BIT Chapters 17-0 1. K w = [H + ][OH ] = 9.5 10 14 [H + ] = [OH ] =.1 10 7 ph = 6.51 The slutin is neither acidic nr basic because the cncentratin f the hydrnium in equals the cncentratin f the hydride in.. The cncentratin f OH is 4.7 10 7 g L 1 The mlar cncentratin f OH is: [OH ] = 7 4.7 10 g OH 1 ml OH 1 L slutin 17.01 g OH poh = lg[oh ] = lg(.76 10 8 M) = 7.56 ph = 14 poh = 14 7.56 = 6.44 The water is acidic.. (a) C 6 H 5 OH(aq) + H O(l) C 6 H 5 O (aq) + H O + (aq) (b) K a = (c) C H O 6 5 C H OH 6 5 H O =.76 10 8 M [C 6 H 5 OH] = 0.550 M since the amunt f dissciatin is negligible [H O + ] = lg(ph) = lg(5.07) = 8.51 10 6 M [C 6 H 5 O ] = 8.51 10 6 M since all f the H O + cmes frm the dissciatin f the phenl K a = C H O 6 5 C H OH 6 5 H O pk a = lg K a = lg(1. 10 10 ) = 9.89 14 14 = 6 6 8.51 10 8.51 10 1 10 1 10 (d) K b = = 7.69 10 5 10 Ka 1. 10 pk b = lg K b = lg(7.69 10 5 ) = 4.11 0.550 = 1. 10 10 4. (a) pk b = 14 pk a = 14 11.68 =. (b) C 7 H SO + H O C 7 H SO H + OH The slutin will be basic K b = C H SO H 7 C H SO 7 OH = 4.79 10 4.79 10 = [C 7 H SO ] [C 7 H SO H] [OH ] I 0.010 C + + E 0.010 + + 0.010 will be large cmpared t 0.010 M. Therefre we need t slve it either by successive apprimatins r the quadratic equatin. = 4.9 10 = [OH ] poh =.1 ph = 11.69
BIT Chapters 17-0 5. Let Cd stand fr cdeine Cd + H O CdH + + OH K b = CdH 1.6 10 6 = Cd OH 0.0115 = 1.6 10 6 [Cd] [CdH + ] [OH ] I 0.0115 C + + E 0.0115 is large cmpared t 0.0115, therefre slve fr using either the quadratic equatin r successive apprimatins = 1.7 10 4 = [OH ] poh =.86 ph = 14 ph = 10.14 6. CH NH (aq) + H O CH NH + (aq) + OH (aq) K b = CHNH CHNH OH 7. pk b =.6 pk a f its cnjugate acid = 14 pk b = 14.6 = 10.64 8. (a) M ascrbic acid slutin =.1 g H C H O 1 ml H C H O (b) H C 6 H 6 O 6 HC 6 H 6 O 6 + H O + K a1 = HC 6 H 6 O 6 C 6 H 6 O 6 + H O + K a = 6 6 6 6 6 6 0.15 L 176.1 g H C H O HC H O 6 6 6 6 6 6 H C H O 6 6 6 H O 6 6 6 C H O HC H O 6 6 6 H O = 0.14 M H C 6 H 6 O 6 = 8.0 10 5 = 1.6 10 1 Since all f the H O + will cme frm the first equilibrium reactin, it can be calculated as fllws: 8.0 10 5 = 8.0 10 5 = [H C 6 H 6 O 6 ] [HC 6 H 6 O 6 ] [H O + ] I 0.14 C + + E 0.14 HC H O 6 6 6 H C H O 0.14 6 6 6 H O Slve fr =.7 10 = [H O + ] ph = lg[h O + ] = lg(.7 10 ) =.47 [C 6 H 6 O 6 ] = 1.6 10 1 [HC 6 H 6 O 6 ] = [H O + ] in the K a equatin, these cancel
BIT Chapters 17-0 9. ph = pk a + lg H 4.50 = 4.74 + lg H 0.4 = lg H 10 0.4 = H H = 0.575 10. (a) C H O + H + HC H O (b) HC H O + OH C H O + H O 11. K a = 1.8 10 5 = 0.10 HO 0.15 [H O + ] =.7 10 5 ph = 4.57 C H O HC H O = 1.8 10 5 H O [HC H O ] final = (0.15 0.04)M = 0.11 M [C H O ] final = (0.10 + 0.04)M = 0.14 M 0.14 HO 0.11 [H O + ] = 1.4 10 5 ph = 4.85 = 1.8 10 5 The change in ph is 0.8 ph units. 1. (a) Prpanic acid, pk a = 4.89 and its salt, sdium prpinate (b) This prblem is best slved using tw equatins and tw unknwns: ph = pk a + lg ml ml H 5.10 = 4.89 + lg 0.005 H 0.005 5.00 = 4.89 + lg H
