[HCO ] log = M. + log = log = log [CH

Transcription

1 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 = log = + [NH ] 0.5 M [CHCOO ] 11. ph = pk + log [CH COOH] [CHCOO ] [CH COOH] 0.58 [CH COO ] 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 chnge x - + x + x equilibrium 1.00 x - x x [H O ][CH NH ] + + [CHNH ] K = 11 (x)( x) (x)(0.80). 10 = (1.00 x) 1.00 x [H O ] M ph log( ) 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 chnge mol finl mol

2 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 chnge x - + x + x equilibrium 0.9 x - x x 11 (x)( x) (x)(0.87). 10 = (0.9 x) 0.9 x [H O ] M; ph log( ) 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 chnge mol finl mol 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 chnge x - + x + x equilibrium 1.11 x - x x x [H O ] M ph log( ) 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

3 1.. At equivlence point, only NH 4 +, concentrtion of M (1/ dilution). NH + 4 (q) + H O(l) H O + (q) + NH (q) [NH + 4 ] H O [H O + ] [NH ] initil ~0 0 chnge x - + x + x equilibrium x - x x K x K b = K w for cid/conjugte bse pir; K b for NH is 1.8 x x 10 [NH ][H O ] (x)(x) x K 5.6 x x 10 [NH ] x x = 5. x 10 6 = [H O + ] ph = 5.8 At equivlence point, only C H O with concentrtion of 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 0 chnge x - + x + x equilibrium x - x x K x K b = K w for cid/conjugte bse pir; K for HC H O is 1.8 x x 10 [HC H O ][OH ] (x)(x) x K 5.6 x x 10 [C H O ] x x = 5. x 10 6 = [OH ] poh = 5.8; ph = 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.

4 .. 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 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

5 4. Zn(OH) (s) Zn + (q) + OH (q) Zn(OH) [Zn + ] [OH ] initil chnge - + x + x equilibrium - x x K sp = 1.8 x = [Zn + ] [OH ] = (x)(x) = 4x x = 1.7 x 10 5 M; [OH ] = x =. x 10 5 M poh = 4.48; ph = CCO (s) C + (q) + CO (q) CCO [C + ] [CO ] initil chnge - + x + x equilibrium x x K sp = 8.7 x 10 9 = [C + ][CO ] = ( x)(x) 0.050x x = 1.7 x 10 7 M = [CCO ] mole CCO g CCO 5. x 10 g CCO x 10 x 0.0 L x = L mole CCO 50.. PbBr (s) Pb + (q) + Br (q) PbBr [Pb + ] [Br ] initil 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 chnge - + x + x equilibrium - x x K sp = 8.9 x 10 6 = [Pb + ][Br ] = (x)(0.0 + x) (0.0) x x =. x 10 4 M = [PbBr ]

6 c. PbBr (s) Pb + (q) + Br (q) PbBr [Pb + ] [Br ] initil chnge - + x + x equilibrium x x K sp = 8.9 x 10 6 = [Pb + ][Br ] = (0.0 + x)(x) (0.0)4x x =. x 10 M = [PbBr ] mole CCl mole Cl 10.0 g CCl x x g CCl 1 mole CCl 1.00 L = 0.18 M Cl AgCl(s) Ag + (q) + Cl (q) AgCl [Ag + ] [Cl ] initil chnge - + x + x equilibrium - x x K sp = 1.6 x = [Ag + ][Cl ] = (x)( x) 0.18x x = 8.9 x 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 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 g CCO 1 mole CCO 5 9. x 10 x.0 L x = g/ kettle 1 kettle 116 g CCO x = g CCO 6. x 10 kettles

7 8. Ag CO (s) Ag + (q) + CO (q) Ag CO [Ag + ] [CO ] initil 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 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 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 x 10 7 M = 1.1 x 10 7 M poh = 6.96; ph = BSO 4 (s) B + (q) + SO 4 (q) BSO 4 [B + ] [SO 4 ] initil chnge - + x + x equilibrium - x x K sp = 1.1 x = [B + ][SO 4 ] = (x)(x) = x x = 1.0 x 10 5 M = [BSO 4 ] mole BSO L.88 g BSO 1 mole BSO x 10 x 5.0 L x = B(NO ) is too soluble (no K sp ) g BSO 4

8 99.. strong cid HX = H O +, strong bse = OH H O + + OH H O 1 K = = K w x 10 b. NH + H O + NH H O 1 K NH 1.8 x 10 K = = = = K NH K 1.0 x 10 5 b w 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 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 x = 1.8 x 10 x 1.8 x 10 x = x PbSO 4 (s) Pb + (q) + SO 4 (q) PbSO 4 [Pb + ] [SO 4 ] initil 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 = L HO 1 g HO 1000 ml HO 1 mole Pb 1 + x 10 g Pb exceeds limit of 0.05 g

9 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 ~ chnge x - + x + x equilibrium x - x x K = 1.8 x 10 [H O ][C H O ] (x)( x) x(0.500) [HC H O ] x x = 1.8 x 10 5 M = [H O + ]; ph = 4.74 [CHO ] or ph = pk + log log [HC H O ] 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 ~ chnge x - + x + x equilibrium x - x x K = 1.8 x 10 [H O ][C H O ] (x)( x) x(0.0500) [HC H O ] x x = 1.8 x 10 5 M = [H O + ]; ph = 4.74 [C H O ] ph = pk + log log 4.74 [HC H O ] 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 0 chnge x - + x + x equilibrium x - x x K = 1.8 x 10 [H O ][C H O ] (x)(x) x [HC H O ] x x =.0 x 10 M = [H O + ]; ph =.5

10 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 0 chnge x - + x + x equilibrium x - x x K = 1.8 x 10 [H O ][C H O ] (x)(x) x [HC H O ] x x = 9.5 x 10 4 M = [H O + ]; ph = 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 pk 9.0