CHAPTER 9 SUPPLEMENTARY SOLUTIONS 22 MONOPROTIC ACID-BASE EQUILIBRIA ] + [ + H 3 NCH 2 CO 2 H] = 0.05 M

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1 CHAPTER 9 SUPPLEMENTARY SOLUTIONS MONOPROTIC ACIDBASE EQUILIBRIA S91. [H ] [ H 3 NCH CO H] [OH ] [H NCH CO ] S9. [H ] 3[Al 3 ] [AlOH ] [Al(OH) ] [K ] [OH ] [Al(OH ) 4 ] S93. [H NCH CO ] [ H 3 NCH CO ] [ H 3 NCH CO H] 0.05 M S94. [Al 3 ] [AlOH ] [Al(OH) ] [Al(OH) 3 (aq)] [Al(OH) 4 [K ] 0.45 M ] M S Pertinent reaction: (Hg ) 3 [Co(CN) 6 ] (s) 3Hg Co(CN) 3 6. Charge balance: [Hg ] 3[Co(CN) 3 6 ] 3. Mass balance: [Hg ] 3[Co(CN) 3 6 ] 4. Equilibrium constant: K sp [Hg ] 3 [Co(CN) 3 6 ] Solve: If [Hg ] x, then [Co(CN) 3 6 ] 3 x. Putting these values into the solubility product gives (x) 3 ( 3 x) K sp x [Hg ] M S96. (a) M t [M ] [ML ] (b) L t [L ] [ML ] [HL] (c) [M ] [H ] [ML ] [L ] [OH ] (d) K K a [ML ] [M ][L ] [H ][L ] [HL] (1) () Equation () gives [L ] [HL] when [H ] M. Putting [L ] [HL] into the L t equation gives L t 0.1 M [L ] [ML ] [L ] [ML ]. Substituting this last expression for [L ] and the expression [M ] M t [ML ] 0.1 [ML ] into Equation (1) gives (0.10 [ML [ML ] ])( [ML Solve quadratic equation ]) [ML ] 0.10 M

2 Chapter 9: Supplementary Solutions 3 Equation (1) gives [M ] [ML ] K[L ] ( )(0.050) M S97. (a) [HF][OH ] [F ] ph 3.00 [HF] 1.5 [F ] [OH ] M Mass balance: [F ] [HF].5 [F ] [Mg ] [F ] 0.80[Mg ] Substitution into solubility product gives [Mg ][F ] [Mg ](0.80[Mg ]) [Mg ] M [F ] 0.80[Mg ] M [HF] 1.5 [F ] M (b) [MgF ] [Mg ][F K [MgF ] ( )[Mg ][F ] ] ( )[ ][ ] M (c) Mass balance: total F (total Mg) [F ] [HF] [MgF ] [Mg ] [MgF ] [F ] [HF] [Mg ] [MgF ] S98. [A ] [HA(aq)][OH ] Putting in [HA(aq)] M and [OH ] M [A ] M [HA] [A ] 0.54 M. S99. (a) Mass balance: [Sr ] [SO 4 ] [HSO 4 ] (1) Charge balance: [Sr ] [H ] [SO 4 ] [HSO 4 ] [OH ] () (b) If ph.50, [OH ] M [HSO 4 ][OH ] [SO 4 ] [HSO 4 ] [SO 4 ] Putting this expression for [HSO 4 ] into the mass balance gives [Sr ] [SO 4 ] [SO 4 ] [SO 4 ] Putting this expression for [SO 4 ] into the solubility product gives [Sr ][SO 4 ] [Sr ] [Sr ] 1.31 K sp [Sr ] M. 0

