Chpter 17: 7, 9, 11, 1, 15, 1, 7,, 9, 4, 49, 50, 51, 79, 81, 8, 87, 9, 99, 105, 11, 117 [HCO ] 7. ph = pk + log [H CO ] [HCO ] [H CO ] log = 1.6 [HCO ] = 0.04 [HCO ] [NH ] 0.15 M 9. ph = pk + log = 9.5 + log = + [NH ] 0.5 M 4 8.88 [CHCOO ] 11. ph = pk + log [CH COOH] [CHCOO ] [CH COOH] 0.58 [CH COO ] 4.50 4.74 log [CH COOH] 1. CH NH + (q) + H O(l) H O + (q) + CH NH (q) [CH NH + ] H O [H O + ] [CH NH ] initil 1.00-0 0.80 chnge x - + x + x equilibrium 1.00 x - x 0.80 + x [H O ][CH NH ] + + [CHNH ] K = 11 (x)(0.80 + x) (x)(0.80). 10 = (1.00 x) 1.00 x [H O ].88 10 11 M ph log(.88 10 11 ) 10.54. The strong bse will completely rect with the cidic component of the rection CH NH + (q) + OH - (q) CH NH (q) + H O(l) [CH NH + ] OH - [CH NH ] [H O] initil mol 1.00 0.070 0.80 - chnge mol 0.070 0.070 + 0.070 - finl mol 0.9 0.00 0.87 -
Re-estblish equilibrium, solve for ph of buffer, the volume of solution is 1.0 L CH NH + (q) + H O(l) H O + (q) + CH NH (q) [CH NH + ] H O [H O + ] [CH NH ] initil 0.9-0 0.87 chnge x - + x + x equilibrium 0.9 x - x 0.87 + x 11 (x)(0.87 + x) (x)(0.87). 10 = (0.9 x) 0.9 x [H O ].46 10 11 M; ph log(.46 10 11 ) 10.61 b. The strong cid will completely rect with the bsic component of the rection CH NH (q) + H O (q) CH NH + (q) + H O(l) [CH NH ] H O [CH NH ] [H O] initil mol 0.80 0.11 1.00 - chnge mol 0.11 0.11 + 0.11 - finl mol 0.69 0.00 1.11 - Re-estblish equilibrium; solve for ph of buffer, the volume of solution is 1.0 L CH NH + (q) + H O(l) H O + (q) + CH NH (q) [CH NH + ] H O [H O + ] [CH NH ] initil 1.11-0 0.69 chnge x - + x + x equilibrium 1.11 x - x 0.69 + x x [H O ].70 10 11 M ph log(.46 10 11 ) 10.4 15. H A(q) + H O(l) H O + (q) + HA (q) K 1 = 1.1 x 10 - HA (q) + H O(l) H O + (q) + A (q) K 1 =.5 x 10-6 Need to convert ech K to pk vlue; the closest one to 5.80 would be the best equilibrium for the buffer. pk 1 = log(1.1 x 10 - ) =.96 pk = log(.5 x 10-6 ) = 5.60 Bsed on these pk vlue, the nd equilibrium is best N A/NHA combintion
1.. At equivlence point, only NH 4 +, concentrtion of 0.050 M (1/ dilution). NH + 4 (q) + H O(l) H O + (q) + NH (q) [NH + 4 ] H O [H O + ] [NH ] initil 0.050 - ~0 0 chnge x - + x + x equilibrium 0.050 x - x x K x K b = K w for cid/conjugte bse pir; K b for NH is 1.8 x 10 5 1.0 x 10 [NH ][H O ] (x)(x) x K 5.6 x 10 1.8 x 10 [NH ] 0.050 x 0.050 14 + 10 5 4 x = 5. x 10 6 = [H O + ] ph = 5.8 At equivlence point, only C H O with concentrtion of 0.050 M (1/ dilution). C H O (q) + H O(l) OH (q) + HC H O (q) [C H O ] H O [H O + ] [HC H O ] initil 0.050 - ~0 0 chnge x - + x + x equilibrium 0.050 x - x x K x K b = K w for cid/conjugte bse pir; K for HC H O is 1.8 x 10 5 1.0 x 10 [HC H O ][OH ] (x)(x) x K 5.6 x 10 1.8 x 10 [C H O ] 0.050 x 0.050 14 10 5 x = 5. x 10 6 = [OH ] poh = 5.8; ph = 8.7 7. An indictor is typiclly wek orgnic cid. The indictor will rect with the bse component of the titrtion. If lrge mount of indictor is used, it will rect with significnt mount of the bse ffecting the equivlence point of the titrtion.
