Chem 1515 Problem Set #12 Spring 2001 Name TA Name Lab Section # ALL work must be shown to receive full credit. Due at the beginning of lecture on Wednesday, April 25, 2001. PS12.1. Calculate the ph of a 0.250 M H 3 PO 4. Calculate the [PO 4 3 ] in the solution. H 3 PO 4 (aq) ä H (aq) H 2 PO 4 (aq) initial.250 1 x 10-7 0 change -x x x x = [H 3 PO 4 ] diss equilibrium.250 - x 1 x 10-7 x 0x K a = [H ][ H 2 PO 4 ] [H 3 PO 4 ] 7.53 x 10-3 x =.250 - x 7.53 x 10-3 (0.250 x) = x 2 x 2 7.53 x 10-3 x 1.88 x 10 3 = 0 solving the quadratic equation x = b ± b2-4ac 2a x = -7.53 x 10-3 ± (7.53 x 10-3 ) 2-4(1)( 1.88 x 10 3 ) 2(1) x = -7.53 x 10-3 ± 8.70 x 10-2 2 Use only the positive root x = 3.98 x 10-2 M = [H ] = [ H 2 PO 4 ] H 2 PO 4 (aq) ä H (aq) HPO 4 2 (aq) initial 3.98 x 10-2 3.98 x 10-2 0 change -x x x x = [H 2 PO 4 ] diss equilibrium 3.98 x 10-2 - x3.98 x 10-2 x 0x K a = [H ][HPO 4 2 ] [ H 2 PO 4 ] 6.2 x 10-8 = 3.98 x 10-2 x x 3.98 x 10-2 - x assume x << 3.98 x 10-2 6.2 x 10-8 = 3.98 x 10-2 x 3.98 x 10-2 6.2 x 10-8 M = x = [HPO 4 2 ], the [H ] = 3.98 x 10-2 M CHEM 1515 1 Spring 2001
HPO 4 (aq) ä H (aq) PO 4 3 (aq) initial 6.2 x 10-8 3.98 x 10-2 0 change -x x x x = [HPO 2 4 ] diss equilibrium 6.2 x 10-8 - x3.98 x 10-2 x 0x K a = [H ][PO 4 3 ] [HPO 4 2 ] 4.2 x 10-13 = 3.98 x 10-2 x x 6.2 x 10-8 - x assume x << 3.98 x 10-2 4.2 x 10-13 = 3.98 x 10-2 x 6.2 x 10-8 6.5 x 10-19 M = x The [H ] is still 3.98 x 10-2 M so the ph = 1.40. The [PO 4 3 ] = 6.5 x 10-19 M. PS12.2. Predict the products of the following neutralization reactions. a) HCl(aq) NaOH(aq) NaCl(aq) H 2 O(aq) b) 2HNO 3 (aq) Ba(OH) 2 (aq) Ba(NO 3 ) 2 (aq) 2H 2 O(aq) c) 2NaOH(aq) H 2 CO 3 (aq) Na 2 CO 3 (aq) 2H 2 O(aq) d) 2NH 3 (aq) H 2 SO 4 (aq) NH 4 ) 2 SO 4 (aq) e) HC 6 H 5 O(aq) NaOH(aq) NaC 6 H 5 O(aq) H 2 O(aq) f) HCN(aq) KOH(aq) KCN(aq) H 2 O(aq) PS12.3. Given a solution containing the following ions (neglect the counterion for the moment), write a reaction (with water) and indicate whether the ion acts as an acid or as a base. a) F (aq) H 2 O(l) ä HF(aq) OH (aq) b) ClO 2 (aq) H2 O(l) ä HClO 2 (aq) OH (aq) c) NO 2 (aq) H2 O(l) ä HNO 2 (aq) OH (aq) d) NH 4 (aq) ä NH 3 (aq) H (aq) e) CH 3 NH 3 (aq) ä CH 3 NH 2 (aq) H (aq) f) C 2 H 5 NH 3 (aq) ä C 2 H 5 NH 2 (aq) H (aq) CHEM 1515 2 Spring 2001