3 Chapter 9: Supplementary Solutions 4 S910. (a) Mass balance: [Sr ] [SrSO 4 ] [SO 4 ] [HSO 4 ] [SrSO 4 ] The net mass balance is unchanged from the previous problem because [SrSO 4 ] cancels on both sides of the equation. (b) The procedure from the previous problem is still entirely valid because the mass balance and the equilibria are unchanged. The only addition is the ion pair whose concentration is [SrSO 4 ] K[Sr ][SO 4 ] K[Sr ] [Sr ] M 0 Fraction in ion pair [SrSO 4 ] [SrSO 4 ] [Sr ] S911. (a) Mass balance: [Ca ] [C O 4 ] [HC O 4 ] [H C O 4 ] Charge balance: [Ca ] [H ] [C O 4 ] [HC O 4 ] [OH ] (b) If ph.30, [OH ] M [HC O 4 ][OH ] [C O 4 ] [H C O 4 ][OH ] [HC O 4 ] K b1 [HC O 4 ] [C O 4 ] K b [H C O 4 ] 0.090[HC O 4 ] 8.1[C O 4 ] Substituting into the mass balance gives [Ca ] [C O 4 ] [C O 4 ] 8.1[C O 4 ] 99.1[C O 4 ] Substituting into the solubility product gives [Ca ][C O 4 ] [Ca ] [Ca ] [Ca ] M S91. (a) Mass balance: [Zn ] 3{[AsO 4 3 ] [HAsO 4 ] [H AsO 4 ] [H 3 AsO 4 ]} Charge balance: [Zn ] [H ] 3[AsO 4 3 ] [HAsO 4 ] [H AsO 4 ] [OH ] (b) ph 6.00 [OH ] [HAsO 4 ][OH ] [AsO 3 4 ] [H AsO 4 ][OH ] [HAsO 4 ] K b1 [HAsO 4 ] [AsO 4 3 ] K b [H AsO 4 ] 9.1[HAsO 4 ] [AsO 4 3 ]

4 Chapter 9: Supplementary Solutions 5 [H 3 AsO 4 ][OH ] [H AsO 4 ] K b3 [H 3 AsO 4 ] [H AsO 4 ] 480[AsO 4 3 ] Putting these expressions into the mass balance gives [Zn ] 3[AsO 3 4 ]( ) [AsO 3 4 ] [AsO 3 4 ] [Zn ] Substituting into the solubility product gives [Zn ]: [Zn ] 3 { [Zn ]} K sp [Zn ] M S913. Concentrations in ML solution A B C D E F G H 1 K ph [H] b [M] [L] [ML] [HL] 1.0E08 0 1E00 1.0E05 9.5E03 9.5E08 9.0E0 9.5E03 3 Ka 1E0 1.0E03 1.0E03 9.9E07 9.9E0 9.9E E05 5 1E05.0E00 4.5E05.E05 1.0E01.E05 5 M(total) 6 1E06 1.1E00 3.3E05 3.0E05 1.0E01 3.0E E08 1.0E00 3.E05 3.E05 1.0E01 3.E08 7 L(total) 10 1E10 1.0E00 3.E05 3.E05 1.0E01 3.E E1 1.0E00 3.E05 3.E05 1.0E01 3.E E14 1.0E00 3.E05 3.E05 1.0E01 3.E14 10 C 10^B 11 D 1$A$*$A$8$A$*$A$6(C/$A$4) 1 E(DSqrt(D^4*$A$*$A$6*(1C/$A$4)))/(*$A$) 13 F ($A$6/E1)/$A$ 14 G $A$*E*F 15 H F*C/$A$4

5 CHAPTER 10: SUPPLEMENTARY SOLUTIONS 6 POLYPROTIC ACIDBASE EQUILIBRIA S101. (a) ph log [H ] log ( ) 3.30 (b) ph log (K w /[OH ]) log ( ) / ( ) S10. [OH ] [H ] [(CH 3 ) 4 N ] [H ] [H ] [OH ] K w [H ] ([H ] ) [H ] M ph log [H ] 7.38 [OH ] total K w /[H ] M Fraction of [OH ] from H O M M [OH ] from H O [H ] M 0.17 S103. (a) For M HBr, ph log [H ] γ H log[(0.050)(0.86)] 1.37 (b) For M NaOH, ph log A H log( K w A OH ) log K w [OH ] γ OH log (0.050)(0.81) S104. N: NH NO 3 pyridine pyridinium nitrate N: H O NH OH K b K w /K a pk b 8.77 S105. Let x [H ] [A ] and x [HA] x x M ph log x 3.0 α x 4 F