BIT Chapters 17-0 0.1 = lg 0.11 = lg 0.005 H 0.005 H 10 0.1 = 10 0.11 = 0.005 H 0.005 H [ ] = 10 0.11 [H] 10 0.1 = 0.11 10 H 0.005 H 0.005 10 0.1 ([H] 0.005) = 10 0.11 [H] + 0.005 1.6[H] 0.008109 = 1.88[H] + 0.005 0.4[H] = 0.0111 [H] = 0.095 M [ ] = 10 0.11 (0.095 M) = 0.05056 M (c) Fr 65 ml (0.65 L) f slutin: Fr determining the mass f prpinic acid needed: g C H 6 O = 0.65 L 0.095 ml C H O 74.09 g C H O 6 6 1 L 1 ml C H O Fr determining the mass f sdium prpinate: g NaC H 5 O = 0.65 L NaC H 5 O Frm the abve calculatin: The cncentratin f prpinic acid is 0.095 M. The cncentratin f sdium prpinate is 0.05056 M. 6 0.05056 ml NaC H O 96.07 g NaC H O 5 5 1 L 1 ml NaC H O = 1.8 g C H 6 O 5 =.04 g 1. (a) Neutral (b) cidic (c) cidic (d) Basic 14. pk b = 14 pk a = 14 4.87 = 9.1 K b = H OH = 7.41 10 10 ll f the acid has been cnverted int the salt at the equivalence pint. (0.115 M)(50.00 ml) = (0.100 M)( ml) = 57.50 ml ttal vlume is 107.50 ml [ ] = 0.0549 M [H] = [OH ] =
BIT Chapters 17-0 15. K b1 = 7.41 10 10 = 0.0549 = 6.96 10 6 M = [OH ] poh = 5.01 ph = 8.799 Thyml blue r phenlphthalein wuld be gd indicatrs. K 1.0 10 w K 1.6 10 a HC H O 6 6 6 6 6 6 C H O OH 14 1 = 6. 10 [C 6 H 6 O 6 ] [HC 6 H 6 O 6 ] [OH ] I 0.050 C + + E 0.050 = 0.050 = 6. 10 is t large t make a simplifying assumptin, therefre slve fr either using the quadratic equatin r by successive apprimatins. = 0.015 = [OH ] poh = 1.8 ph = 1.18 16. Since half f the acid is neutralized the cncentratin f the acid is equal t the cncentratin f its cnjugate base, the pk a can be determined: ph = pk a + lg H.56 = pk a + lg 1 pk a = ph =.56 K a = 10 pka = 10.56 =.75 10 4 17. ph = pk a + lg H pk a = 14 pk b pk b = lg K b = lg 1.8 10 5 pk a = 14 4.74 = 9.6 9.6 = 9.6 + lg H [ ] = [H] [NH + 4 ] = 0.100 befre the additin f NaOH. In rder fr the tw cncentratins t be equal, half as much NaOH must be added: (0.100 L NH + 4 )(0.100 M NH + + 4 slutin) = 0.0100 ml NH 4 (0.0100 ml NH + 4 )(0.5) = 0.00500 ml NaOH 40.00 g NaOH 0.00500 ml NaOH = 0.00 g NaOH 1 ml NaOH 18. K sp = [g + ] [CrO 4 ] [g + ] = (6.5 10 5 M) = 1.0 10 4 M [CrO 4 ] = 6.5 10 5 M K sp = (1.0 10 4 ) (6.5 10 5 )
