Ans:The plot of pH vs. added base should have the general shape of those shown in Fig. 2-17, p. 59, with the midpoint of the titration (inflection point) at pH 3.2. The ratio of A–to HA is 3 when 0.75 equivalents of base have been added. From the Henderson-Hasselbalch equation, the pH at this point can be calculated: pH = pKa+ log[acid]base][conjugate= 3.2 + log 3 = 3.2 + 0.48 = 3.68 33.Buffering against pH changes in biological systems Page: 61–62 Difficulty: 2 What is the pH of a solution containing 0.2 M acetic acid (pKa= 4.7) and 0.1 M sodium acetate? Ans:pH = pKa+ log [acid]base][conjugate= 4.7 + log (0.1/0.2) = 4.7 – 0.3 = 4.4 34.Buffering against pH changes in biological systems Page: 61–62 Difficulty: 2 You have just made a solution by combining 50 mL of a 0.1 M sodium acetate solution with 150 mL of 1 M acetic acid (pKa= 4.7). What is the pH of the resulting solution? Ans:pH = pKa+ log [acid]base][conjugate= 4.7 + log (5/150) = 4.7 – 1.48 = 3.22
Chapter 2 Water 2035.Buffering against pH changes in biological systems Page: 61–62 Difficulty: 2 For a weak acid with a pKaof 6.0, show how you would calculate the ratio of acid to salt at pH 5. Ans:36.Buffering against pH changes in biological systems Page: 61-62 Difficulty: 3 Suppose you have just added 100 mL of a solution containing 0.5 mol of acetic acid per liter to 400 mL of 0.5 M NaOH. What is the final pH? (The pKaof acetic acid is 4.7.) Ans:Addition of 200 mmol of NaOH (400 mL 0.5 M) to 50 mmol of acetic acid (100 mL 0.5 mM) completely titrates the acid so that it can no longer act as a buffer and leaves 150 mmol of NaOH dissolved in 500 mL, an [OH–] of 0.3 M. Given [OH–], [H+] can be calculated from the water constant: [H+][OH–] = 10–14[H+] = 10–14M2/ 0.3 M pH is, by definition, log (1/[H+]) pH = log (0.3 M /10–14M2) = 12.48. 37.Buffering against pH changes in biological systems Page: 61-62 Difficulty: 2 A weak acid HA, has a pKaof 5.0. If 1.0 mol of this acid and 0.1 mol of NaOH were dissolved in one liter of water, what would the final pH be? Ans:Combining 1 mol of weak acid with 0.1 mol of NaOH yields 0.9 mol of weak acid and 0.1 mol of salt. pH = pKa+ log [acid]base][conjugate= 5.0 + log (0.1/0.9) = 4.05 38.Water as a reactant Page: 65 Difficulty: 1 Give an example of a biological reaction in which water participates as a reactant and a reaction in which it participates as a product. Ans:See p. 65, for examples such as condensation and hydrolysis reactions. 39.The fitness of the aqueous environment for living organisms Pages: 65–66 Difficulty: 1 If ice were denser than water, how would that affect life on earth? Chapter 2 Water 21Ans:Ice formed at the surface of bodies of water would sink, hence streams, ponds, lakes, and so on would freeze from the bottom up. With a reservoir of ice at the bottom they would be perpetually cold, and in the limit they would freeze solid, precluding life as we know it.
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