6 Chapter 10: Supplementary Solutions 7 S106. NH N H K a F x x x x K a x ph 3.15 S107. BH B H K a x x x 0.00 x x M ph log x 6.17 [B] x M [BH ] 0.00 M S108. HA H A K a (10.36 )(10.36 ) pk a log K a 3.70 S109. (a) HA H A K a x x x x x M ph 5.69 α x F (b) For F M, the ph must be very close to K a [H ][A ] F [A ] ( )[A ] F [A ] [A ] M α x F S1010. A H O AOH H F x x x F x x M ph 3.55

7 Chapter 10: Supplementary Solutions 8 S1011. B H O BH OH K b x x x x x ph log (K w /x) α x F S101. B H O BH OH K b x x x x x ph log (K w /x) 8.57 [B] [(CH 3 CH ) NH] M [BH ] M S1013. OCl H O HOCl OH 0.06 x x x K b K w /K a K w /( ) x K b x [OH ] M ph log (K w /[OH ]) 9.97 α x F S1014. HCO H O HCO H OH F x x x K b K w /K a For F 10 1 and 10 M, we solve the equation / (F x) K b. The fraction of association is α x F. This gives x M for F 10 1 M ( α ) and x M for F 10 ( α ). For F 10 1 M, ph 7.00 and [OH ] 10 7 M. K b [BH ][OH ] [B] [BH ] 10 7 [B] [BH ] 10 7 K b [B] Putting this relation between [BH ] and [B] into the definition of fraction of association gives α [BH ] [BH ] [B] 10 7 K b [B] 10 7 K b [B] [B] 10 7 K b 10 7 K b

8 Chapter 10: Supplementary Solutions 9 S1015. B H O BH OH x x x F α ph [OH ] M K b x ( ) S1016. α x F Since F M, x M. K b F x ( ) S1017. B H O BH OH x x x K b K w /K a x α x F K b x ph log (K w /x) S1018. The pk a values are (a) , (b) 9.44, (c) 5.96 and (d) Since pk a for 3nitrophenol is closest to 8.5, buffer (d) will have the greatest buffer capacity at ph 8.5. S1019. ph pk a log [A ] [HA] 3.46 log (5.13 g)/(11.16 g/mol) (.53 g)/( g/mol) 3.59 S100. ph log [CH 3NH ] [CH 3 NH 3 ] ph [CH 3 NH ]/[CH 3 NH 3 ] S101. pk a pk b 0.17 ph 0.17 log [IO 3 ] [HIO 3 ]

9 Chapter 10: Supplementary Solutions 30 ph [HIO 3 ]/[IO ] S10. (a) ph pk a log [tris] [trish ] log (10.0 g)/( g/mol) (10.0 g)/( g/mol) 8.19 (b) trish OH tris H O Initial mmol: Final mmol: ph log S103. B H BH Initial mmol: x Final mmol: x x log x x x mmol ml required ( mmol)/(0.113 mmol/ml) 4.68 ml S104. (a) 50.0 ml of M buffer requires mol NaOAc. H O.95 g OAc H HOAc Initial mol: Final mol: x x ph pk a log [OAc ] [HOAc] log x x x mmol 90.9 ml HCl (b) 1. Calibrate the ph electrode and meter at 5 C.. Dissolve.95 g NaOAc. H O in ~100 ml H O at 5 C. 3. While measuring the ph, add enough HCl to bring the ph to 5.00 at 5 C. 4. Dilute to exactly 50 ml at 5 C.

