CHAPTER 11 AQUEOUS EQUILIBRIA 11.1 (a) When solid sodium acetate is added to an acetic acid solution, the concentration of HO decreases because the equilibrium HC H O (aq) H O(l) H O (aq) C H O (aq) shifts to the left to relieve the stress imposed by the increase of [CHO ] (Le Chatelier s principle). (b) When HCl is added to a benzoic acid solution, the percentage of benzoic acid that is deprotonated decreases because the equilibrium C H COOH(aq) H O(l) H O (aq) C H CO (aq) 6 5 6 5 shifts to the left to relieve the stress imposed by the increased Chatelier s principle). (c) When solid NH4Cl is added to an ammonia solution, the concentration of OH decreases because the equilibrium 4 [HO ] NH (aq) H O(l) NH (aq) OH (aq) shifts to the left to relieve the stress imposed by the increased [NH 4 ] (Le Chatelier s principle). Because [ OH ] decreases, [HO ] increases and ph decreases. (Le 11. (a) a [H O ][A ] [A ] [HA] [HA] = ;pa = phlog If [ A ] = [HA], then p = ph. a ph = p =.08, = 8. a a 4 (b) 1 Let = [lactate ion] = [L ] and y = [HO ] SM-14
Concentration (mol L ) HL(aq) H O(l) H O (aq) L (aq) 1 initial change y y y equilibrium y y y a [H O ][L ] ( y)( y ) ( y)( ) [HL] ( y) ( ) = = = 8. 4 y = 4 1 (8. ) 1.7 mol L [HO ] ph.77 11.5 In each case, the equilibrium involved is HSO (aq) H O(l) H O (aq) SO (aq) 4 4 HSO (aq) and SO (aq) 4 4 are conjugate acid and base; therefore, the ph calculation is most easily performed with the Henderson-Hasselbalch equation: [base] [SO 4 ] ph = pa log = pa log [acid] [HSO 4 ] (a) 1 0.5 mol L ph = 1.9 log 1.6, poh 14.00 1.6 1.8 1 = = = 0.5 mol L (b) (c) 1 0. mol L ph = 1.9 log 1., poh 1.78 1 = = 0.50 mol L ph = p = 1.9, poh = 1.08 a See solution to Eercise 11.. 11.7 0.56 g NaF 1 mol NaF = 0.17 mol L 0.050 L 41.99 g NaF Concentration 1 (mol L ) HF(aq) HO(l) HO (aq) F (aq) initial 0.40 0 0.17 1 SM-15
change equilibrium 0.40 0.17 a [H O ][F ] ( )(0.17 ) ( )(0.17) [HF] (0.40 ) (0.40) = = =.5 4 4 1 8. mol L [HO ] 4 ph = log[ho ] = log(8. ) =.09 change in ph =.09 1.9 = 1.16 11.9 (a) HCN(aq) HO(l) HO (aq) CN (aq) total volume = 0 ml = 0.0 L 1 moles of HCN = 0.000 L 0.050 mol L = 1.5 mol HCN 1 moles of NaCN = 0.0700 L 0.00 mol L =.1 mol NaCN 1.5 mol initial [HCN] 0 = = 1.5 mol L 0.0 L 1.1 mol initial [CN ] 0 = =.1 mol L 0.0 L Concentration 1 1 (mol L ) L HCN(aq) H O(l) H O (aq) CN (aq) initial 1.5 0 change.1 equilibrium1.5.1 [H O ][CN ] ( )(.1 ) ( )(.1 ) a = = = 4.9 [HCN] (1.5 ) (1.5 ) 1.5 mol L [HO ] ph = log[ho ] = log(.5 ) = 9.46 (b) The solution here is the same as for part (a), ecept for the initial concentrations: SM-16
1 0.0400 L 0.00 mol L [HCN] 0 = = 1. mol L 0.0 L 1 0.0600 L 0.050 mol L [CN ] 0 = =.0 mol L 0.0 L ( )(.0 ) a = 4.9 = 1 1 (1. ) = [H O ] =.0 mol L 1 ph = log(.0 ) = 9.71 (c) [HCN] 0 = [NaCN] 0 after miing; therefore, ()[NaCN] = 4.9 = = = [H O ] 0 a [HCN] 0 ph = pa = log(4.9 ) = 9.1 11.11 In a solution containing HClO(aq) and ClO (aq), the following equilibrium occurs: HClO(aq) H O(l) L H O (aq) ClO (aq) The ratio [C lo ]/[HClO] is related to ph, as given by the Henderson- Hasselbalch equation: [ClO ] ph = pa log, [HClO] or [ClO ] log = ph pa = 6.50 7.5 = 1.0 [HClO] [ClO ] = 9. [HClO] 11.1 The rule of thumb we use is that the effective range of a buffer is roughly within plus or minus one ph unit of the (a) p a =.08; ph range, 4 (b) p a = 4.19; ph range, 5 p a of the acid. Therefore, SM-17
(c) p a = 1.68; ph range, 11.5 1.5 (d) p a = 7.1; ph range, 6 8 (e) p b = 7.97, p a = 6.0; ph range, 5 7 11.15 Choose a buffer system in which the conjugate acid has a p a close to the desired ph. Therefore, (a) HClO and NaClO, p a =.00 (b) NaHPO4 and Na HPO 4, p a = 7.1 (c) CHClCOOH and NaCHClCO, p a =.85 (d) Na HPO4 and NaPO 4, p a = 1.68 11.17 (a) a HCO (aq) H O(l) L CO (aq) H O (aq) [HO ][CO ], p a.5 [HCO ] = = [CO ] = ph pa log [HCO ] [CO ] log ph pa 11.0.5 0.75 = = = [HCO ] [CO ] = [HCO ] 5.6 (b) [CO ] = 5.6 [HCO ] = 5.6 0.0 mol L = 0.56 mol L 1 1 moles of mass of CO = moles of CO = 0.56 mol L 1 L = 0.56 mol 1 CO 18.1 g CO = 0.56 mol = 77 g CO 1molCO (c) 1 [CO ] 0.0 mol L [HCO ] = = = 1.8 mol L 5.6 5.6 moles of 1 HCO = moles of HCO 1.8 = mol L 1 1 L = 1.8 mol SM-18