.. CuBr(s) Cu + (q) + Br (q) K sp = [Cu + ][Br ] b. ZnC O 4 (s) Zn + (q) + C O 4 (q) K sp = [Zn + ][ C O 4 ] c. Ag CrO 4 (s) Ag + (q) + CrO 4 (q) K sp = [Ag + ] [CrO 4 ] d. Hg Cl (s) Hg + (q) + Cl (q) K sp = [Hg + ][Cl ] e. AuCl (s) Au + (q) + Cl (q) K sp = [Au + ][Cl ] f. Mn (PO 4 ) (s) Mn + (q) + PO 4 (q) K sp = [Mn + ] [PO 4 ] 9. MnCO (s) Mn + (q) + CO (q) MnCO [Mn + ] [CO ] initil - 0 0 chnge - + x + x equilibrium - x x K sp = [Mn + ][CO ] = (x)(x) = x x = [MnCO ] = 4. x 10 6 M x = (4. x 10 6 ) = K sp = 1.8 x 10 11
4. Zn(OH) (s) Zn + (q) + OH (q) Zn(OH) [Zn + ] [OH ] initil - 0 0 chnge - + x + x equilibrium - x x K sp = 1.8 x 10 14 = [Zn + ] [OH ] = (x)(x) = 4x x = 1.7 x 10 5 M; [OH ] = x =. x 10 5 M poh = 4.48; ph = 9.5 49. CCO (s) C + (q) + CO (q) CCO [C + ] [CO ] initil - 0.050 0 chnge - + x + x equilibrium - 0.050 + x x K sp = 8.7 x 10 9 = [C + ][CO ] = (0.050 + x)(x) 0.050x x = 1.7 x 10 7 M = [CCO ] mole CCO 100.086 g CCO 5. x 10 g CCO 7 6 1.7 x 10 x 0.0 L x = L mole CCO 50.. PbBr (s) Pb + (q) + Br (q) PbBr [Pb + ] [Br ] initil - 0 0 chnge - + x + x equilibrium - x x K sp = 8.9 x 10 6 = [Pb + ][Br ] = (x)(x) = 4x x = 1. x 10 M = [PbBr ] b. PbBr (s) Pb + (q) + Br (q) PbBr [Pb + ] [Br ] initil - 0 0.0 chnge - + x + x equilibrium - x 0.0 + x K sp = 8.9 x 10 6 = [Pb + ][Br ] = (x)(0.0 + x) (0.0) x x =. x 10 4 M = [PbBr ]
c. PbBr (s) Pb + (q) + Br (q) PbBr [Pb + ] [Br ] initil - 0.0 0 chnge - + x + x equilibrium - 0.0 + x x K sp = 8.9 x 10 6 = [Pb + ][Br ] = (0.0 + x)(x) (0.0)4x x =. x 10 M = [PbBr ] 51. 1 mole CCl mole Cl 10.0 g CCl x x 110.984 g CCl 1 mole CCl 1.00 L = 0.18 M Cl AgCl(s) Ag + (q) + Cl (q) AgCl [Ag + ] [Cl ] initil - 0 0.18 chnge - + x + x equilibrium - x 0.18 + x K sp = 1.6 x 10 10 = [Ag + ][Cl ] = (x)(0.18 + x) 0.18x x = 8.9 x 10 10 M = [AgCl] 79. A solubility equilibrium is n equilibrium between n ionic solid (rectnt) nd its components ions (products) in solution. Only (d) represents solubility equilibrium. 81. CCO (s) C + (q) + CO (q) CCO [C + ] [CO ] initil - 0 0 chnge - + x + x equilibrium - x x K sp = 8.7 x 10 9 = [C + ][CO ] = (x)(x) = x x = 9. x 10 5 M = [CCO ] mole CCO L 100.086 g CCO 1 mole CCO 5 9. x 10 x.0 L x = 0.019 g/ kettle 1 kettle 116 g CCO x = 0.019 g CCO 6. x 10 kettles
8. Ag CO (s) Ag + (q) + CO (q) Ag CO [Ag + ] [CO ] initil - 0 0 chnge - + x + x equilibrium - x x K sp = 8.1 x 10 1 = [Ag + ] [CO ] = (x) (x) = 4x x = 1. x 10 4 M = [Ag CO ] mole Ag CO L 75.744 g Ag CO 1 mole Ag CO 5 1. x 10 x = 0.05 g AgCO 87. Al(OH) (s) Al + (q) + OH (q) Al(OH) [Al + ] [OH ] initil - 0 0 chnge - + x + x equilibrium - x x K sp = 1.8 x 10 = [Al + ] [OH ] = (x)(x) = 7x 4 x =.9 x 10 9 M; [OH ] = x = 