10 Chapter 10: Supplementary Solutions 31 S105. The solution will have a ph near pk a ( ), so we can neglect [H ] relative to [OH ] in Equations 100 and 101: [HA] F HA [OH ] [A ] F A [OH ] B H O BH OH K b K b [BH ][OH ] [B] (F BH [OH ])[OH ] F B [OH ] ( [OH ])[OH ] [OH ] [OH ] M [B] [OH ] M [BH ] [OH ] M S106. (a) Assume that the activity coefficients of neutral species are 1. Equilibria: B H O BH OH K b [BH ][OH ] [B] (a) H O H OH K w [H ][OH ] (b) Charge balance: [H ] [BH ] [OH ] (c) Mass balance: F [B] [BH ] (d) Substitute [BH ] [OH ] [H ] from the charge balance, and [B] F [BH ] F [OH ] [H ] from the mass and charge balances into the K b equilibrium. This gives K b ([OH ] [H ]) [OH ] F [OH ] [H. ] Now replace [H ] by K w /[OH ] to get K b ([OH ] K w /[OH ]) [OH ] F [OH ] K w /[OH. ] This equation can be rearranged to: K b F K b [OH ] K b K w /[OH ] [OH ] K w. Multiplying both sides by [OH ] and rearranging gives a cubic polynomial: [OH ] 3 K b [OH ] (K b F K w )[OH ] K b K w 0 (b) ph of 10 3 M B (K b 10 5 ) is 9.978; ph of 10 5 M B (K b 10 9 ) is

11 Chapter 10: Supplementary Solutions 3 A B C D E 1 Solving cubic polynomial for ph of B 3 Kb Guess for ph [H] [OH] Value of polynomial E05 9 1E E14 5 Kw 10 1E E E E10 3.E E13 7 F E10 7.9E E E E10 9.5E05.675E E4 D4^3 $A$4*D4^ ($A$4*$A$8$A$6)*D4 $A$4*$A$6 S107. At 5 C, K a increases with increasing temperature, so the reaction must be endothermic, by Le Châtelier's principle. At 45 C, K a decreases with temperature, so the reaction must be exothermic. S108. Equilibria: HA H A K a [H ][A ] [HA] (a) H O H OH K w [H ][OH ] (b) Charge balance: [H ] [A ] [OH ] (c) Mass balance: F [A ] [HA] (d) Substitute [A ] [H ] [OH ] from the charge balance, and [HA] F [A ] F [H ] [OH ] from the mass and charge balances into the K a equilibrium. This gives K a [H ] ([H ] [OH ]) F [H ] [OH. ] Now replace [OH ] by K w /[H ] to get K a [H ] ([H ] K w /[H ]) F [H ] K w /[H ] This equation can be rearranged to: K a F K a [H ] K a K w /[H ] [H ] K w. Multiplying both sides by [H ] and rearranging gives a cubic polynomial: [H ] 3 K a [H ] (K a F K w )[H ] K a K w 0.

12 Chapter 10: Supplementary Solutions 33 S109. A B C D 1 Solving cubic polynomial for ph of HA 3 Ka Guess for ph [H] Value of polynomial E E13 5 Kw E E E E E14 7 F 4.3 5E E E E E D4 C4^3 $A$4*C4^ ($A$4*$A$8$A$6)*C4 $A$4*$A$6 S1030. We will assume that the activity coefficients of neutral species are 1. Equilibria: HA H A K a [H ]γ H [A ]γ A [HA] (a) H O H OH K w [H ]γ H [OH ]γ OH (b) Charge balance: [H ] [A ] [OH ] (c) Mass balance: F [A ] [HA] (d) Substitute [A ] [H ] [OH ] from the charge balance, and [HA] F [A ] F [H ] [OH ] from the mass and charge balances into the K a equilibrium. This gives K a [H ]γ H ([H ] [OH ])γ A F [H ] [OH ]. Now replace [OH ] by K w /[H ]γ H γ OH to get K a [H ]γ H ([H ] K w /[H ]γ H γ OH )γ A F [H ] K w /[H ]γ H γ OH. This equation can be rearranged to: K a F K a [H ] K a K w /[H ]γ H γ OH [H ] γ H K w γ A /γ OH. Multiplying both sides by [H ] and rearranging gives a cubic polynomial: [H ] 3 γ H K a [H ] K a F K wγ A γ OH [H ] K ak w γ H γ OH 0

