1) B 2) C 3) B 4)` E 5) B 6) E 7) C 8)` D 9) D 10) E 11) B 12) D. Bronsted-Lowry Bronsted-Lowry Bronsted-Lowry Bronsted-Lowry

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2 AP Chem Test Double Replacement Reactions and Titrations Name and Date Multiple Choice Record your answers here. 1) B 2) C 3) B 4)` E 5) B 6) E 7) C 8)` D 9) D 10) E 11) B 12) D 13) Identify the Bronsted-Lowry acid-base pairs in the equilibrium of ammonia in water. Label the ACID and BASE NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH (aq) K b = 1.8x10-5 = Bronsted-Lowry Bronsted-Lowry Bronsted-Lowry Bronsted-Lowry CONJUGATE CONJUGATE BASE ACID ACID BASE Accepts Donates Donate Accepts Proton (H+) Proton (H+) Proton (H+) Proton (H+) in the reverse reaction a) Use this equation to justify the statement, Ammonia is a weak base. The presence of hydroxide ions in the product suggests base. The reversible reaction suggests a weak base. Though technically all reaction are reversible. A better way to make this determination is be given the equilibrium constant. b) Write the K b for this weak base. The K b is 1.8x10-5. How does this show that position of equilibrium and indicate that this is a weak base. [NH + 4 ][ OH ] [NH 3 ] Since the Kb is much less than one the ratio of products (the ions including OH-) to reactants (the molecules) is less than one so the bottom of the fraction must be larger. So, this is a weak base. c) Describe equilibrium. At equilibrium the rate of forward reaction is equal to the rate of the reverse reaction. NH 3 is reacting with water at the same rate as the ammonium ions collide with hydroxide ions to form ammonia and water again. Since the rates are equal the conditions ratio of products reactants remains constant and thus an equilibrium constant may be calculated.

3 14) Identify the Bronsted-Lowry acid-base pairs in the equilibrium of hydrofluoric acid in water. Label the ACID and BASE HF (aq) + H 2 O (l) H 3 O + (aq) + F (aq) Bronsted-Lowry Bronsted-Lowry Bronsted-Lowry Bronsted-Lowry CONJUGATE CONJUGATE ACID BASE ACID BASE Accepts Donates Donate Accepts Proton (H+) Proton (H+) Proton (H+) Proton (H+) in the reverse reaction a) Use this equation to justify the statement, Hydrofluoric acid is a weak acid. The presence of hydrogen ions in the product suggests acid. The reversible reaction suggests a weak acid, though technically all reaction are reversible. A better way to make this determination is to be given the equilibrium constant. b) Write the Ka for this weak acid. The Ka is 3.55 x How does this show that position of equilibrium and indicate that this is a weak acid? [H + ][ F ] K a = 1.8x10-5 = [HF] Since the Kb is much less than one the ratio of products (the ions including OH-) to reactants (the molecules) is less than one so the bottom of the fraction must be larger. So, this is a weak base

4 15) Hydrofluoric acid is titrated with lithium hydroxide. a) Write the balanced equation for this reaction. HF (aq) + LiOH (aq) LiF (aq) + H 2 O (l) b) Write the total ionic equation. HF (aq) + Li + (aq) + OH - (aq) + Li + (aq) + F - (aq) + H 2 O (l) c) Write the net ionic equation. HF (aq) + OH - (aq) + F - (aq) + H 2 O (l) 16) HNO 3 (aq) + H 2 O NO 3 - (aq) + H 3 O + (aq) K eq = a very large number How does this show whether nitric acid is a strong or weak acid? The Keq is much greater than one, therefore the top of the fraction (H + and NO 3 ions) must be much larger than the bottom of the fraction (HNO 3 molecules). Since almost all the molecules break up into the ions the acid is strong acid with hydrogen ion concentration equal to or greater than the concentration of the acid. [H + ][ NO 3 ] [HNO 3 ] 17) If 25.0 ml of a standard 0.05 M HCl solution is required to neutralize 20.0 ml of a solution of Sr(OH) 2, what is the concentration of the base? (Start by writing a balanced equation!) 2 HCl + Ba(OH) 2 BaCl 2 (aq) + 2 H 2 O (l) K a = very large 25 ml HCl soln 1L 0.05 mol HCl 1 mol Ba(OH) mol Ba(OH) ml 1L 2 mol HCl mol Ba(OH) ml 0.03 mol Ba(OH) 2 20 ml Ba(OH) 2 soln 1L L of sol n 18) How many ml of a 3M NaOH solution are required to completely neutralize 20.0 ml of 1.5M H 2 SO 4? (Start by writing a balanced equation!) H 2 SO NaOH Na 2 SO 4 (aq) + 2 H 2 O (l) 25 ml H 2 SO 4 soln 1L 1.5 mol H 2 SO 4 1 mol NaOH 1 L NaOH sol n 1000 ml 6.3 ml 1000 ml 1L 2 mol H 2 SO 4 3 mol NaOH 1L 19) Calculate the mass of aluminum hydroxide required to completely react with 20.0 ml of 0.45M HCl. (HINT: write the balanced equation and then use stoichiometry.) 3 HCl + Al(OH) 3 AlCl H 2 O (l) 20 ml HCl soln 1L 0.45 mol HCl 1 mol Al(OH) g Al(OH) g. Al(OH) ml 1L 3 mol HCl 1 mol Al(OH) 3

