Exam4_Solutions_Alternative

# A 100 ml sample of 0 2 m nh 3 solution is titrated to

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A 100 mL sample of 0 . 2 M NH 3 solution is titrated to the equivalence point with 50 mL of 0 . 4 M HCl. What is the final [H 3 O + ]? The ionization constant of NH 3 is 1 . 8 × 10 5 . 1. 1 . 00 × 10 7 M 2. 3 . 70 × 10 11 M 3. 1 . 10 × 10 3 M 4. 6 . 09 × 10 6 M 5. 8 . 61 × 10 6 M correct Explanation: V NH 3 = 100 mL [NH 3 ] = 0 . 2 M V HCl = 50 mL [HCl] = 0 . 4 M K b = 1 . 8 × 10 5 Initially, n NH 3 = (0.1)(0.20) = 0.02 mol n HCl = (0.05)(0.4) = 0.02 mol The titration reaction is NH 3 + HCl NH + 4 + Cl ini, mol 0 . 02 0 . 02 Δ, mol 0 . 02 0 . 02 0 . 02 0 . 02 fin, mol 0 0 0 . 02 0 . 02 At the equivalence point NH + 4 , which hy- drolyses and Cl , a spectator ion, are in the solution. The total volume is 150 mL, giving [NH + 4 ] = 0 . 02 mol 0 . 150 L = 0 . 133333 M NH + 4 + H 2 O NH 3 + H 3 O + ini, M 0 . 133333 0 0 Δ, M x x x eq, M 0 . 133333 x x x K a = K w K b = [NH 3 ][H 3 O + ] [NH + 4 ] 1 × 10 14 1 . 8 × 10 5 = x 2 0 . 133333 M x x 2 0 . 133333 M x = [H 3 O + ] = 8 . 60663 × 10 6 M ( 0 . 133) 007 10.0 points For a redox reaction to be SPONTANEOUS under standard conditions, the following is true 1. Δ G < 0 K > 1 E cell > 0 correct 2. Δ G < 0 K < 1 E cell < 0 3. Δ G > 0 K < 1 E cell > 0 4. Δ G > 0 K > 1 E cell < 0 Explanation: ΔG is negative for a spontaneous reaction. K = [products] [reactants] , so when the equilibrium fa- vors the products (making the reaction spon- taneous), K > 1. Also, E 0 = 0 . 05916 log K n . So when K > 1, E 0 is positive. 008 10.0 points K sp for iron(III) iodate (Fe(IO 3 ) 3 ) is 1 . 0 × 10 14 . We mix two solutions, one containing Fe 3+ and one containing IO 3 . At the instant of mixing, [Fe 3+ ] = 10 4 M and [IO 3 ] = 10 5 M. Which of the following statements is true?

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Version 070 – Exam 4 – Sutcliffe – (52410) 3 1. No precipitate forms because Q sp is greater than K sp . 2. None of the statements is true. 3. A precipitate forms because Q sp is greater than K sp . 4. No precipitate forms because Q sp is less than K sp . correct 5. A precipitate forms because Q sp is less than K sp . Explanation: 009 10.0 points A catalyst 1. is used up in a chemical reaction. 2. changes the value of Δ G of the reaction. 3. does not take part in the reaction in any way. 4. is always a solid. 5. lowers the activation energy of the reac- tion. correct Explanation: A catalyst speeds up a chemical reaction by providing an alternate mechanism which re- quires a lower energy of activation. Although the catalyst takes part in the reaction it is not used up. Catalysts may be in solid, liquid, gaseous or aqueous phase and only a small amount is used. 010 10.0 points Consider the voltaic cell Pt | H 2 (1 atm) | H + (? M) || Cl (1 M) | AgCl(s) | Ag 2 H + + 2 e H 2 E 0 = 0.00 V AgCl + 1 e Ag + Cl E 0 = 0.222 V If the measured cell potential for the cell is 0.430 volts, what is the pH of the solution? 1. less than 1.00 2. 3.75 3. 3.52 correct 4. 4.00 5. 0.253 Explanation: 011 10.0 points NOTE: Use a=1 for this question. For a first-order reaction, after 230 s, 33% of the reactants remain. Calculate the rate constant for the reaction. 1. 0.00209 s 1 2. 0.00174 s 1 3. 207 s 1 4. 0.00482 s 1 correct 5. 0.000756 s 1 Explanation: 012 10.0 points How many mL of a 0.001 M chloride solution must be added to a 100 mL solution of 7 . 2 × 10 5 M Ag + solution for AgCl to precipitate?
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