13 CHAPTER 11: SUPPLEMENTARY SOLUTIONS 34 ACIDBASE TITRATIONS S111. CH OH CH OH H N C H C O H O H 3 N C HCO OH K b1 K w /K a CH OH CH OH H 3 N C H CO H O H 3 N C HCO H OH K b K w /K a S11. (a) x K 1 x M [H ] [HA ] ph 3.00 [H A] x M [A ] K [HA ] [H ] M (b) [H ] K 1 K F K 1 K w K 1 F M ph 7.00 [HA ] M [H A] [H ][HA ] K M [A ] K [HA ] [H ] M (c) x K w K x [OH ] [HA ] M ph [A ] x M [H A] [H ][HA ] K M ph [H A] [HA ] [A ] M H A M NaHA M Na A S113. N NH Cl (BH Cl ) (BH Cl ) is the intermediate form of a diprotic system, with F M, K and K [H K 1 K F K 1 K w ] K 1 F M ph 7.53 [BH ] M [B] K [BH ] [H ] 4 M

14 Chapter 11: Supplementary Solutions 35 [BH ] [BH ][H ] K M S114. Charge balance: [H ] [Na ] [BH ] [Cl ] [HA ] [A ] [OH ] Mass balances: [Cl ] F 1 [Na ] F F [H A] [HA ] [A ] F 3 [B] [BH ] Equilibria : [H ]γ H [OH ] γ OH K w K 1 [H ]γ H [HA ] γ HA / [H A] γ H A K [H ] γ H [A ] γ A / [HA ] γ HA K b [BH ] γ BH [OH ] γ OH / [B]γ B S115. Tartaric acid: H A pk pk (a) At ph 3.00, there is a mixture of H A and HA. H A OH HA Initial mmol: x Final mmol: x x log x x x mmol 3.77 ml KOH. (b) At ph 4.00, there is a mixture of HA and A. We must add mmol ( ml) of KOH to convert H A into HA. Then we need to add more KOH: HA OH A H O Initial mmol: x Final mmol: x x x log x x mmol.370 ml KOH Total KOH ml

15 Chapter 11: Supplementary Solutions 36 S116. Malonic acid H A pk pk (a) At ph 6.00, there is a mixture of A and HA. A H HA Initial mmol:.775 x Final mmol:.775 x x log.775 x x x mmol.19 ml HCl. (b) At ph 3.0, there is a mixture of H A and HA. We first add.775 mmol ( ml) of HCl to convert A into HA. Then we need to add more HCl HA H H A Initial mmol:.775 x Final mmol:.775 x x log.775 x x x mmol.05 ml HCl Total HCl ml S117. Oxalic acid H A pk pk 4.66 At ph 3.0, there is a mixture of A and HA. We begin with 5.00 g mol A. The reaction of H A with A creates moles of HA : H A A HA Initial mmol: x Final mmol: x x log x.31 g oxalic acid. x mol S118. NH 3 NH 3 NH 3 NH CHCH CO H K 1 K CHCH CHCH CO CHCH CO CO H K 3 CO H CO CO CO Aspartic acid