BIT Chapters 17-0 K sp = 1.1 10 1 19. Mg(OH) (s) Mg + (aq) + OH (aq) K sp = 5.6 10 1 = [Mg + ][OH ] 5.6 10 1 = [][] = 1.1 10 4 =. 10 4 = [OH ] poh = lg[oh ] = lg(. 10 4 ) =.66 ph = 14.66 = 10.4 0. Fe(OH) (s) Fe + (aq) + OH (aq) k sp = 4.9 10 17 = [Fe + ][OH ] ph = 10.00 poh = 14 10.00 = 4.00 [OH ] = 10 4.00 = 1 10 4 4.9 10 17 = [Fe + ][1 10 4 ] 4.9 10 9 M = [Fe + ] 9 + 4.9 10 ml Fe 1 ml Fe(OH) 89.86 g Fe(OH) g L 1 = 1 L slutin + 1 ml Fe 1 ml Fe(OH) 1. (e) = 4.40 10 7 g L 1. (a) This is a limiting reagent prblem. (b) 1 ml Pb(NO ml KI needed = ) 1.04 g Pb(NO ) 1. g ml KI 1000 ml KI slutin 1 ml Pb(NO ) 0.500 ml KI 0.0 ml KI slutin is supplied. KI is in ecess. 1 ml Pb(NO g PbI = ) 1.04 g Pb(NO ) 1. g 1 ml PbI 461.0 g PbI 1 ml Pb(NO ) 1 ml PbI = 1.45 g PbI = 1.6 ml KI slutin The cncentratin f the spectatr ins: [K + ] = (0.500 M K + )(0.0 ml slutin) 50.0 ml = 0.00 M K + [NO ] = (0.100)()(0.0 ml slutin) 50.0 ml = 0.1 M NO [I ]: Ttal mles f I = (0.500 M I )(0.000 L slutin) = 0.0100 ml I ml I used = 1.04 g Pb(NO ) 1 ml Pb(NO ) ml I + 1. g 1 ml Pb = 6. 10 ml I [I ] = (0.0100 ml I 6. 10 ml I )/(0.050 L slutin) = 0.0740 M I [Pb + ]: This will be the amunt f Pb + that is in slutin after the slid PbI reaches equilibrium PbI (s) Pb + (aq) + I (aq)
BIT Chapters 17-0 [Pb + ] [I ] I 0.0740 M C + + E 0.0740 + K sp = 9.8 10 9 = [Pb + ][I ] = ()(0.0740 + ) << 0.0740 9.8 10 9 = ()(0.0740) = 1.79 10 6 M = [Pb + ]. This is a simultaneus equilibrium prblem. CuCO (s) Cu + (aq) + CO (aq) K sp = [Cu + ][CO ] =.5 10 10 Cu + (aq) + 4NH (aq) Cu(NH ) 4 + (aq) K frm = Cu NH Cu 4 4 NH = 1.1 10 1 CuCO (s) + 4NH (aq) K verall = Cu(NH ) 4 + (aq) + CO (aq) Cu NH CO 4 NH 4 =.75 10 ll f the Cu(NH ) + 4 cmes frm the CuCO [Cu(NH ) + 1 ml CuCO 4 ] = 1.00 g CuCO 1 = 8.09 10 1.6 g CuCO 1.00 L slutin s the Cu + is used t frm Cu(NH ) + 4, the [CO ] = [Cu(NH ) + 4 ] = 8.09 10.75 10 = Cu NH CO 4 NH 4 = 8.09 10 8.09 10 4 = 0.014 M NH 0.014 ml NH ml NH = 1.00 L slutin 1 L slutin =0.014 ml NH 4. ssume the slutin is saturated with CO and therefre the cncentratin f H CO is 0.00 M H CO H + + CO K = K a1 K a = (4. 10 7 )(5.6 10 11 ) =.4 10 17 = H CO H CO First, calculate the [H + ] at which thepbco will begin t precipitate: PbCO (s) Pb + (aq) + CO (aq) K sp = 7.4 10 14 = [Pb + ][CO ] [Pb + ] = 0.010 M [CO ] = (7.4 10 14 )/(0.010) = 7.4 10 1 M [H + ] = ph =.50 1 1 17 17.4 10 H CO.4 10 0.00 1 CO 7.4 10 =.1 10 4 Nw, determine the [H + ] at which the BaCO will begin t precipitate: BaCO (s) Ba + (aq) + CO (aq) K sp =.6 10 9 = [Ba + ][CO ]
BIT Chapters 17-0 [Ba + ] = 0.010 M [CO ] = (.6 10 9 )/(0.010) =.6 10 7 M [H + ] = ph = 5.78 1 1 17 17.4 10 H CO.4 10 0.00 7 CO.6 10 = 1.7 10 6 The ph range is between.50 and 5.78, abve 5.78, BaCO will begin t precipitate. 