1 mass HCO 1.8 = mol 0.1 g mol = 1.8 g HCO (d) [CO ] = 5.6 [HCO ] moles of 1 HCO = moles HCO = 0.0 mol L 0.0 L = 1.00 mol. Because the final total volume is the same for both HCO and CO, the number of moles of CO required is 5.6 1.00 mol= 5.6 mol. Thus, 5.6 mol volume of CO solution = 1 0.00 mol L = = 0.8 L.8 ml 11.19 (a) [CH CO ] = (see Eercise 11.) ph pa log [CH COOH] 0.0 ph = pa log = 4.75 0.0 (initial ph) final ph: (0.00 L) (0.950 1 mol L ) = 9.50 mol NaOH (strong base) produces from CHCOOH 9.50 mol CHCO 1 0.0 mol L 0.0 L = 1.00 mol CHCOOH initially 1 0.0 mol L 0.0 L = 1.00 mol CHCO initially After adding NaOH: (1.00 9.5 ) mol [CHCOOH] = = 5 mol L 0.1 L (1.00 9.5 ) mol 1 [CHCO ] = = 1.77 mol L 0.1 L 1 1 1.77 mol L ph = 4.75 log 4.75 1.5 6. 1 = = 5 mol L 1 1 (b) 1 (0.000 L)(0.0 mol L ) =.00 mol HNO (strong acid) produces.00 mol CHCOOH from CHCO. SM-19
After adding HNO [see part (a) of this eercise]: (1.00.00 ) mol [CHCOOH] = = 1.00 mol L 0. L (1.00.00 ) mol 1 1 [CHCO ] = = 6.7 mol L 0. L 1 6.7 mol L ph = 4.75 log 4.75 0.17 4.58 1 1 = = 1.00 mol L ph = 0.17 1 11.1 14 7 (b) ph at stoichiometric point = 7.0 (a) initial ph = 1 0 0 0 11. HCl(aq) NaOH(aq) HO(l) Na (aq) Cl (aq) (a) V L 0.1 mol NaOH = ( )(5.0 ml) 1mL 1L 1 HCl 1molHCl 1LHCl 1 mol NaOH 0.150 mol HCl 9.17 L HCl = (b) (c) 9.17 L = 0.018 L volume = (0.050 0.018) L = 0.04 L 0.1 mol NaOH 1 mol Na 1 [Na ] = (0.050 L) 1 L 1 mol NaOH 0.04 L = 0.065 mol L 1 SM-0
0.150 mol (d) number of moles of HO (from acid) = (0.000 L) 1L =.00 mol HO number of moles of OH (from base) 0.1 mol Na = (0.050 L) 1L =.75 mol OH ecess H O = (.00.75) mol =.5 mol H O 4 4.5 mol 1 [HO ] = = 5.6 mol L 0.0450 L ph log(5.6 ).5 = = 11.5 The moles of OH - added are equivalent to the number of moles of HA present: -1 ( )( ) 0.050 mol L 0.050 L = 0.018 mol OH, 0.018 mol of HA were present in solution. 4.5 g molar mass = = 4 g mol 0.018 mol -1 0.0695 mol HCl 11.7 mass of pure NaOH = (0.04 L HCl) 1LHCl 1 mol NaOH 40.00 g NaOH 00 ml 1 mol HCl 1 mol NaOH 5.0 ml = 1.14 g 1.14 g percent purity = 0% = 71.7% 1.59 g 11.9 (a) poh = log(0.1) = 0.959, ph = 14.00 0.959 = 1.04 (b) 0.1 mol initial moles of OH (from base) = (0.050 L) 1L =.75 mol OH SM-1
0.150 mol 4 moles of HO added = (0.0050 L) = 7.5 mol HO 1L ecess OH (.75 0.75) mol.00 mol OH = =.00 mol [OH ] = = 0.067 mol L 0.00 L poh = log(0.067) = 1.17, ph = 14.00 1.17 = 1.8 1 (c) moles of H O added = 7.5 mol = 1.50 mol H O 4 ecess OH = (.75 1.50) mol = 1.5 mol OH 1.5 mol [OH ] = = 0.06 mol L 0.05 L poh = log(0.06) = 1.44, ph = 14.00 1.44 = 1.56 (d) ph = 7.00 V HCl (.75 mol NaOH) 1 mol NaOH 0.150 mol HCl 1 1molHCl = = 0.018 L 1LHCl 0.150 mol 1 (e) [HO ] = (0.0050 L) 1 L (0.050 0.018 0.0050) L = 0.016 mol L ph = log(0.016) = 1.80 (f) 1 0.0 L 0.150 mol [ HO ] = = 0.08 mol L 0.05 L 1 L 1 ph = log(0.08) = 1.55 11.1 (a) Concentration mol L 1 ( ) CH COOH(aq) H O(l) H O (aq) CH CO (aq) initial 0. 0 0 change equilibrium 0. [H O ][CH CO ] [CH COOH] 0. 0. 5 a = = 1.8 = SM-
= 1.8 6 = = 1 1. mol L [HO ] initial ph = log(1. ) =.89 (b) moles of CHCOOH = (0.050 L)(0. M) =.50 mol CHCOOH moles of NaOH = (0.00 L)(0. M) = 1.0 mol OH After neutralization, 1.50 mol CHCOOH 4.9 mol L 1 = CH COOH 0.050 L 1.0 mol CHCO 0.050 L Then consider equilibrium, Concentration (mol L ) initial = 1.86 mol L CHCO a [HO ][CHCO ] = [CH COOH] 1 CH COOH(aq) H O(l) H O (aq) CH CO (aq) 4.9 0 change equilibrium 4.9 (4.9 ).86.86 5 ( )(.86 ) 1.8 = ; assume and negligible. 5 1 5 [HO ] = =.7 mol L and ph = log(.7 ) = 4.56 (c) Because acid and base concentrations are equal, their volumes are equal at the stoichiometric point. Therefore, 5.0 ml NaOH is required to reach the stoichiometric point and 1.5 ml NaOH is required to reach the half stoichiometric point. (d) At the half stoichiometric point, (e) 5.0 ml; see part (c) (f) The ph is that of 0.050 M NaCHCO. ph = p = 4.75 a SM-
Concentration initial 0.050 0 0 change equilibrium 0.050 1 (mol L ) HO(l) CHCO (aq) CHCOOH(aq) OH (aq) 1.00 0.050 0.050 14 w b = = = 5.6 = 5 a 1.8 =.8 11 = = 6 1 5. mol L [OH ] poh = 5.8, ph = 14.00 5.8 = 8.7 11. (a) b Concentration [NH 4 ][OH ] = = 1.8 [NH ] 5 ol L 1 (m ) H O(l) NH (aq) NH (aq) OH (aq) initial 0.15 0 0 change equilibrium 0.15 5 = 1.8 0.15 0.15 [OH ] = = 1.6 mol L 1 poh =.80, initial ph = 14.00.80 = 11.0 (b) initial moles of NH = (0.0150 L)(0.15 mol L ) =. mol NH 1 1 moles of HCl = (0.0150 L)(0. mol L ) = 1.5 mol HCl (. 1.5 ) mol NH 0.000 L 1.5 mol HCl 0.000 L 4 Then consider the equilibrium: = 1.7 mol L NH = 1 1 5.0 mol L HCl 5.0 mol L NH4 SM-4