16 Chapter 11: Supplementary Solutions 37 N H 3 CHCH CH CH N HC CO H N H 3 N H N H K 1 CHCH CH CH N HC N H N H CO K N H CHCH CH CH N HC N H K 3 N H CHCH CH CH N HC N H CO Arginine N H CO N H S119. (a) H 3 His pk H His pk 6.0 HHis Histidine pk His For M histidine, [H ] ph log [HHis] [H His ] K K 3 F K K w K F [H His ] [HHis] M (b) For M H His Cl, [H ] K 1 K F K 1 K w K 1 F M S1110. NH 3 ph log [HHis] [H His ] CHCH CH CH NHC CO NH NH H Arg found in Arginine. HCl [H His ] [HHis] 13 For H 3 Arg pk pk pk [H ] K 1 K (0.01 0) K 1 K w K 1 (0.01 0) M ph 5.58 [H Arg ][H ] [H 3 Arg ] K 1 [H 3 Arg ] 6 (0.01 0)( ) M [HArg][H ] [H Arg ] K [HArg] ( )(0.01 0) M

17 Chapter 11: Supplementary Solutions 38 [Arg ][H ] [HArg] K 3 [Arg ] ( )( ) M S1111. (a) Since pk , BH is predominant at ph 11 and B is predominant at ph 1. (b) (c) log log [B] [BH ] [B] [BH ] [B] [BH ] [B] [BH ] S111. (a) HSO 3 (b) HSO 3 (c) HSO 3 (d) SO 3 pk S1113. H 3 Cit H Cit pk HCit pk Cit 3 At ph 5.00, HCit is dominant. S1114. Fraction in form HA α HA [H ] [H ] K a Fraction in form A α A K a [H ] K a [A ] [HA] S1115. Fraction in form BH α BH [H ] [H ] [H ]K 1 K 1 K, where K and K α BH S1116. K K ph 5.00 ph 6.00 α H A [H ] [H ] [H ]K 1 K 1 K α HA α A [H ]K 1 [H ] [H ]K 1 K 1 K K 1 K [H ] [H ]K 1 K 1 K

18 Chapter 11: Supplementary Solutions 39 S1117. K K ph: α H A α HA α A S1118. Isoelectric ph pk 1 pk 7.36 Isoionic ph K 1 K (0.010) K 1 K w K 1 (0.010) M ph 7.36 S1119. (a) K e ( S /R) ( H /RT) S R 3.3 (±0.53) and H ( ± 0.15) 10 RT T J Using the value R mol. K gives: S [ 3.3 (±0.53)][ ] 7 (±4) J mol. K H [ 1.44 (±0.15) 10 3 ][ ] 1 (±1) kj mol (b) K [SO 3H ] [HSO 3 ] e [3.3 (±0.53) 1.44 (±0.15) 103 (1/T)] For T 98 K, K e[3.3 (±0.53) (1 440 (±150)) / 98] e3.3 (±0.53) 4.83 (±0.50) K e 1.60 (±0.73) (±?) For propagation of uncertainty, we consider K e x in Table 31: e K K e x e K Ke x (4.953)(±0.73) ±3.6 K 5.0 (±3.6)

19 CHAPTER 1: SUPPLEMENTARY SOLUTIONS 40 EDTA TITRATIONS S11. 0 ml: ph log ( ) ml: [H ] M ph ml: ml: ml: ml: Equivalence point ph ml: [OH ] M ph ml: 11.4 S1. 0 ml: HA H A x 10 4 x M ph.66 1 ml: ph pk a log [A ] [HA] 4.00 log ml: ph 4.00 log ml: ph 4.00 log ml: ph 4.00 log ml: A H O HA OH pk b ( ) x x x x x [OH ] ph ml: [OH ] (0.500 M ) M ph ml: [OH ] 1 56 (0.500 M ) M ph S13. V e 5.0 ml pk a ml: HA A HA x x M ph 4.3 To find the volume at which ph pk a 1, we set up a HendersonHasselbalch equation: pk a 1 pk a log [A ] [HA] [A ]/[HA] 0.1, or [A ] 1/11 and [HA] 10/11 of total.