PS12.4. PS12.5. PS12.6. PS12.7. Can you make any generalizations about the acid-base character of the ions in Problem #12.3? If so, state them. Generally anions act as bases reacting with water to form [OH ]. Cations act as acids reacting with water to form [H ]. Indicate an acid and a base which could react, in a neutralization reaction, to form each of the following salts. In some cases water will be present as another product. a) NaOH(aq) HC6H7O6(aq) NaC6H7O6 (aq) H2O(l) b) KOH(aq) HClO(aq) KClO(aq) H2O(l) c) (CH3)2NH(aq) HNO 3 (aq) (CH3)2NH2NO3(aq) d) NH3(aq) HBr(aq) NH4Br(aq) e) KOH(aq) HCl(aq) KCl(aq) H2O(l) f) 2NH3(aq) H2SO4(aq) (NH4)2SO4(aq) If each salt in Problem 12.5 is added to water, indicate whether the resulting solution is, basic or neutral. a) NaC6H7O6(aq) Na (aq) C6H7O6 (aq) C6H7O6 (aq) H 2 O(l) ä HC6H7O6(aq) OH (aq) b) KClO(aq) K (aq) ClO (aq) ClO (aq) H 2 O(l) ä HClO(aq) OH (aq) c) (CH3)2NH2NO3(aq) (CH3)2NH2 (aq) NO 3 (aq) (CH3)2NH2 (aq) ä (CH3)2NH2(aq) H (aq) d) NH 4 Br(aq) NH 4 (aq) Br (aq) NH 4 (aq) ä NH 3 (aq) H (aq) basic basic e) KCl(aq) K (aq) Cl (aq) neutral f) (NH 4 ) 2 SO 4 (aq) 2NH 4 (aq) SO4 2 (aq) NH 4 (aq) ä NH 3 (aq) H (aq) Calculate the ph of the following salt solutions a) 0.355 M NaC 6 H 7 O 6 H 2 O C 6 H 7 O 6 ä OH HC 6 H 7 O 6 initial 0.355 0 0 change x x x x = [ClO ] R final 0.355x x x K b = 1 x 1014 [HC 6 H 7 O 6 ][OH ] 8 x 10 5 = [ C 6 H 7 O 6 ] 1.25 x 10 10 = [x][x] [0.355 - x] x<<<0.355 6.6 x 10 6 = x = [OH ] poh = 5.18 ph = 8.82 CHEM 1515 3 Spring 2001
b) 0.777 M C 5 H 5 NHClO 4 C 5 H 5 NH ä H C 5 H 5 N initial 0.777 0 0 change x x x x = [NH 4 ] D equilibrium 0.777 x x x c) 0.0345 M KCl K a = 1 x 1014 1.7 x 10 9 = [NH 3][H ] [NH 4 ] 5.88 x 10 6 [x][x] = [0.777 - x] x<<<0.777 2.14 x 10 3 M = x = [H ] ph = 2.67 d) 0.411 M KC 3 H 5 O 2 ph = 7.00 H 2 O C 2 H 3 O 2 ä OH HC 2 H 3 O 2 initial 0.411 0 0 change x x x x = [C 2 H 3 O 2 ] R final 0.411x x x K b = 1 x 1014 1.8 x 10 5 = [HC 2H 3 O 2 ][OH ] [C 2 H 3 O 2 ] 5.56 x 10 10 = [x][x] [0.411 - x] x<<<0.411 1.51 x 10 5 = x = [OH ] ph = 9.18 CHEM 1515 4 Spring 2001
PS12.7. Continued e) 1.00 M NaHSO 4 HSO 4 ä H SO 4 2 initial 1.00 0 0 change x x x x = [HSO 4 ] R final 1.00 x x x K a = 1.2 x 10 2 = [H ][SO 4 2 ] [HSO 4 ] 1.2 x 10 2 [x][x] = [1.00 - x] 1.2 x 10 2 (1.00 x) = x 2 x 2 1.2 x 10 2 x 1.2 x 10 2 = 0 solving the quadratic equation x = b ± b2-4ac 2a x = -1.2 x 102 ± (1.2 x 10 2 ) 2-4(1)(1.2 x 10 2 ) 2(1) x = -1.2 x 102 ± 0.219 2 Use only the positive root x = 0.104 M = [H ] ph = 0.983 PS12.8. In the series of oxyacids, XOH, OXOH, and O 2 XOH, list the acids in order of increasing acid strength. Justify your answer. XOH < OXOH < O 2 XOH As the number of oxygen atoms bonded to the central atom, X, increases, the more electron density is removed from the OH bond. The smaller electron density decreases the strength of the OH bond resulting in a larger ionization of the proton, increasing the acidity of the substance. CHEM 1515 5 Spring 2001