Chemistry 12 Assignment 9 Solutions

Chemistry 12 Assignment 9 - Solutions Fall 2009 Chapter 8: Applications of Aqueous Equilibria Chapter 10: Spontaneity, Entropy, and Free Energy Chapter 8: 145, 47, 53, 54, 55, 71, 123, 131 Acid Base Titrations 45. This is a strong acid (HCIO,) titrated by a strong base (KOH). Added OH from the strong base will react completely with the H' present from the strong acid to produce H,O. a. Only strong acid present. [F°] = 0.200 M; pH - 0.699 b. mmol OH added - 10.0 mL * 0.100 mmol OH * = 1.00 mmol OFT mmol IT' present - 40.0 mL × 0.200 mmol H = 8.0 mmol H* Note: The units mmoles are usually easier numbers to work with. The units for molarity are moles/L but are also equal to mmoles/mL. H* + OH → н,о Before Before 2,00 mmol 1.00 mmol -1.00 mmol Reacts completely After 7.00 mmol The excess HT determines the pH. [H"even 7.00 mmol H • = 0.140 M; pH = 0.854 40.0 mL - 10.0 mL c. mmol OH added - 40.0 mL. × 0.100 M - 4.00 mmol OH: H* + OH: → HO Before 8.00 mmol 4.00 mmol After 4.00 mmol [Наси " 4.00 mmo! - = 0.0500 M; pH - 1.301 (40.0 + 40.0) mL d. mmol OH added - 80.0 mL. x 0.100 M - 8.00 mmol OH; This is the equivalence point since we have added just enough OH to react with all the acid present. For a strong acid-strong base titration, pH - 7.00 at the cquivalence point since only neutral species are present (K*, CIO,, H20). e. mmol OH added = 100.0 mL × 0.100 M = 10.0 mmol OH: H,O Before After 8.00 mmol 10.0 mmol 2.0 mmol Past the equivalence point, the pH is determined by the excess OH present. 2.0 mmol (OH Jen* (40.0 + 100.0) mL - =0.014 М; pOH = 1.85; pH - 12.15 Chemistry 12 47. Assignment 9 - Solutions This is a weak acid (HC,H,O,) titrated by a strong base (KOH). a. Only weak acid is present. Solving the weak acid problem: HC,H,0, + с,н,о; Initial 0.200 M x mol/L HC,H,O, dissociates to reach equilibrium Change Equil. 0.200 - x K, - 1.8 × 10' = 0.200 - x x = [H] =19 × 10' M 0.2001 pH = 2.72; Assumptions good. b. The added OH' will react completely with the best acid present, HC,H,O,, mmol HC,H,O, present = 100.0 mL x 0.200 mmol HC, H,02 - 20.0 mmol HC,H,O, mmol OF added = 50.0 mL x 0.100 mmol OH - = 5.00 mmol OH: Fall 2009 HC,H,O, Before 20.0 mmol Change -5.00 mmol OH 5.00 mmol -5.00 mmol Сн,о; + н,О +5.00 mmol Reacts completely After 15.0 mmol 5.00 mmol After reaction of all the strong base, we havo a buffer solution containing a weak acid (HC,H,O,) and its conjugate base (C,H,O;). We will use the Henderson-Hasselbaich equation to solve for the pH. [C,H,02 ] pH -pk, + log - - log (1.8 × 10*) + log [HC,H,02) 5.00 mmol/V., 15.0 mmol/V, where V, - total volume pH = 4.74 + log 5.00 = 4.74 + (-0.477) - 4.26 15.0 Note that the total volume cancels in the Henderson-Hasselbalch equation. For the [base]/(acid] term, the mole ratio equals the concentration ratio since the components of the buffer are always in the same volume of solution. mmol OH added - 100.0 mL, × 0.100 mmo! OH/mL = 10.0 mmol OH*; The same amount (20.0 mmol) of HC,H,0, is present as before (it never changes). As before, let the OH react to completion, then see what is remaining in solution after this reaction. Before After HGH,0, 20.0 mmol 10.0 mmol сно; + н,0 10.0 mmol 10.0 mmol A buffer solution results after reaction. Since [C,H,O,] = [HC,H,O,] » 10.0 mmol/total volume, then pH - pK,. This is always true at the halfway point to equivalence for a weak acid/strong base titration, pH - pK, pH = -log (1.8 × 10**) = 4.74 Chemistry 12 Assignment 9 - Solutions Fall 2009 d. mmol OF added = 150.0 mL x 0.100 M = 15.0 mmol OH. Added OH reacts completely with the weak acid. Before After HC,H,0, 20.0 mmol 5.0 mmol + GH,O; + H,0 15.0 mmol 15.0 mmol We have a buffer solution after all the OH reacts to completion. Using the Henderson- Hasselbalch equation: [C, H,O,] pH = 4.74 + log = 4.74 + log [нс,н,0,) ph - 4.74 + 0.48 * 5.22 15.0 mmol 5.0 mmol (Total volume cancels, so we can usc mol ratios. mmol OH added = 200.00 mL × 0.100 M- 20.0 mmol OH*; As before, let the added OH react to completion with the weak acid, then see what is in solution after this reaction. HC,H,0, CH,O; + H,0 Before After 20.0 mmol 20.0 mmol 20.0 mmol This is the equivalence point. Enough OH has been added to exactly neutralize all the weak acid present initially. All that remains that affects the pH at the equivalence point is the conjugate base of the weak acid, C,H,O,. This is a weak base equilibrium problem. сно, + но р нс,н,0, OH K, - K, _ 10x 10-14 K, 1.8 × 10-3 - 5.6 × 10-30 Initial 20.0 mmol/300.0 mL x mol/L C,H,0, reacts with HO to reach equilibrium tx Change Equil. 