20 Chapter 1: Supplementary Solutions 41 Volume of OH V e 11.7 ml Similarly, when ph pk a 1, volume of OH 10V e ml When volume of OH 1 V e 1.5 ml, ph pk a V e 5.0 ml: HA is converted to A : A H O HA OH K b (0.05) x x x x [OH ] M ph V e 30.0 ml: [OH ] 5 80 (0.100 M ) M ph S14. HA OH A H O Initial mmol: Final mmol: pk a log pk a 3.7 S15. 0 ml: B H O BH OH x x x x [B] 1 ml: ph pk BH log [BH ].5 ml: ph pk BH x [OH ] M ph log ml: ph log ml: ph log ml: BH H O B H (0.050) x x x x x [H ] M ph ml: [H ] (0.500 M ) M ph ml: [H ] 1 56 (0.500 M ) M ph.05

21 Chapter 1: Supplementary Solutions 4 S16. Titration reaction: CH 3 CH CO H CH 3 CH CO H At the equivalence point, moles of HCl moles of NaA V e ml. 0 ml: A H O HA OH x x x x 1/4 V e : ph pk a log [A ] [HA] 1/ V e : ph pk a 4.87 K b K w K x M ph log K w a x log /4 V e : ph pk a log V e : HA H A F x x x F HA ( ) M x K a x M ph V e : [H ] V e / V ( ) M ph.58 e S17. V e1 5 ml V e 10 ml 0 ml: B H O BH OH x x x x x [OH ] M ph [B] 1 ml: ph pk BH log [BH ].5 ml: ph pk BH ml: ph pk BH log ml: ph pk BH log ml: [H K 1 K F K 1 K w ] K 1 F M ph 7.00 [B] 5. ml: ph pk BH log [BH ] log log ( ) log ml: ph pk BH log

22 Chapter 1: Supplementary Solutions ml: ph pk BH ml: ph pk BH log ml: ph pk BH log ml: BH BH H x x x x x M ph ml: [H ] M ph ml: [H ] M ph.09 1 ml: [H ] M ph 1.79 S18. V e1 5 ml V e 10 ml 0 ml: H A H HA x x x x x M ph ml: ph pk 1 log [HA ] [H A] 5.00 log ml: ph pk ml: ph pk 1 log ml: ph pk 1 log ml: [H K 1 K ( ) K 1 K w ] K 1 ( ) 5. ml: ph pk log ph ml: ph pk log ml: ph pk ml ph pk log ml: ph pk log

23 Chapter 1: Supplementary Solutions ml: A H O HA OH x x x K b1 K w K x K x [OH ] M ph ml: [OH 0. ] (0.500 M) M ph ml: [OH ] 1 61 M ph ml: [OH ] 6 M ph 1.1 S19. Titration reaction: NH H NH 3 0 ml: B H O BH OH x x x x ml: ph pk a log pk a V e 10.0 ml K b K w K a K b x [OH ] M ph [B] [BH ] 4 ml: ph log ml: ph log ml: ph log log ml: BH B H K a 5 35 (0.100) x x x x M x M ph ml: [H ] 37 (0.50 M) M ph 1.87 S110. (a) NH 3 CHCH OH OH CO (b) V ml of tyrosine will require NH CHCH CO OH V.50 V ml of KOH. The formal

24 Chapter 1: Supplementary Solutions 45 concentration of product will be V V.50 V ( ) M. [H ] K K 3 ( ) K K w K M ph 9.81 S111. Leucine is HL. The reaction is: HL OH L H O Initial mmol: 7.7 x Final mmol: 7.7 x x x ph 8.00 pk log 7.7 x x 0.18 mmol 0.18 mmol mmol/ml.97 ml of NaOH S11. (a) NH 3 NH NH 3 NH NH 3 NH H H CHCH CHCH CHCH N N N CO CO H CO H H HA H H 3 A A (b) pk pk 6.0 pk H 3 A H A HA A ph 1 ( ) 3.86 At ph 3.00, HA has been converted entirely to H A, and some H A has been converted to H 3 A. H A H H 3 A Initial mmol: 1.00 x Final mmol: 1.00 x x ph pk 1 log [H A ] [H 3 A ] log 1 x x x mmol We need mmol HClO 4. Volume mmol mmol/ml 1.0 ml S113. The initial color will be blue. The color will change near the first equivalence point from blue to green. After the first equivalence point, the color will turn yellow and will remain yellow.