Concentration 1 (mol L ) H O(l) NH (aq) NH (aq) OH (aq) initial 4.7 5.0 0 change equilibrium.7 5.0 b [NH 4 ][OH ] = = 1.8 [NH ] 5 ()(5.0 ) ; assume that and are negligible = (.7 ) 6 1 [OH ] = = 9.7 mol L and poh = 5.01 Therefore, ph = 14.00 5.01 = 8.99 (c) At the stoichiometric point, moles of NH = moles of HCl 1 (0.15 mol L NH )(0.0150 L) 1 volume HCl added = = 0.05 L HCl 0. mol L HCl Therefore, halfway to the stoichiometric point, volume HCl added =.5/ = 11.5 ml (d) At half stoichiometric point, Therefore, ph = 14.00 4.75 = 9.5 poh = pb and poh = 4.75 (e).5 ml; see part (c) (f) 4 L NH (aq) H O(l) H O (aq) NH (aq) The initial moles of NH have now been converted to moles of NH 4 in a (15.5 = 7.5) ml volume:.5 mol 1 4 = = [NH ] 0.060 mol L 0.075 L 1.00 = = = 5.6 14 w a 5 b 1.8 Concentration ol L 1 (m ) NH (aq) H O(l) H O (aq) NH (aq) 4 initial 0.060 0 0 change SM-5
equilibrium 0.060 a = 5.6 = 0.060 0.060 = [H O ] = 5.8 mol L 6 1 6 ph = log(5.8 ) = 5.4 11.5 At the stoichiometric point, the volume of solution will have doubled; therefore, the concentration of CHCO will be 0. M. The equilibrium is Concentration 1 (mol L ) CH CO (aq) H O(l) HCH CO (aq) OH (aq) initial 0. 0 0 change equilibrium 0. 1.00 = = = 5.6 14 w b 5 a 1.8 [HCH CO ][OH ] b = = = [CHCO ] 0. 0. = = 6 1 7.5 mol L [OH ] 5.6 6 poh log(7.5 ) 5.1, ph 14.00 5.1 8.88 = = = = From Table 11., we see that this ph value lies within the range for phenolphthalein, so that indicator would be suitable; the others would not. 11.7 Eercise 11.1: thymol blue or phenolphthalein; Eercise 11.: methyl red or bromocresol green. 11.9 (a) To reach the first stoichiometric point, we must add enough solution to neutralize one H on the H AsO. 4 To do this, we will require 1 0.0750 L 0.17 mol L 0.0 mol OH. = The volume of base required will be given by the number of moles of base required, divided by SM-6
the concentration of base solution: 0.0750 L 0.17 mol L 1 0.75 mol L 1 = 0.074 L or 7.4 ml (b) and (c) To reach the second stoichiometric point will require double the amount calculated in (a), or 74.8 ml, and the third stoichiometric point will be reached with three times the amount added in (a), or 11 ml. 11.41 (a) The base HPO is the fully deprotonated form of phohorous acid HPO (the remaining H attached to P is not acidic). It will require an equal number of moles of HNO to react with HPO, in order to reach the first stoichiometric point (formation of HPO ). The value will be given by 0.055 L 0.158 mol L 1 0.55 mol L 1 = 0.00 L or.0 ml (b) To reach the second stoichiometric point would require double the amount of solution calculated in (a), or 44.0 ml. 11.4 (a) This value is calculated as described in Eample.1. First we calculate the molarity of the starting phohorous acid solution: 0.1 g = 1 81.99 g mol 1 0.0500 L 0.098 mol L. We then use the first acid dissociation of phohorous acid as the dominant equilibrium. The is 1.0. Let H P a1 Concentration 1 (mol L ) represent the fully-protonated phohorus acid. HP(aq) HO(l) HP(aq) HO(aq) initial 0.098 0 0 change final 0.098 SM-7
[H O ][HP ] a = = [HP] 1.0 1.0 = = 0.098 0.098 If we assume << 0.098, then the equation becomes = (1.0 )(0.098) =.98 = 1.7 4 Because this value is more than % of 0.0400, the full quadratic solution should be undertaken. The equation is = (1.0 )(0.098 ) or (1.0 ) (.98 ) = 0 4 Using the quadratic formula, we obtain = 0.01. ph = 1.89 (b) First, carry out the reaction between phohorous acid and the strong base to completion: H P(aq) OH (aq) HP (aq) H O(l) moles of moles of 1 HP (0.098 mol L = )(0.0500 L) = 1.49 mol 1 4 OH = (0.00500 L)(0.175 mol L ) = 8.76 mol 8.75 mol OH will react completely with 1.49 mol HP 4 to give 8.75 mol HP with 6.1 moles of HP remaining. 4 6.1 mol [HP] = = 0.09 mol L 0.0565 L 8.75 mol 1 [HP ] = = 0.0155 mol L 0.0565 L Concentration 1 (mol L ) 1 H P(aq) H O(l) HP (aq) H O (aq) initial 0.09 0.0155 0 change final 0.09 0.0155 SM-8
The calculation is performed as in part (a): (0.0155 ) 1.0 = 0.09 = 1.08 ph = 1.96 (c) moles of HP = 1.49 moles of OH = 8.75 8.75 = 1.75 4 4 Following the reaction between H P and OH, 0 mol of H P remain and 4.6 mol OH remain. [OH ] = 4.7 M, [H O ] =.9 M. 1 1.49 mol 1.49 mol of HP remain, [HP ] = =.48 M Concentration 0.060 L 1 (mol L ) initial change final HP H O H P H O.48.48.9 1 0.9 1 The calculation is performed as in part (a): 0 = (.48 )(.9 1 ) 16 Using the quadratic equation, we find = 5.6, 1 1 [HO] =.9, and ph = log(.9 ) = 11.6 11.45 (a) The reaction of the base Na HPO 4 with the strong acid will be taken to completion first: HPO (aq) H O (aq) H PO H O(l) 4 4 Initially, moles of HPO = moles of H O = 0.0500 L 0.75 mol L = 0.018 mol 1 4 SM-9
