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Chapter 17 Additional Aspects of Aqueous Equilibria

Chapter 17 Additional Aspects of Aqueous Equilibria. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Troy Wood University of Buffalo Buffalo, NY  2006, Prentice Hall.

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Chapter 17 Additional Aspects of Aqueous Equilibria

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  1. Chapter 17Additional Aspects of Aqueous Equilibria Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Troy Wood University of Buffalo Buffalo, NY  2006, Prentice Hall

  2. What is [H+] of a 0.050 M HF solution dissolved in 0.20 M NaF? Ka(HF) = 6.8  104. 1.7  104 M 3.4 104 M 6.8  104 M 1.4  103 M 2.7  103 M

  3. Correct Answer: 1.7  104 M 3.4 104 M 6.8  104 M 1.4  103 M 2.7  103 M The solution to problem is on the following slide.

  4. Correct Answer: -  4 K [HF] (6.8 10 )(0.050) + = = a [H ] - (0.20) [F ] + - =  4 [H ] 1.7 10

  5. Calculate the pH of a 0.50 M solution of sodium acetate in 0.050 M acetic acid (HOAc). pKa(HOAc) = 4.74. 3.74 4.24 4.74 5.24 5.74

  6. Correct Answer: 3.74 4.24 4.74 5.24 5.74 [base] = + pH p K log a [acid] [0.50] = + pH 4.74 log [0.050]

  7. Calculate the pH of a buffer solution containing 0.25 moles sodium acetate and 0.30 moles acetic acid (HOAc) to which 0.20 moles HCl are added. pKa(HOAc) = 4.74. 3.74 4.24 4.74 5.24 5.74

  8. Correct Answer: 3.74 4.24 4.74 5.24 5.74 [base] = + pH 4.74 log [acid] - [0.25 0.20] = + pH 4.74 log + [0.30 0.20] [0.05] = + = pH 4.74 log 3.74 [0.50]

  9. What is the pH of an aqueous solution to which 51.0 mL of 0.10 M NaOH have been added to 50.0 mL 0.10 M HCl? 2.00 2.70 3.00 3.30 4.00

  10. Correct Answer: 2.00 2.70 3.00 3.30 4.00 The solution to problem is on the following slide.

  11. Correct Answer: Moles H+ = (0.10 M)(0.0500 L) = 0.0050 Moles OH = (0.10 M)(0.0490 L) = 0.0049 - (0.0050 0.0049) + = =  3 [H ] 1.0 10  + (0.050 0.049) pH = log(1.0  103) = 3.00

  12. What is the pH of an aqueous solution to which 26.0 mL of 0.10 M NaOH have been added to 50.0 mL 0.050 M HCl? 10.00 10.70 11.00 11.30 12.00

  13. Correct Answer: 10.00 10.70 11.00 11.30 12.00 The solution to problem is on the following slide.

  14. Correct Answer: Moles H+ = (0.050 M)(0.050 L) = 0.0025 Moles OH = (0.10 M)(0.0260 L) = 0.0026 - (0.0026 0.0025) - = =  3 [OH ] 2.0 10  + (0.026 0.025) pOH = log(2.0  103) = 2.70 pH = 14.00  2.70 = 11.30

  15. What is the molar solubility of BaF2 if its Ksp = 1.0  106? 3.1  103 M 1.0 102 M 6.3  103 M 8.0  103M

  16. Correct Answer: 3.1  103 M 1.0 102 M 6.3  103 M 8.0  103M The solution to problem is on the following slide.

  17. Correct Answer: Ksp = [Ba2+][F]2 Let x = [Ba2+]; [F] = 2x. Ksp = (x)(2x)2 = 4x3 x = (Ksp /4)1/3 x = 6.3  103

  18. What is the molar solubility of CaF2 in 0.010 M NaF? Ksp = 3.9  1011? 9.5  106 M 1.9 107 M 3.9  107M 3.9  109M

  19. Correct Answer: 9.5  106 M 1.9 107 M 3.9  107M 3.9  109M The solution to problem is on the following slide.

  20. Correct Answer: Ksp = [Ca2+][F]2 Let x = [Ca2+]; [F] = 0.010 + 2x. Ksp = (x)(0.010)2 x = Ksp /(0.010)2 x = 3.9  107

  21. Which of the following substances will not be more soluble in acidic solution than basic solution? The sinkhole below is a dire result of such “solubility.” AgCl Ni(OH)2 CaCO3 BaF2

  22. Correct Answer: AgCl Ni(OH)2 CaCO3 BaF2 Of these, only AgCl does not produce a basic anion that will react with H+.

  23. Calculate [Ag+] when a NaCN solution is added to 0.10 M AgNO3 (and the equilibrium [CN] = 0.10 M. Kf (Ag(CN)2) = 1  1021? 1  1017 M 1  1018 M 1  1019 M 1  1020M 1  1021M

  24. Correct Answer: 1  1017 M 1  1018 M 1  1019 M 1  1020M 1  1021M The solution to problem is on the following slide.

  25. Correct Answer: Kf = [Ag(CN)2]/[Ag+][CN]2 Assume: Ag+ initially is converted to Ag(CN)2 [Ag+] = [Ag(CN)2]/Kf [CN]2 [Ag+] = (0.10)/[1 1021](0.1)2 [Ag+] = 1 1018

  26. Upon addition of 6 M HCl to an aqueous solution, a precipitate forms. Which of the following cannot be in the precipitate? AgCl Hg2Cl2 PbCl2 CuCl2

  27. Correct Answer: AgCl Hg2Cl2 PbCl2 CuCl2 Chlorides of Ag+, Hg22+, and Pb2+ are insoluble.

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