25 Chapter 1: Supplementary Solutions 46 S114. (a) red (b) yellow (c) yellow (d) purple S115. There is not an abrupt change in ph at V e. The color of the indicator would change very gradually. S116. NHBr pk a 5.9 K a V e : ph pk a log V e : This is a M pyridine solution : N H O NH OH x x x K b 5.90 K w x x [H ] K w x ph 8.96 (0.01)(0.10 M ) V e 1.01 V e : This solution contains (.01) V M [OH ] ph e Two possibilities are cresol red (orange red) and phenophthalein (colorless red). S117. [B] [BH 3 ( g) /(78.16 g/mol) ] L Absorbance ε B [B] ε BH [BH ] M ( 860)[B] (937)( [B]) [B] M, [BH ] M ph pk a log [B] [BH ] ( ) log S118. (a) First equivalence point : ph 1 (pk 1 pk ) 3.46 Second equivalence point : We need to estimate the formal concentration of A at this point. Volume of NaOH needed to react with 1 g of acid is (1 g/ g/mol) M ml. [A ] g/l g/mol M. A H O HA OH x x x

26 Chapter 1: Supplementary Solutions x K w x M ph 8.4 (b) nd. (There will be little break in the curve near the 1 st end point.) (c) Thymolphthalein first trace of blue. S119. One mole of borax reacts with two moles of H. [HNO 3 ]( L) g g/mol [HNO 3 ] M S10. Derivation of equation for titration of dibasic B with strong acid (HCl): Charge balance: [H ] [BH ] [BH ] [OH ] [Cl ] φ fraction of titration C av a C b V b Substitutions: [Cl C a V a ] V a V b [BH ] α BH C bv b V a V b [BH ] α C bv b BH V a V b Putting these expressions into the charge balance gives Now multiply by V a V b : [H ] α BH C b V b V a V b α BH C b V b V a V [OH ] b C a V a V a V b [H ]V a [H ]V b α BH C b V b α BH C b V b [OH ]V a [OH ]V b C a V a Collect V a and V b terms: V a ([H ] [OH ] C a ) V b ([OH ] [H ] α BH C b α BH C b ) V a V b (α BH α BH )C b [H ] [OH ] C a [H ] [OH ] Finally, multiply both sides by 1/C b 1/C a φ C av a C b V b α BH α BH [H ] [OH ] C b 1 [H ] [OH ] C a

27 Chapter 1: Supplementary Solutions 48 S11. Derivation of fractional composition equations with activity coefficients: K HA H A [H ]γ H [A a ]γ A K a [HA]γ HA Mass balance: F [HA] [A ] [A ] F [HA] Substitute for [A ] in the equilibrium expression: K a [H ]γ H (F [HA])γ A [HA]γ HA [HA] [H ]γ H γ A F [H ]γ H γ A K a γ HA α HA [HA] F [H ]γ H γ A [H]γ H γ A K a γ HA To derive an expression for α A, substitute [HA] F [A ] into the equilibrium expression: K a [H ]γ H [A ]γ A (F [A [A ] ])γ HA K a Fγ HA [H ]γ H γ A K a γ HA α A [A ] F K a γ HA [H ]γ H γ A K a γ HA S1. rate D m [T] r tanθ ( m mol L 6 mol /s) ( m)(tan 3.0 ) L 3 s m 0.84 fmol/s In a year, the total volume delivered would be 16 mol s d day mol. s h day year This much titrant is contained in mol L 0.66 µl mol/l The equivalent volumetric flow rate is 0.66 µl per year!