Because this reaction proceeds with no ecess base or acid, we are dealing with a solution that can be viewed as being composed of The problem then becomes one of estimating the ph of this solution, which can be done from the relationship ph = (p p ) 1 1 a1 a ph = (.1 7.1) = 4.66 HPO 4. (b) This reaction proceeds as in (a), but there is more strong acid available, so the ecess acid will react with HPO 4 to produce Addition of the first 50.0 ml of acid solution will convert all the into HPO 4. HPO 4 : HPO. 4 HPO 4 The additional 5.0 ml of the strong acid will react with H PO (aq) H O (aq) H PO (aq) H O(l) 4 4 0.018 mol HPO 4 will react with 0.006 88 mol HO to give 0.0069 mol HPO4 with 0.069 mol HPO 4 be in ecess. The concentrations will 0.0069 mol [HPO 4] = [HPO 4 ] = = 0.055. The appropriate relationship 0.15 L to use is then Concentration 1 (mol L ) HPO 4(aq) HO(l) HPO 4 (aq) HO (a q) initial 0.055 0.055 0 change final 0.055 0.055 a1 [HPO 4 ][HO ] = [H PO ] 4 Because the equilibrium constant is not small compared to 0.055, the full quadratic solution must be calculated: = 0.055 7.6 (0.055 ) SM-0
0.06 4. = 0 4 = 1.6 ph = log(1.6 ) =.80 (c) The reaction of Na HPO 4 with strong acid goes only halfway to completion. 0.75 mol of HPO will react with 4 1 (0.05 L 0.75 mol L ) 6.9 = mol HCl to produce 6.9 mol HPO and leave 6.9 HPO 4 unreacted. 4 6.9 mol 0.075 L = 0.09 mol L Concentration 1 (mol L ) 1 HPO 4 (aq) HO(l) HPO 4 (aq) HO(aq) intial 0.09 0.09 0 change final 0.09 0.09 a [HPO ][H O ] = 6. = 8 4 [HPO 4 ] [0.09 ][HO ] = 6. [0.09 ] assuming << than 0.09 = [H O ] = 6. 8 8 ph = log(6. ) = 7.1 8 11.47 (a) The solubility equilibrium is AgBr(s) Ag (aq) Br (aq). 7 1 [Ag ] = [Br ] = 8.8 mol L = S = solubility 7 7 1 = [Ag ][Br ] = (8.8 )(8.8 ) = 7.7 (b) The solubility equilibrium is PbCrO (s) Pb (aq) CrO (aq) 4 4 [Pb ] = 1. mol L = S, [CrO ] = 1. mol L = S 7 1 7 1 4 7 7 = [Pb ][CrO ] = (1. )(1. ) = 1.7 4 14 SM-1
(c) The solubility equilibrium is Ba(OH) (s) Ba (aq) OH (aq) 1 [Ba ] = 0.11 mol L = S, 1 [OH ] = 0. mol L = S = [Ba ][OH ] = (0.11)(0.) = 5. (d) The solubility equilibrium is 1 [Mg ] = 1. mol L = S, MgF (s) Mg (aq) F (aq) 1 [F ] =.4 mol L = S = [Mg ][F ] = (1. )(.4 ) = 6.9 9 11.49 (a) Equilibrium equation: AgS(s) Ag (aq) S (aq) = [Ag ] [S ] = ( S) ( S) = 4S = 6. 51 S = 1. mol L 17 1 (b) Equilibrium equation: CuS(s) Cu (aq) S (aq) = [Cu ][S ] = S S = S = 1. 6 S = 1.1 mol L 18 1 (c) Equilibrium equation: CaCO (s) Ca (aq) CO (aq) = [Ca ][CO ] = S S = S = 8.7 S = 9. mol L 5 1 9 11.51 Tl CrO (s) Tl (aq) CrO (aq) 4 4 [CrO ] = S = 6. mol L 5 1 4 [Tl ] = S = (6. ) mol L 5 1 = [Tl ] [CrO ] = ( S) ( S) 4 = [(6. )] (6. ) = 1.0 5 5 1 1 11.5 (a) Concentration (mol L ) A gcl(s) Ag (aq) Cl (aq) initial 0 0.0 change S S equilibrium S S 0.0 SM-
= [Ag ][Cl ] = ( S) ( S 0.0) = 1.6 Assume S in (S 0.0) is negligible, so 0.0 S = 1.6 S 1 = 8.0 mol L = [Ag ] = molar solubility of AgCl in 0.15 M NaCl 1 (b) Concentration (mol L ) Hg Cl (s) Hg (aq) Cl (aq) initial 0 0.150 change S S equilibrium S 0.150 S = [Hg ][Cl ] = ( S) (S 0.150) = 1. 18 Assume S in (S 0.150) is negligible, so 0.150S = 1. 18 18 1 S = 8.7 mol L = [Hg ] = molar solubility of in 0. 5 Μ NaCl HgCl 1 (c) Concentration ( mol L ) PbCl (s) Pb (aq) Cl (aq) initial 0 0.05 = 0.05 change S S equilibrium S S 0.05 S S 5 = [Pb ][Cl ] = ( 0.05) = 1.6 S may not be negligible relative to 0.05, so the full cubic form may be required. We do it both ways: 5 For S 0.0 S (0.005 S) (1.6 ) = 0, the solution by standard methods is S = 1 4.6 mol L. If S had been neglected, the answer would have been the same, 4.6, to within two significant figures. (d) Concentration 1 (mol L ) Fe(OH) (s) Fe (aq) OH (a q) initial.5 0 change S S equilibrium.5 S S SM-
S S 14 = [Fe ][OH ] = (.5 ) ( ) = 1.6 Assume S in ( S.5 ) is negligible, so 4 S (.5 ) S = 1.6 1 = 1.6 1. S = 1. mol L = molar solubility of Fe(OH) in.5 FeCl 6 1 M 11.55 (a) Ag (aq) Cl (aq) AgCl(s) Concentration 1 (mol L ) Ag Cl initial 0 change 0 1.0 5 equilibrium 1.0 5 = = = 5 [Ag ][Cl ] 1.6 ( )(1.0 ) = [Ag ] = 1.6 mol L 5 1 (b) mass AgNO 5 1.6 mol AgNO 169.88 g AgNO 1µ g = (0.0 L) 6 1L 1molAgNO g.7 µ g AgNO = 11.57 (a) Ni (aq) OH (aq) Ni(OH) (s) Concentration 1 (mol L ) Ni OH initial 0.060 0 change 0 equilibrium 0.060 = [Ni ][OH ] = 6.5 = (0.060)( ) 18 [OH ] = = 1.0 mol L 8 1 8 poh = log(1.0 ) = 8.00, ph = 14.00 8.00 = 6.00 (b) A similar set up for [Ni ] = 0.00 Μ SM-4