5. The less sluble substance is PbS. We need t determine the minimum [H + ] at which NiS will precipitate. + Ni HS (0.100)(0.1) spa + + K = = = 40 (frm Table 18.) H [H ] + (0.10)(0.1) [H ] = = 0.016 40 ph = lg[h + ] = 1.80. t a ph lwer than 1.80, PbS will precipitate and NiS will nt. t larger values f ph, bth PbS and NiS will precipitate. We als need t determine the [H + ] at which PbS will start t precipitate + Pb HS (0.100)(0.1) 7 spa + + K = = = 10 (frm Table 18.) H [H ] + (0.10)(0.1) [H ] = = 18 7 10 ph = lg[h + ] =.6. ny acid in water will precipitate the PbS. The ph range is.6 t 1.80 t allw the PbS t precipitate withut the NiS. 6. The less sluble substance is SnS. We need t determine the minimum [H S] at which FeS will precipitate. + spa + Fe H S (0.10) H S K = = = 600 (frm Table 18.) H [1 10 ] (600)(1 10 ) HS = = 6 10 (0.1) The cncentratin f Sn + can nw be determined. + + Sn HS Sn 6 10 5 spa + K = = = 1 10 (frm Table 18.) H [1 10 ] + 5 (1 10 )(1 10 ) 9 Sn = = 1.7 10 (6 10 ) 7. (a) MS(s) M + (aq) + S (aq) K sp = [M + ][S ] = 4.0 10 9 MS(s) + H O M + (aq) + HS (aq) + OH (aq)
BIT Chapters 17-0 K spa = M H H S = (4.0 10 9 )(10 1 ) = 4.0 10 8 (b) (c) 0.0 = 4.0 10 8 = 6.0 10 5 M MS wuld be cnsidered an acid insluble sulfide making M a grup in. Grup ins frm insluble sulfides in base. 8. NO(g) NO (g) + N O(g) G = {1 ml G f [NO (g)] + 1 ml G f [N O(g)]} { ml G f [NO(g)]} G = {1 ml (51.84 kj/ml) + 1 ml (10.6 kj/ml)} { ml ( 86.69 kj/ml)} G = 416 kj G = RTlnK P 4.16 10 5 J = (8.14 J ml 1 K 1 )(98 K)(ln K P ) ln K P = 168 K P = 1. 10 7 K P = K c (RT) ng 1. 10 7 = K c [(0.081 L atm ml 1 K 1 )(98)] 1 K c =.9 10 7 9. CH 4 (g) + Cl (g) CH Cl(g) + HCl(g) f f ΔG = { ΔG [CH Cl(g)] + ΔG [HCl(g)]} { ΔG [CH 4 (g)] + ΔG = {1 ml ( 58.6 kj/ml) + 1 ml ( 95.7 kj/ml)} {1 ml ( 50.79 kj/ml) + 1 ml (0 kj/ml)} ΔG = 10.08 kj = 1.0 10 5 J f ΔG f [Cl (g)]} G = RTlnK P 1.0 10 5 J = (8.14 J ml 1 K 1 )(47 K)(ln K P ) ln K P = 6.19 K P =.4 10 11 K P = K c (RT) ng.4 10 11 = K c [(0.081 L atm ml 1 K 1 )(47)] 0 K c =.4 10 11 CH Cl HCl Kc.4 10 CH Cl 11 4 1 ml 4.56 g 16.04 g CH4 0.14M.00 L
BIT Chapters 17-0 1 ml 8.67 g 70.91 g Cl 0.0611M.00 L Sme CH 4 is lst t frm CH Cl and sme Cl is lst t frm HCl. [CH 4 ] [Cl ] [CH Cl] [HCl] I 0.14 0.0611 C + + E 0.14 0.0611 + + Kc.4 10 0.14 0.0611 11 The equilibrium lies s far t the right that yu d nt need t slve the equatin. Cl is the limiting reagent and thus will be cmpletely cnsumued in the reactin. t equilibrium the fllwing cncentratins eist: [Cl ] = 0 ( actually it is slightly greater than zer but etremely small) [CH 4 ] = (.0 L 0.14 M.0 L 0.0611 M)/.0 L = 0.0809 M [CH Cl] = [ HCl] = 0.0611 M 0. First calculate the atmizatin energy fr the frmatin f C H 6, and the atmizatin energy fr the C H and H. The difference is the energy required fr the reactin. Then calculate the amunt f heat required fr 5.0 g f C H 6. atmizatin