8.6 x 10 9 M pure wter hs 1.0 x 10 7 M OH so; totl [OH ] = 8.6 x 10 9 M + 1.0 x 10 7 M = 1.1 x 10 7 M poh = 6.96; ph = 7.04 9. BSO 4 (s) B + (q) + SO 4 (q) BSO 4 [B + ] [SO 4 ] initil - 0 0 chnge - + x + x equilibrium - x x K sp = 1.1 x 10 10 = [B + ][SO 4 ] = (x)(x) = x x = 1.0 x 10 5 M = [BSO 4 ] mole BSO L.88 g BSO 1 mole BSO 5 4 4 1.0 x 10 x 5.0 L x = B(NO ) is too soluble (no K sp ) 4 0.01 g BSO 4
99.. strong cid HX = H O +, strong bse = OH H O + + OH H O 1 K = = K w 14 1.0 x 10 b. NH + H O + NH 4 + + H O 1 K NH 1.8 x 10 K = = = = K NH K 1.0 x 10 5 b + 14 4 w 9 1.8 x 10 c. HC H O + OH C H O + H O 1 K HCHO 1.8 x 10 5 K = = = = 14 K b CHO K w 1.0 x 10 9 1.8 x 10 d. HC H O + NH C H O + NH 4 + K = K HC H O x 1 1 = K HC H O x K NH x + b K NH4 Kw 1 1.0 x 10 5 5 = 1.8 x 10 x 1.8 x 10 x = 14 4. x 10 105. PbSO 4 (s) Pb + (q) + SO 4 (q) PbSO 4 [Pb + ] [SO 4 ] initil - 0 0 chnge - + x + x equilibrium - x x K sp = 1.6 x 10 8 = [Pb + ][SO 4 ] = (x)(x) = x x = 1. x 10 4 M = [Pb + ] mole Pb 1 ml H O 1 L H O 07. g Pb 1. x 10 x 1 x 10 g H O x x x = + + 4 6 + L HO 1 g HO 1000 ml HO 1 mole Pb 1 + x 10 g Pb exceeds limit of 0.05 g
11. For the buffer HC H O (q) + H O(l) C H O (q) + H O + (q) [HC H O ] H O [H O + ] [C H O ] initil 0.500 - ~0 0.500 chnge x - + x + x equilibrium 0.500 x - x 0.500 + x K = 1.8 x 10 [H O ][C H O ] (x)(0.500 + x) x(0.500) [HC H O ] 0.500 x 0.500 + 5 x = 1.8 x 10 5 M = [H O + ]; ph = 4.74 [CHO ] 0.500 or ph = pk + log 4.74 + log [HC H O ] 0.500 4.74 HC H O (q) + H O(l) C H O (q) + H O + (q) [HC H O ] H O [H O + ] [C H O ] initil 0.0500 - ~0 0.0500 chnge x - + x + x equilibrium 0. 500 x - x 0.0500 + x K = 1.8 x 10 [H O ][C H O ] (x)(0.0500 + x) x(0.0500) [HC H O ] 0.0500 x 0.0500 + 5 x = 1.8 x 10 5 M = [H O + ]; ph = 4.74 [C H O ] 0.0500 ph = pk + log 4.74 + log 4.74 [HC H O ] 0.0500 or For the cetic cid solution HC H O (q) + H O(l) C H O (q) + H O + (q) [HC H O ] H O [H O + ] [C H O ] initil 0.500 - ~0 0 chnge x - + x + x equilibrium 0.500 x - x x K = 1.8 x 10 [H O ][C H O ] (x)(x) x [HC H O ] 0.500 x 0.500 + 5 x =.0 x 10 M = [H O + ]; ph =.5
HC H O (q) + H O(l) C H O (q) + H O + (q) [HC H O ] H O [H O + ] [C H O ] initil 0.0500 - ~0 0 chnge x - + x + x equilibrium 0.0500 x - x x K = 1.8 x 10 [H O ][C H O ] (x)(x) x [HC H O ] 0.0500 x 0.0500 + 5 x = 9.5 x 10 4 M = [H O + ]; ph =.0 117.. Point 1: first buffer region: H A nd HA Point : first equivlence point: HA Point : second buffer region: HA nd A Point 4: second equivlence point: A Point 5: beyond equivlence points: OH, A b. pk 1 is equl to the ph t hlfwy to the first equivlence point, nd pk is equl to the ph t hlfwy to the second equivlence point. pk 1 4.8 pk 9.0