Chemistry 12 Assignment 7 Solutions

Chemistry 12 Assignment 7-Solutions Fall 2009 32. Chapter 7: 31, 32, 34, 39, 47, 49, 67, 113 Chapter 8: 26, 27, 31, 103 Solutions of Acids 31. Strong acids are assumed to completely dissociate in water, e.g., HCl(aq) + H₂O(I)→ H₂O"(aq) + Cl(aq) or HCl(aq) + H'(aq)+Cl(aq). a. A 0.10 MHCl solution gives 0.10 MH and 0.10 MCI since HCl completely dissociates. The amount of H' from H₂O will be insignificant. pH--log [H]-log (0.10) = 1.00 b. 5.0 MH is produced when 5.0 MHCIO, completely dissociates. The amount of H' from H₂O will be insignificant. pH-log (5.0)0.70 (Negative pH values just indicate very concentrated acid solutions.) C. 1.0x 10" MH is produced when 1.0 x 101 MHI completely dissociates. If you take the negative log of 1.0 × 10 this gives pH=11.00. This is impossible! We dissolved an acid in water and got a basic pH. What we must consider in this problem is that water by itself donates 1.0×10 MH. We can normally ignore the small amount of H from H₂O except when we have a very dilute solution of an acid (as in the case here). Therefore, the pH is that of neutral water (pH=7.00) since the amount of HI present is insignificant. 50.0 mL. con. HCl soln x 1.198 x mL 20.0 mL. con. HNO, soln x 38 g HC 100 g con. HCl soln 1.42 g x mL X I mol HCI 36.5 g 70. g HNO, 1 mol HNO, X 100 g soln 63.0 g HNO, =0.32 mol HNO, HCl(aq)+H+(aq) + Cl(aq) and HNO, (aq) → H+(aq) +NO, (aq) (Both are strong acids.) So we will have 0.62+0.32-0.94 mol of H in the final solution. [H]= [OH] = 0.94 mol 1.00 L K [H] =0.94 M; pH-log [H"]=-log (0.94) = 0.027 -0.03 1.0 x 10-14 0.94 = 1.1×10 M 0.62 mol HCl 34. At pH 2.000, [H]=100001.00×10° M, At pH=4.000, [H]104.000 1.00×10 M mol H present = 0.0100 L x 0.0100 mol H 1.00×10 mol H L Let V total volume of solution at pH 4.000: 1.00×10 mol/L 4 Volume of water added 1.00 L-0.0100 L 0.99 L=990 mL 1.00 x 10 mol H V V 1.00 L Chemistry 12 Assignment 7-Solutions Fall 2009 39. This is a weak acid in water. We must solve a weak acid problem. Let HBz = C,H,CO₂H. 0.56 g HBz x I mol HBz 122.1 g 4.6x10 mol; [HBz], -4.6 × 10 M HBz H + Bz Initial 4.6 x 10 M ~0 0 x mol/L HBz dissociates to reach equilibrium Change ← -x 4.6 x 10°-x +x Equil. K,6.4×10 [H] [Bz"] [HBz] = 4.6 x 10x 4.6 x 10 x=[H]=5.4 x 10; Check assumptions: x 100 = 5.4 x 10 4.6 x 103 4.6 x 10 ×100-12% using the quadratic formula or the method of successive approximations (see Appendix 1.4 of text). Using successive approximations: x² 6.410, x 5.1 x 10 Assumption is not good (x is 12% of 4.6 x 10). When assumption(s) fail, we must solve exactly (4.6x10-5.4 x 10) (4.6 x 10-5.1 x 10) 6.4x10, x 5.1 x 10 M (consistent answer) So: x= [H] [Bz] = [CH,CO₂] 5.1 x 10 M [HBz]-[CH,CO₂H]=4.6×10-x-4.1×10 M pH = -log (5.1 x 10) 3.29; pOH-14.00 pH=10.71; [OH]-10-10711.9 × 10" M 47. In all parts of this problem, acetic acid (HC₂H₂O₂) is the best weak acid present. We must solve a weak acid problem. HC₂H₂O, H+ C₂H₂O Initial 0.50 M -0 0 x mol/L HC₂H₂O, dissociates to reach equilibrium +x x= [H] [C,H,O,1-3.0 x 10 M Assumptions good. Percent dissociation- [H] [HC₂H₂O₂) x100-3.0 x 10 x 100 = 0.60% 0.50 Change Equil. 