 PO 4 + H 2 O #$ H 3 O + + H 2 PO 4 K 1 = 10-2.13 so [H 3 O + ] = 10-2.13 x 0.20 = 0.0385 M Considering the second equilibrium H 2 PO 4 + H 2 O #$ H 3 O + + HPO 4 K 2 = 10-7.13 Initial 0.0385 0.0385 0 so Q=0 <K shift right Change -x x x Eq. 0.00385 x 0.0385 + x x K 2 = 10-7.13 = 7.41 x 10-8 = [0.0385 + x][ x] assuming x << 0.0385.0385-x x = 7.41 x 10-8 M So [H 3 O + ] at this point is 0.0385 M +7.41 x 10-8 M = 0.0385 M, contribution from the second step is negligible! Since there is a 10 5 difference in the Ka 1 and Ka 2, one can neglect the contribution from the second and the third dissociation So [H 3 O + ] total = 0.0385 M ph = 1.41