gives = 1.5 8 8 poh = log(1.5 ) = 7.8 ph = 14.00 7.8 = 6.18 11.59 1mL = = = 0 drops 5 1 drop 0.05 ml 0.05 L 5 L 5 1 7 7 and (5 L)(0.0 mol L ) = 5 mol NaCl = 5 mol Cl (a) Q Ag (aq) Cl (aq) AgCl(s), [Ag ][Cl ] = 1 7 (0.0 L)(0.0040 mol L ) 5 mol = = 0.0 L 0.0 L 7 Will precipitate, because Q ( ) > (1.6 ) 7 (b) Q Pb (aq) Cl (aq) PbCl (s), [Pb ][Cl ] = 1 7 (0.00 L)(0.0040 mol L ) 5 mol = = 1 0.0 L 0.0 L 11 Will not precipitate, because Q (1 ) < (1.6 ) 11 5 11.61 (a) [Ni(OH)] < [Mg(OH)] < [Ca(OH)] This is the order for the solubility products of these hydroides. Thus, the order of precipitation is (first to last): Ni(OH), Mg(OH), Ca(OH). (b) [Ni(OH) ] = 6.5 = [Ni ][OH ] 18 6.5 [OH ] = = 6.5 0.00 8 [OH ] = 8.1 18 15 poh = log[oh ] = 7.09 ph 7 [Mg(OH) ] = 1.1 = [Mg ][OH ] 11 1.1 [OH ] = = 1.0 0.00 11 4 4 poh log(1.0 ) 4.00 ph 14.00 4.00.00, ph = = = = = 6 [Ca(OH) ] 5.5 = [Ca ][OH ] SM-5
6 5.5 [OH ] = = 7.4 0.00 poh = log(7.4 ) = 1.1 ph = 14.00 1.1 = 1.87, ph 1 11.6 The values are MgF BaF MgCO BaCO 6.4 9 1.7 6 1.0 5 8.1 9 The difference in these numbers suggests that there is a greater solubility difference between the carbonates, and thus this anion should give a better separation. Because different numbers of ions are involved, it is instructive to convert the reaction is values into molar solubility. For the fluorides the MF (s) M (aq) F (aq) Change = ( ) Solving this for MgF gives 0.001 M and for BaF gives 0.0075 M. For the carbonates: MCO (s) M (aq) CO (aq) = Solving this for MgCO gives 0.00 M and for BaCO gives 9.0 5 M. Clearly, the solubility difference is greatest between the two carbonates, and CO is the better choice of anion. 7 11.65 Cu(IO ) ( = 1.4 ) is more soluble than 1 Pb(IO ) ( =.6 ) so Cu will remain in solution until SM-6
essentially all the Pb(IO ) has precipitated. Thus, we epect very little Pb Cu(IO ) to be the left in solution by the time we reach the point at which begins to precipitate. The concentration of IO at which Cu begins to precipitate will be 7 = [Cu ][IO ] = 1.4 = [0.00][IO ] given by 1 [IO ] = 0.01 mol L 1 The concentration of Pb in solution when the [IO ] = 0.01 mol L is given by = [Pb ][IO ] =.6 = [Pb ][0.01] 1 [Pb ] = 1.8 mol L 9 1 11.67 (a) ph = 7.0; [OH ] = 1.0 mol L 7 1 Al (aq) OH (aq) Al(OH) (s) [Al ][OH ] = = 1.0 S ( ) = 1.0 7 1.0 S = = = 1 1 1 1 1.0 mol L [Al ] = molar solubility of Al(OH) at ph = 7.0 (b) ph = 4.5; poh = 9.5; [OH ] =. mol L 1 [Al ][OH ] = = 1.0 S (. ) = 1.0 1.0 S = = = 9. 5 1.1 mol L [Al ] = molar solubility of Al(OH) at ph = 4.5 (c) ph = 7.0; [OH ] = 1.0 mol L 7 1 Zn (aq) OH (aq) Zn(OH) (s) [Zn ][OH ] = =.0 17 SM-7
S (1.0 ) =.0 7 17 17.0 S = = = 14 1.0 1.0 mol L [Zn ] = molar solubility of Zn(OH) at ph = 7.0 (d) ph = 6.0; poh = 8.0; [OH ] = 1.0 mol L 8 1 17 [Zn ][OH ] =.0 = S (1.0 ) =.0 8 17 17.0 S = = = = 16 1.0 1 1.0 0.0 mol L [Zn ] = molar solubility of Zn(OH) at ph = 6.0 11.69 CaF (s) Ca (aq) F (aq) = 4.0 11 F (aq) H O(l) HF(aq) OH (aq) 11 b (F ) =.9 (a) Multiply the second equilibrium equation by and add to the first equilibrium: CaF(s) HO(l) Ca (aq) HF(aq) OH (aq) = = (4.0 )(.9 ) =.4 w (b) The calculation of 11 11 b is complicated by the fact that the anion of the salt is part of a weak base-acid pair. If we wish to solve the equation algebraically, then we need to consider which equilibrium is the dominant 7 one at ph = 7.0, for which [H O ] = 1. To determine whether F either), consider the base hydrolysis reaction: F (aq) H O(l) HF(aq) OH (aq) or HF is the dominant ecies at this ph (if b =.9 11 b [HF][OH ] = [F ].9 11 = 7 [HF][1 ] [F ] 11 [HF].9 = = 7 [F ] 1 4 SM-8
Given that the ratio of HF to F 4 is on the order of to 1, the F ecies is still dominant. The appropriate equation to use is thus the original one for the of CaF (s) : CaF (s) Ca (aq) F (aq) = 4.0 11 = [Ca ][F ] 4.0 = ( ) = 4 =. 11 4 molar solubility =. mol L 4 1 (c) At 1 11 1 ph =.0, [HO ] = 1 mol L and [OH ] = 1 mol L. Under these conditions b [HF][OH ] = [F ].9 11 = 11 [HF][1 ] [F ] 11 [HF].9 = = 11 [F ] 1 [HF] = [F ] As can be seen, at ph.0 the amounts of F the protonation of F [Ca ] = [F ] [HF] and HF are comparable, so to form HF cannot be ignored. The relation stoichiometry of the dissolution equilibrium. [Ca ] = [F ] [F ] [Ca ] = 4[F ] [Ca ] = [F ] Using this with ( [F ]) [F ] = 4.0 is required by the mass balance as imposed by the relationship: 11 [F ] = 4.0 11 [F ] = 4 4 4 1 [Ca ] = ()( ) = 4 mol L SM-9
