SQSM-07

SQSM-07 - Chapter 7 Mixtures of Acids and Bases 1 What is a...

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Chapter 7 Mixtures of Acids and Bases 7-1 1. What is a common ion and what is the common-ion effect? Common ions are involved in an equilibrium but have more than one source. The common-ion effect is the shift of the equilibrium away from the common ion so as to reduce the effect of the common ion. 3. What is a buffer and how does it function? A buffer is a mixture of a weak acid and its conjugate base in comparable and appreciable amounts. Buffers convert strong acids and bases into the weak acids and bases, which minimizes the effect of the addition of a strong acid or base. 5. Explain why a solution that is 1.4 mM HF and 6.4 mM KF is not a good buffer. The concentrations are fairly low, so this buffer would not have a very good buffer capacity. 7. Explain why a solution of a strong acid and its conjugate base is not a buffer. Use a solution of 0.1 M HCl and 0.1 M KCl as an example. The solution is well protected against additional base because H 3 O 1+ would react with the strong base and convert it to water. However, there is no protection against additional acid because Cl 1- is too weak a base to react with H 3 O 1+ . 9. Of the acid-base pairs listed in the Appendix C, which would be the best to prepare buffers at the following pH’s: Choose the conjugate acid-base pair in which the pK a of the acid is closest to the desired pH. a) pH = 1.5 H 2 SO 3 /HSO 3 1- pK a = 1.82 b) pH = 7.0 HSO 3 1- /SO 3 2- pK a = 7.00 c) pH = 12.0 HPO 4 2- /PO 4 3- pK a = 12.32 11. What is the pH of a solution that is 0.16 M NH 3 and 0.43 M NH 4 Cl? Use Equation 7.2 and the pK a of NH 4 1+ from Appendix C: b a a C 0.16 pH = pK + log =9.25 + log = 8.82 C 0.43 13. What is the pH of a solution made by dissolving 7.6 g of KNO 2 to 750 mL of 0.11 M HNO 2 ? Use Equation 7.2 and the pK a of HNO 2 from Appendix C × × 2 2 b 2 a 5 2 b a a 5 1 mol KNO 0.11 mol HNO n = 7.6 g KNO = 0.089 mol base ; n = 0.75 L solution = 0.082 mol acid 85.1 g KNO L solution n 0.089 pH = pK + log =3.40 + log = 3.43 n 0.82 15. How many grams of potassium acetate must be added to 2.5 L of 0.250 M acetic acid to prepare a pH = 4.26 buffer? We employ the Henderson-Hasselbalch equation to a buffer solution with pH = 4.26. K a = 1.8 x 10 -5 for acetic acid (Appendix C), so pK a = 4.74 = + -0.48 b b b a a a a a n n n pH pK log log = pH - pK = 4.26 - 4.74 = -0.48 = 10 = 0.33 n n n Thus, n b = 033 n a . We are given that n a = 2.5 x 0.250 = 0.63 mol, so n b = (0.33)(0.63) = 0.21 mol Potassium acetate is CH 3 COOK and has a molar mass of 98.1 g/mol. Consequently, (0.21 mol)(98.1 g/mol) = 20. g of potassium acetate should be dissolved 17. How many milliliters of 6.0 M NaOH must be added to 0.50 L of 0.20 M HNO 2 to prepare a pH = 3.86 buffer? (0.50 L)(0.20 M) = 0.10 mol HNO 2 , so the reaction table for the mixing of the weak acid and strong base is HNO 2 + OH 1- NO 2 1- + H 2 O In 0.10 x 0 mol -x -x +x mol Eq 0.10 – x ~0 x mol Solve Equation 7.2 for the n b /n a ratio 0.46 b b a a a 1- n n log = pH - pK = 3.86 - 3.40 = 0.46 = 10 = 2.9 n n x 0.29 = 2.9 x = 0.29 - 2.9 x, so x = = 0.074 mol = 74 mmol OH 0.1 - x 3.9

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Mixtures of Acids and Bases 7-2 19. Use the data in Appendix C to determine the equilibrium constants for the following reactions?
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