energy 1 = (6 ml C H B.E.) + (1 ml C C B.E.) = (6 ml 41 kj ml 1 ) + (1 ml 48 kj ml 1 ) = 80 kj atmizatin energy = ( ml C H B.E.) + (1 ml C C B.E.) + (1 ml H H B.E.) = ( ml 41 kj ml 1 ) + (1 ml 960 kj ml 1 ) + (1 ml 46 kj ml 1 ) = 0 kj atmizatin energy 1 atmizatin energy = 80 kj 0 kj = 600 kj 600 kj are absrbed. kj fr 5.0 g = 5.0 g 1 ml CH6 600 kj 6.04 g C H 1 ml C H 6 6 = 576 kj 1. (a) mle e = 0.0 min 60 s 1.00 C 1 ml e 4 1 min 1 s 9.65 10 C = 1.4 10 ml e mle OH = 1.4 10 ml e 1 ml e 1 ml OH [OH 1.4 10 ml OH ] = 0.50 L NaCl slutin poh = lg[oh ] = 1.0 ph = 14 poh ph = 14 1.0 = 1.697 = 0.0497 M OH = 1.4 10 ml OH
BIT Chapters 17-0 (b) mle e 60 s 5.00 C 1 ml e = 10.0 min 4 1 min 1 s 9.65 10 C =.11 10 ml e 1 ml H mle H =.11 10 ml e ml e = 1.56 10 ml H ml H = 1.56 10 1 1 ml H 0.081 L atm ml K 7 K 1000 ml 1 atm 1 L = 48 ml. (a) Eternal circuit ( ) electrn flw (+) Fe Salt Bridge Cu Fe + (aq) nde Cu + (aq) Cathde (b) (c) (d) (e) Fe(s) + Cu + (aq) Cu(s) + Fe + (aq) Fe Fe + Cu + Cu E cell = E substance reduced E substance idized Ecell E + E Cu Fe E cell = 0.4 V ( 0.44 V) = 0.78 V e = 50.0 h 600 s 0.10 C 1 ml e 1 h s 96,500 C = 1.87 10 1 ml e 1 ml Fe ml Fe + = 1.87 10 1 ml e ml e = 9.5 10 ml Fe + The change in cncentratin f Fe + will be + 9.5 10 ml Fe = +0.95 M Fe + 0.100 L slutin The final cncentratin f Fe + will be: 1.00 M + 0.95 M Fe + = 1.94 M Fe + The change in cncentratin f Cu + will be + 9.5 10 ml Cu = 0.95 M Cu + 0.100 L slutin The final cncentratin f Cu + will be: 1.00 M 0.95 M Cu + = 0.065 M Cu +
BIT Chapters 17-0 + RT Fe cell cell E = E ln nf Cu -1-1 (8.14 J ml K )(98 K) E cell =0.78 V ln -1 (96,500 C ml ) = 0.78 V 0.0184(.96) E cell = 0.76 V 1.94 0.065. (a) gcl(s) + Ni(s) g(s) + Cl (aq) + Ni + (aq) (b) E cell = E reductin E idatin = 0. V ( 0.57 V) = 0.479 V 4. E cell 0.80 V 0.4 V 0.46 V + RT Cu cell cell E = E ln nf g -1-1 (8.14 J ml K )(98 K) 0.00 E =0.46 V ln (96,500 C ml ) 0.100 cell -1 = 0.46 V 0.0184(.996) V = 0.4 V cell ptential drp f 10% wuld be 0.04 V s the new ptential wuld be 0.78 V. -1-1 (8.14 J ml K )(98 K) 0.00 0.78 = 0.46 V ln (96,500 C ml ) 0.100 0.08 V = 0.0184 V ln 0.00 0.100 6.86 0.00 e 59.65 0.100-1 = 0.099 M This is the cncentratin increase that wuld ccur during the time perid that the cell ptential decreased by 10 %. Each cell has a vlume f 15 ml. ml Cu + = 0.099 M 0.15 L = 4.99 10 - ml Cu +
BIT Chapters 17-0 t = q/i t = 4.99 10 ml Cu 0.10 ml e 96,500 C ml Cu 1 ml e 966 s r.67 hr Fr a 15 ml cell, the cpper electrde wuld lse: 6.55 g 0.00499 ml Cu 0.17 g 1 ml Cu The silver electrde wuld gain: 107.87 g 0.00499 ml g 1.08 g 1 ml Cu The ttal mass f the cell remains unchanged accrding t the law f cnservatin f mass-energy.