The solubility is about double that at ph = 7.0. 11.71 AgBr(s) L Ag (aq) Br (aq) = 7.7 1 Ag (aq) CN (aq) Ag(CN) (aq) = 5.6 f AgBr(s) CN (aq) Ag(CN) (aq) Br (aq) = 4. 8 4 Hence, Concentration [Ag(CN) ][Br ] = = 4. [CN ] 1 (mol L ) AgBr(s) CN (aq) Ag(CN) (aq) Br (aq) initial 0. 0 0 change S S S equilibrium 0. S S S 4 (.1 ) (4. S) 1.0 mol L molar solubility of AgBr 4 [Ag(CN) ][Br ] S = = 4. [CN ] (0. S) S 0. S 1.04S =.1 = 4. =.1 S = S = = 4 11.7 The two salts can be distinguished by their solubility in NH. The equilibria that are pertinent are AgCl(s) NH (aq) Ag(NH ) (aq) Cl (aq) = f =.6 AgI(s) NH (aq) Ag(NH ) (aq) I (aq) = f =.4 9 For eample, let s consider the solubility of these two salts in solution: 1.00 M NH For AgCl [Ag(NH ) ][Cl ] = =.6 [NH ] SM-40
Concentration 1 (mol L ) AgCl(s) HN (aq) Ag(NH ) (aq) Cl (aq) initial 1.00 0 0 change final 1.00 [Ag(NH ) ][Cl ] = =.6 [NH ] [ ][ ] [1.00 ] [1.00 ] = =.6 0.051 = 1.00 = 0.046 0.046 mol AgCl will dissolve in 1.00 L of aqueous solution. The molar mass of AgCl is 1 14. g mol ; this correonds to 0.046 1 1 mol L 1 14. g mol = 6.5 g L. For AgI, the same calculation gives 5 1 = 4.9 mol L. The molar mass of AgI is 1 4.77 g mol, giving a solubility of 5 1 1 1 4.9 mol L 4.77 g mol = 0.0 g L. Thus, we could treat a 0. g sample of the compound with 0.0 ml of 1.00 M NH. The AgCl would all dissolve, whereas practically none of the AgI would. Note: AgI is also slightly yellow in color, whereas AgCl is white, so an initial distinction could be made based upon the color of the sample. 11.75 In order to use qualitative analyses, the sample must first be dissolved. This can be accomplished by digesting the sample with concentrated HNO and then diluting the resulting solution. HCl or HSO 4 could not be used, because some of the metal compounds formed would be insoluble, whereas all of the nitrates would dissolve. Once the sample is SM-41
dissolved and diluted, an aqueous solution containing chloride ions can be introduced. This should precipitate the Ag as AgCl but would leave the bismuth and nickel in solution, as long as the solution was acidic. The remaining solution can then be treated with will precipitate but NiS will not. Once the HS. In acidic solution, BiS has been precipitated, the ph of the solution can be raised by addition of base. Once this is done, NiS should precipitate. BiS 11.77 The suggested reaction is: C H CH CH(CH )NH H O C H CH CH(CH )NH H O, 6 5 6 5 p a =.89 a 11 [C6H5CHCH(CH )NH ][HO ] = 1.88 = [C H CH CH(CH )NH ] 6 5 1.7 Given a ph of 1.7, [HO ] 0.00. Substituting this value into the equation above we obtain the ratio: 11 [C6H5CHCH(CH )NH ] 1.88 = = 6.45 [C H CH CH(CH )NH ] 0.00 6 5 = = [Base form] 11.79 The relation to use is ph = pa log. The a value for [Acid form] 5 acetic acid is 1.8 and p = 4.74. Because we are adding acid, the a ph will fall upon the addition and we want the final ph to be no more than 0.0 ph units different from the initial ph, or 4.54. [Base form] 4.54 = 4.74 log [Acid form] [Base form] 0.0 = log [Acid form] [Base form] = 0.6 [Acid form] We want the concentration of the base form to be 0.6 times that of the acid form. We do not know the initial number of moles of base or acid SM-4
forms, but we know that the two amounts were equal. Let C = initial number of moles of acetic acid and the initial number of moles of sodium acetate. The number of moles of HO to be added (in the form of HCl(aq)) is 1 0.001 00 L 6.00 mol L = 0.006 00 mol. The total final volume will be 0. L. C 0.006 00 0. L C 0.006 00 0. L = 0.6 C 0.006 00 = 0.6 C 0.006 00 C 0.006 00 = 0.6(C 0.006 00) C 0.006 00 = 0.6 C 0.00 78 0.7 C = 0.009 78 C = 0.06 The initial buffer solution must contain at least 0.06 mol acetic acid and 0.06 mol sodium acetate. The concentration of the initial solution will then be acetate. 0.06 mol 0.0 L = 0.60 M in both acetic acid and sodium 11.81 It stands to reason that if the solution desired has a higher ph than the one available, we should add conjugate base (sodium benzoate). The solution available on the shelf has a [A - ]/[HA] ratio of: [A ] [A ] ph =.95 = pa log = 4.19 log [HA] [HA] [A ] = = [HA].954.19 0.575 0.00 M If [A ] = 0.00 M as advertised on the label, then [HA] = = 0.48 M 0.575 The desired solution will have a [A - ]/[HA] ratio of: SM-4