0.50-x K-1.8 × 105 [H] [C,H,O₂] [HC₂H₃O₂] 0.50-x 0.50 Chemistry 12 Assignment 7-Solutions Fall 2009 b. The set-up for solutions b and e are similar to solution a except the final equation is slightly different, reflecting the new concentration of HC₂H₂O₂. K₁ = 1.8 × 10- 2 0.050 x 0.050 x= [H] [C,H,O₂]=9.5×10 M Assumptions good. % dissociation 9.5 x 104 0.050 ×100-1.9% c. K1.8×10= x² 0.0050 0.0050 x= [H] [C₂H₂O] 3.0×10M; Check assumptions. Assumption that x is negligible is borderline (6.0% error). We should solve exactly. Using the method of successive approximations (see Appendix 1.4 of text): 1.8 x 105- x2 0.0050-3.0 x 10 0.0047 x=2.9×10 Next trial also gives x=2.9 x 10. % dissociation 2.9 x 10 5.0 x 10-3 × 100 = 5.8% d. As we dilute a solution, all concentrations are decreased. Dilution will shift the equilibrium to the side with the greater number of particles. For example, suppose we double the volume of an equilibrium mixture of a weak acid by adding water, then: H X 2 Q= 2 2 K 2 Q K, so the equilibrium shifts to the right or towards a greater percent dissociation. e. [H] depends on the initial concentration of weak acid and on how much weak acid dissociates. For solutions a-c the initial concentration of acid decreases more rapidly than the percent dissociation increases. Thus, [H'] decreases. 49. pH-2.77, [H]=1021.7×10 M HOCN Initial 0.0100 Equil. 0.0100-x x=[H]=[OCN] -1.7×10° M; [HOCN] =0.0100-x 0.0100-0.0017 0.0083 M H+ OCN 0 -0 K= [H] [OCN] (1.7×103) -3.5×10 [HOCN] = 8.3 x 10 Chemistry 12 Assignment 7-Solutions Fall 2009 Solutions of Bases 67. 5.0 x 10g x 1 mol 0.0100 L 299.4 g -1.7x10 M=[codeine); Let cod codeine, CHNO Solving the weak base equilibrium problem: cod+ H₂O codH + OH K108.9 x 10" Initial Change 1.7 x 10 M 0 x mol/L. codeine reacts with H₂O to reach equilibrium -0 ← Equil. 1.7x10-x +x X K-8.9×10= 1.7 x 10 x 1.7 x 103 x=3.9×10 Assumptions good. [OH] -3.9×10 M; [H]K/[OH]=2.6×10 M; pH=-log [H]-9.59 Challenge Problems 113. HA H + A K-1.00 × 10 Initial C -0 0 C-[HA], for pH = 4.000 solution Equil. C-1.00×10 1.00 × 10 1.00×10 [H]=1.00×10 M K= (1.00×10)=1.00×10%; Solving: C0.0101 M C-1.00 x 10 The solution initially contains 50.0 × 103 L x 0.0101 mol/L 5.05 × 10 mol HA. We then dilute to a total volume V in liters. The resulting pH = 5.000, so [H]1.00x10. In the typical weak acid problem, x [H], so: HA H + A Initial 5.05 x 10 mol/V Equil. 5.05x10/V-1.00×10 ~0 1.00 x 10 1.00 × 10 0 K- (1.00×10-5) 5.05x10/V-1.00×10-5 1.00×10, 1.00×10-5.05×10/V-1.00×105 V-4.59 L; 50.0 mL are present initially, so we need to add 4540 mL of water. Chemistry 12 Assignment 7-Solutions Fall 2009 Chapter 8: 26, 27, 31, 103 Buffers a. HNO,H+NO, K-4.0x10, pK,-log (4.0×10)-3.40 26. This is a buffer solution. Using the Henderson-Hasselbalch equation: pH = pK,+log (Base) =3.40+log (0.15) -3.40+0.18-3.58 [Acid] 1 mol (0.10) 0.416 mol b. 25.0 g HC,H,O,X 40.0 g NaC₂H₂O, × pH=1 60.05 g I mol 82.03 g PK,+log [Base]; [Acid] 0.416 mol HC,H,O,; [HC,H,O,]" 0.488 mol NaC,H,O₂; [C₂H₂O₂] 0.500 L 0.488 mol 0.500 L ; K-1.8 x 10; pK-log (1.8 x 10)=4.74 = 0.832 M -0.976 M pH-4.74+ log (0.976)4.74+0.0693 = 4.81 (0.832) c. Let's work with mmol in this problem. This requires the use of mmol/ml. for molarity units instead of mol/L (the 10 conversion factors cancel). 50.0 mL x 1.0 mmol ml. 50. mmol HOCI; 30.0 mL x 0.80 mmol NaOH mL -24 mmol OH The strong base reacts with the best acid available, HOCI. The reaction goes to completion. Whenever strong base or strong acid reacts, the reaction is always assumed to go to completion. HOCI + OH OC + H₂O 50. mmol 24 mmol 0 24 mmol OH reacts completely. Change -24 After 26 mmol 0 -24 +24 Before Reacts completely 24 mmol After reaction, the solution contains 26 mmol HOCI and 24 mmol OCl in 250 ml. of solution. This is a buffer solution. Using the Henderson-Hasselbalch equation: pH=pK,+log [Base]; K-3.5×10%; pK-7.46 [Acid] 24 mmol OCI 250 mL 26 mmol HOCI 250 mL pH = 7.46+log 7.46+(-0.035)=7.43