0.0667 -x K, = 5.6 × 10-10 = 0.0667 - 0.0667' *= [OH] = 6.1 × 10* M pOH = 5.21; pH = 8.79; Assumptions good. mmol OH added - 250.0 mL x 0.100 M = 25.0 mmol OH Before After HC,H,0, 20.0 mmol OH 25.0 mmol 5.0 mmol - 20.0 mmo! After the titration reaction, we have a solution containing excess OH and a weak base, C,H,O,. When a strong base and a weak base are both present, assume the amount of OH added from the weak base will be minimal, i.e., the pH past the equivalence point is determined by the amount of excess base. 5.0 mmol [OH ascens" = 0.014 Mf; pOH = 1.85; pH = 12.15 100.0 mL. + 250.0 mL Chemistry 12 Assignment 9 - Solutions Fall 2009 53. a. This is a weak acid/strong base titration. At the halfway point to equivalence, [weak acid] - [conjugate base], so pH - pK, (always for a weak acid/strong base titration). pH = - log (6.4 × 10%) = 4.19 mmol HC,H,0, present - 100. mL x 0.10 M- 10. mmol HC,H,0,. For the equivalence point, 10. mmol of OH must be added. The volume of OH added to reach the equivalence point is: 1 mL. 10. mmol OH x 0.10 mmol OH = 1.0 × 10ª mL OH At the equivalence point, 10. mmol of HC,H,O, is neutralized by 10. mmol of OH to produce 10. mmol of C,H,O,. This is a weak base. The total volume of the solution is 100.0 mL + 1.0 x 10* mL - 2.0 x 10* mL. Solving the weak base equilibrium problem: 1.0 × 10-14 C,H,0, + H0 + HC,H,0, + OH K, - - 1.6 × 10-*® 6.4 × 10-$ Initial 10. mmol/2.0 x 10* mL Equil. 0.050 -x * К, - 1.6 × 10*'° - 0.050 -x 0.050' x- (OH] - 28 x 10€ M pOH = 5.55; pH = 8.45 Assumptions good. b. At the halfway point to equivalence for a weak base/strong acid titration, pHI - pK, since [weak base - (conjugate acid). К. - 10× 10-14 5.6 × 10-1 • - 1.8 × 10; pH - pK, " -log (1.8 × 10'"') - 10.74 For the equivalence point (mmol acid added = mmol base present): mmol C,H,NH, present = 100.0 mL × 0.10 M = 10. mmol C,H,NH, mL H' added = 10. mmol H x - 1 mL 0.20 mmol H: = 50. ml. H* The strong acid added completely converts the weak base into its conjugate acid. Therefore, at the cquivalence point, [CH,NH, *]. = 10. mmol/(100.0 + 50.) mL = 0.067 M. Solving the weak acid equilibrium problem: C,H,NH,* H* + C,HỊNH, Initial 0.067 M Equil. 0.067 -x K, = 1.8 × 10-* = 0.067 -x 0.067 **(H] * 1.1 × 10° M pH - 5.96; Assumptions good. c. In a strong acid/strong base titration, the halfway point has no special significance other than exactly one-half of the original amount of acid present has been neutralized. mmol H° present = 100.0 mL x 0.50 M = 50. mmol H* Chemistry 12 Assignment 9 - Solutions fall 2009 mL. OH added = 25, mmol OH x 1 mL = 1.0 × 10° mL OH 0.25 mmol H° + OH → H,O Before After 50. mmol 25 mmol 25 mmol 25 mmol - - 0.13 Mf, pH - 0.89 (100.0 + 1.0 x 10) mL At the equivalence point of a strong acid/strong base titration, only neutral species are present (Na*,CT, H,O) so the pH - 7.00. 34. At equivalence point: 16.00 mL. × 0.125 mmol/ml. = 2.00 mmol OH' added; There must be 2.00 mmol HX present initially. 2.00 mL. NaOH added - 2.00 mL, * 0.125 mmol/mI. - 0.250 mmol OH*; 0.250 mmol of OH added will convert 0.250 mmol HX into 0.250 mmol X. Remaining HX - 2.00 - 0.250 - 1.75 mmol HX; This is a buffer solution where (H°) - 10**3 - 1.22 × 10* M. Since total volume cancels: К, - Н° (X) - 1.22 × 10- (0250) - 1.74 × 10° Note: We could also solve for K, using the Henderson-Hasselbalch equation. 5S. a 1.00 L × 0.100 mol/L - 0.100 mol HC| added to reach stoichiometric point. The 10.00 & sample must have contained 0. 100 mol of Na. 1000 - 100. /mol b. 500.0 mL. of HCI added represents the halfway point to cquivalence. So, pH - pK, - 5.00 and K, - 1.0 × 10%. At the equivalence point, enough H' has been added to convert all the A' present initially into HA. The concentration of HA at the equivalence point is: [HA), - 0.100 mol - 0.0909 M 1.10 L HA - H° + A K, - 1.0 × 10' Initial 0.0909 M Equil. 0.0909 -x K, - 1.0 × 10* - **I 0.0909 -x 0.0909 * -9,5 × 10* M - [H"]: pH - 3.02 Assumptions good.