[A ] [A ] ph = 4.5 = pa log = 4.19 log [HA] [HA] [A ] = = [HA] 4.54.19 1.45 To achieve this ratio, we must increase [A ]. Fiing [HA] at 0.48 M, the concentration of [A ] needed is: [A ] = 1.45 0.48 M = 0.50 M. Therefore, [A ] needs to increase by: 5.0 M 0.00 M = 0.0 M. To determine the number of grams of conjugate base needed: ( )( ) 0.0 M 0.0 L = 0.00 mol 1 ( )( ) 0.00 mol 144.04 g mol = 4.5 g 11.8 (a) M a = 0.0567, M b = 0.096, V a = 15.0, V b = 0.0 to 50.0 1.0 ph 6.6 1.4 0 5.0 50.0 V b (L) (b) 8.6 ml (c) 1.4 (d) Because this is a titration of a strong acid with a strong base, the ph at the equivalence point will be 7.00. 11.85 The strong acid, HCl, will protonate the HCO ion. 1 4 moles of HCl = 0.0040 L 0.070 mol L =.8 mol HCl (H ) moles of HCO = 0.0600 L 0. mol L = 6.0 mol HCO After protonation, there are 1 (6.0 0.8) mol HCO = 5.7 mol HCO SM-44
4 and.8 mol HCOOH. 5.7 mol [HCO ] = = 8.9 mol L 0.0640 L 4.8 mol 1 [HCOOH] = = 4.4 mol L 0.0640 L Concentration 1 1 (mol L ) HCOOH(aq) HO(l) HO (aq) HCO (aq) initial 4.4 0 8.9 change equilibrium 4.4 8.9 4 [HO ][HCO ] ()(8.9 ) a = 1.8 = = [HCOOH] (4.4 ) 4 (8.9 ) ( ) Assume that and are negligible; so 1.8 = 4.4 = 0. [H O ] = = 8.9 mol L 6 1 6 1 [HCOOH] = (4.4 ) (8.9 ) 4.4 mol L 6 ph = log(8.9 ) = 5.05 11.87 (a) The acidity constant may be found using: a G r RT G = 69.1 kj mol ( 96.46 kj mol ) = 7.15 kj mol a = e r -1-1 -1-1 7150 J mol 8.145 J mol 98 5-1 -1 ( )( ) = 7150 J mol = e = 1.74 (b) Given the acidity constant above, the mass of sodium acetate needed is: p a = a = 4.76 [A ] [A ] ph = pa log = 4.8 = 4.76 log [HA] [HA] -1 SM-45
[A ] = = [HA] 4.84.76 1. Given [HA] = 1.0 M, [A ] = 1. 1.0 M = 1. M. -1 ( )( )( ) and 1. M.5 L 8.04 g mol = 5 g 11.89 Let novocaine = N; N(aq) HO(l) HN (aq) OH (aq) b [HN ][OH ] = [N] p = p p = 14.00 5.05 = 8.95 a w b [N] = ph pa log [HN ] [N] log ph pa 7.4 8.95 1.55 = = = [HN ] Therefore, the ratio of the concentrations of novocaine and its conjugate 1.55 acid is [N]/[HN ] = =.8. 11.91 CHCOOH(aq) HO(l) HO (aq) CHCO (aq) a [HO ][CHCO ] = [CH COOH] [CH CO ] 0.50 = = [CH COOH] 0.150 = 4.75 0.5 = 5.7 (initial ph) ph pa log pa log (a) 1 (0.00 L)(1. mol L ) = 1. mol HCl added (a strong acid) Produces 1. mol CH COOH from CH CO after adding HCl: 1 (0.0 L)(0.150 mol L ) 1. mol [CHCOOH] = 0.1 L 0.1 L = 0.45 mol L 1 1 (0.0 L)(0.50 mol L ) 1. mol [CHCO ] = 0.1 L 0.1 L = 0.45 mol L 1 SM-46
0.45 ph = 4.75 log = 4.75 0.15 = 4.90 (after adding HCl) 0.45 (b) 1 (0.0500 L)(0.095 mol L ) 4.7 = mol NaOH added (a strong base) Produces 4.7 mol CHCO from CHCOOH after adding NaOH: 1 (0.0 L)(0.150 mol L ) 4.7 mol [CHCOOH] = 0.1500 L 0.150 L = 6.9 mol L 1 [CHCO ] = 0.150 L 0.150 L 1 (0.0 L)(0.50 mol L ) 4.7 mol =.6 mol L 1 1 1.6 ph = 4.75 log 4.75 0.7 5.47 (after adding base) = = 6.9 11.9 CaF (s) Ca (aq) F (aq) [Ca ][F ] = 4.0 11 [Ca ](5 ) = 4.0 [Ca ] = 0.016 mol L 5 11 1 The maimum concentration of [Ca ] allowed will be 1 0.016 mol L. 11.95 4 1 5 moles of Ag = (0.0 L)(1.0 mol L ) = 1.0 mol Ag moles of CO = (0.0 L)(1.0 mol L ) = 1.0 mol CO 4 1 5 Ag CO (s) Ag (aq) CO (aq) [Ag ] [CO ] = Q 5 5 1.0 1.0 0.00 0.00 1. 1 = = Q Because Q 1 1 (1. ) is less than (6. ), no precipitate will form. SM-47
11.97 In addition to the reaction correonding to the dissolution of PbF (s): PbF (s) Pb (aq) F (aq) =.7 8 The buffer will provide a source of H O (aq) ions which will allow the reaction: H O (aq) F HF(aq) H O(l) = 1/( a (HF)) = 1/(.5 4 ) =.86 These two coupled reactions give two equilibrium epressions which must be simultaneously satisfied: [HF] 6 [F ] [Pb ] = 1.7 and =.86 [HO ][F ] Given that all fluoride ions come from PbF (s) and wind up as either F (aq) or HF(aq), and that for every one mole of Pb (aq) generated two moles of F (aq) are also produced, we can write a third equation which relates the concentration of the fluoride containing ecies to the concentration of dissolved barium: [Pb ] = ½([F ] [HF]). In the end, the concentration of Pb (aq) will be equal to the solubility of PbF (s). To determine the equilibrium concentration of Pb (aq), we first determine [H O ], which is fied by the buffer system, and then use the three simultaneous equations above to solve for [Pb ] eq. The buffer determines the equilibrium concentration of H O (aq). The initial concentration of H O (aq) and NaCH CO (aq) are: -1 ( 0.055 L)( 0.15 mol L ) [HO ] i = = 0.085 M and 0. L -1 ( 0.045 L)( 0.65 mol L ) [CHCO ] i = = 0.9 M. 0. L To determine their equilibrium concentrations we solve using the familiar method: SM-48