5 AP Chem Test Double Replacement Reactions and Titrations DO NOT WRITE ON THIS PAGE 1) Which of the following acids is a strong acid? (A) H 3 PO 4 (B) HNO 3 (C) H 2 CO 3 (D) H 3 BO 3 (E) H 2 SO 3 2) What volume of molar HCl is required to neutralize 25.0 milliliters of molar Ba(OH) 2? (A) 20.0 ml (B) 30 0 ml (C) 40.0 ml (D) 60.0 ml (E) 80.0 ml 25 ml Ba(OH) 2 soln 1L 0.12 mol Ba(OH) 2 2 mol HCl 1 L HCl soln 1000 ml 40.0 ml 1000 ml 1L 1 mol Ba(OH) mol HCl 1 L 3) At 25 C, aqueous solutions with a ph of 8 have a hydroxide ion concentration, [OH ], of (A) M (B) M (C) M (D) 1M (E) 8M ph = - log [H+] so 10 -ph = [H+] 4) The net ionic equation for the reaction that occurs during the titration of nitrous acid with sodium hydroxide is (A) HNO 2 + Na + + OH NaNO 2 + H 2 O (B) HNO 2 + NaOH Na + + NO 2 + H 2 O (C) H + + OH H 2 O (D) HNO 2 + H 2 O NO 2 + H 3 O + (E) HNO 2 + OH NO 2 + H 2 O Nitrous acid is a weak acid so it is present mostly as molecules in the aqueous solution. NaOH is a strong base and completely dissociates so the ions are separated in the solution. Sodium ions are spectator ions since they are dissociated in solution attracted to water molecules before and after the reaction. 5) When H 2 SO 4 and Ba(OH) 2 are reacted in a double replacement reaction, one of the products of the reaction is a) H 2 d) BaH 2 b) H 2 O e) SO 2 An acid/base neutralization so one of the products is water the hydrogen ions from the acid got together with hydroxide ions from the base c) BaS

6 6) In the double replacement reaction between the weak acid, HC 2 H 3 O 2 and strong base, NaOH, which ion(s) are spectator ions? a) Na +, C 2 H 3 O 2 d) H +, C 2 H 3 O 2 b) Na +, OH e) Na + only similar to nitrous acid reaction (#4) above since acetic acid is a weak acid as well c) OH only 7) What is the concentration of an NaOH solution if it takes ml of a M HCl solution to titrate ml of the NaOH solution? a) M d) M b) M e) M c) M MV = MV 0.1 (16.25) = x (25) Short cut MV = MV only works effective for 1:1 ratio otherwise it can be confusing 16.2 ml HCl soln 1L 0.10 mol HCl 1 mol NaOH mol NaOH 1000 ml 1L 1 mol HCl mol NaOH 1000 ml M NaOH 25 ml NaOH soln 1 L 8) Consider the reaction system, CoO(s) + H 2 (g) Co(s) + H 2 O(g). The equilibrium constant expression is a) [CoO][H 2] [H b) 2 ] c) [Co][H 2O] [H d) 2 O] e) [Co][H 2O] [Co][H 2 O] [H 2 O] [CoO][H 2 ] [H 2 ] [H 2 ] leave out solids and liquids put products on top

7 9) A student pipetted five milliliter samples of hydrochloric acid and transferred each sample to an Erlenmeyer flask, diluted it with distilled water, and added a few drops of phenolphthalein to each. Each sample was then titrated with a sodium hydroxide solution to the appearance of the first permanent faint pink color. The following results were obtained: Volumes of NaOH Solution First Sample ml Second Sample ml Third Sample ml Fourth Sample ml Fifth Sample ml Which of the following is the most probable explanation for the variation in the student's results? (A) The burette was not rinsed with NaOH solution. (B) The student misread a 5 for a 6 on the burette when the first sample was titrated. (C) A different amount of water was added to the first sample. (D) The pipette was not rinsed with the HCI solution. If the pipet were initialed rinsed with water and then not rinsed again with the HCl solution, in the first trial the HCl was diluted by the water and thus less NaOH was required to titrate (E) The student added too little indicator to the first sample. 10) What is the [H + ] when [OH - ] = 8.1 x 10-5? a) 8.1 x 10-5 M b) 3.6 x 10-6 M c) 1.0 x 10-7 M d) 8.1 x 10-5 M e) 1.2 x M Since the Kw for water = [H+][OH-] = 1.0 x then 1.0 x = [H+] 8.1 x 10-5 so the 1.0 x /8.1 x 10-5 = 1.2 x M 11) A sample of lemon juice is found to have a ph of What is the H + concentration of the juice? a) M d) M b) M e) 355 M c) 11.6 M ph = - log [H+] so 10 -ph = [H+] so =

8 12) A sample of milk is found to have a ph of What is the OH - concentration of the milk? a) 2.5 x M d) 4.0 x 10-8 M b) 1.0 x 10-7 M e) 2.5 x 10-7 M c) 5.0 x 10-7 M ph = - log [H+] so 10 -ph = [H+] so = 2.51 x 10-7 Since the Kw for water = [H+][OH-] = 1.0 x then 1.0 x = [H+] 2.51 x 10-7 so the 1.0 x /2.51 x 10-7 = 3.98 x 10-8 M