Concentration 1 (mol L ) CHCOOH(aq) HO(l) HO (aq) CHCO (aq) initial 0 0.085 0.9 change equilibrium 0.085 0.9 a 5 [HO ][CHCO ] (0.085 )(0.9 ) = 1.8 = = [CH COOH] ( ) Rearranging this epression we obtain: = 0.0407 0.745 0. = = 6 Using the quadratic formula we find 0.0849, and [HO ] 7.1 M With this equilibrium concentration of H O (aq) we revisit the three simultaneous equations from above, namely 8 [F ] [Pb] =.7 ; 1 [HO ][F ] ( ) [Pb ] = [F ] [HF] ; and [HF].86. 6 Due to the presence of the buffer, [HO ] = 7.1 and this last equation simplifies to [HF] = [F ] =.0. Rearranging these three simultaneous equations we find: 8.7 [Pb ] =, [HF] = [F ] (.0 ), and [F ] 1 8 ( ) [Pb ] = [F ] [HF].7 [F ] ( ( )) 1 = [F ] [F ].0. Solving this epression for [F ]:.7 0.50 8 [F ] = = 4.14. SM-49
The equilibrium concentration of Pb Therefore, the solubility of PbF (s) is. M (aq) is then: 8 8.7.7 [Pb ] = = =.16 M [F ] (4.14 ) 11.99 The values from Tables 11.5 and 11.7 are 6 Cu, 1. ; Co, 5 ; 9 Cd, 4. All the salts have the same epression for, = [M ][S ], so the compound with the smallest will precipitate first, in this case CuS. 11.1 (1) first stoichiometric point: OH (aq) H SO (aq) H O(l) HSO (aq) 4 4 Then, because the volume of the solution has doubled, [HSO 4 ] = 0. M, Concentration 1 (mol L ) HSO (aq) H O(l) SO (aq) H O (aq) 4 4 initial 0. 0 0 change equilibrium 0. a = 0.01 = 0. 0.001 0.01 = 0.01 0.001 = 0 = = [H O ] = 0.09 mol L 0.01 (0.01) (4)(0.001) 1 ph = log(0.09) = 1.54 () second stoichiometric point: HSO (aq) OH (aq) SO (aq) H O(l). Because the volume of 4 4 the solution has increased by an equal amount, SM-50
Concentration 4 4 OH (aq) 1 (mol L ) SO (aq) H O(l) HSO (aq) initial 0.067 0 change 1 7 equilibrium 0.067 7 1 7 ( )(1 ) b = 8. = 0.067 (5.6 ) (8. ) = 1 14 1 7 (1 ) (5.6 ) = 0 7 14 = = [HSO ] = 1.9 mol L 7 7 14 1 (1 ) (4)(5.6 ) 7 1 4 = = 7 7 7 1 [OH ] (1.0 ) (1.9 ).9 mol L 7 poh = log(.9 ) = 6.54, ph = 14.00 6.54 = 7.46 11. (a) PbCl (s) Pb (aq) Cl (aq) 5 [Pb ][Cl ] = = 1.6 [0.0][Cl ] = 1.6 [Cl ] = 1.6 5 1 [Cl ] = 4.0 mol L ; will precipitate lead (II) ion Ag (aq) Cl (aq) AgCl(s) [Ag ][Cl ] = = 1.6 [0.0][Cl ] = 1.6 8 1 [Cl ] = 1.6 mol L ; will precipitate Ag (b) From part (a), Ag will precipitate first. (c) 1 [Ag ](4.0 ) = 1.6 0 1.6 [Ag ] 4.0 mol L 4.0 9 1 = = SM-51
4.0 (d) percentage A remaining = 0.0 9 g = unprecipitated; virtually 0% of the first cation (Ag ) 5 4.0 % is precipitated 8 11.5 The value for PbF obtained from Table 11.5 is.7. Using this value, the G G = RT ln. of the dissolution reaction can be obtained from G = 1 1 8 (8.14 J mol )(98. )ln (.7 ) G = 4.4 kj mol 1 1 From the Appendices we find that f (F, aq) = 78.79 kj mol and G = 1 f (Pb, aq) 4.4 kj mol. G G = 4.4 kj mol = G (Pb, aq) G (F, aq) G f(pbf, s) G 1 f f = 1 1 1 4.4 kj mol ( 4.4 kj mol ) ( 78.79 kj mol ) (PbF, s) f G (PbF, s) = 45.65 kj mol f 1 11.7 (a) The dissolution reaction for calcium oalate is: CaC O 4 (s) Ca (aq) C O 4 (aq) p 8.59 9 = = =.57 The concentration of each ion at equilibrium is equal to the molar solubility, s. Therefore, = = s s = s = 9 5-1 [Ca ][CO 4 ].57, and 5.07 mol L. (b) Given [Mg ] = 0.00 M and [C O - 4 ] = 0.05 M: Q = [Mg ][C O - 4 ] = (0.00)(0.05) = 7-4 Q is significantly less than, which is given as -4.07 = 8.51-5, indicating that a precipitate will form. 11.9 (a) The amount of CO present at equilibrium may be found using the equilibrium epression for the reaction of interest: SM-5
HO (aq) HCO HO() l CO( aq) [CO ] = 7.9 = [H O ][HCO ] Solving for [CO ] : 7 [CO ] = (7.9 )[H O ][HCO ] 7 Given: [H O ] = = 7.9 M, and [HCO ] = 5.5 µ mol L = 5.5 M 6.1 7-1 6 [CO ] = (7.9 )(7.9 7 7 In 1.0 L of solution there will be.5 )(5.5 ) =.5 mol L 6 18-1 18 mol of CO (aq) (b) Adding 0.65-6 mol of H O (aq) to the equilibrium system in (a) will give an initial [H O ] of 1.44-6 M. To determine the equilibrium concentration of H O (aq) we set up the familiar problem: Concentration 1 (mol L ) H O (aq) HCO (l) H O(l) CO (aq) initial 1.44-6 5.5-6.45-18 change equilibrium 1.44-6 7 = 7.9 = 5.5-6.45-18 18 6 6 ( 1.44 )( 5.5 ) 18 7.8 7.9 = 0 18.45 Rearranging to obtain a polynomial in : Using the quadratic forumla one finds: =.85 6 Giving: [HO ] = 1.4, and ph 5. = 8. The same ph as the initial solution. SM-5