Chemistry 192 Problem Set 6 Spring, 2018 Solutions

1 Chemistry 192 Problem Set 6 spring , 2018 Solutions1. The solubility product of Al(OH)3is 10 15. Calculate the concentrations ofaluminum ions, hydroxide ions and hydronium ions in a saturated aqueous solution ofaluminum :Al(OH)3(s) Al3+(aq)+ 3OH (aq)Ksp= [Al3+][OH ]3= 10 15[Al3+][OH ]initial0 M0 Mchanges3sequilibriumsM3sM27s4= 10 15s= [Al3+] = 10 5M[OH ] = 3s= 10 4M[H3O+] = 10 10 4= 10 112. The solubility product of Ag2S is 10 49. Calculate the molar solubility of silversulfide in :Ag2S(s) 2Ag+(aq)+ S2 (aq)[Ag+][S2 ]initial0 M0 Mchange2ssequilibrium2sMsM1(2s)2s= 4s3= 10 49s= molar solubility = 10 17M3. The solubility of Ag3 AsO4in water is 10 4g mL 1. Calculate the solubilityproduct of silver :Ag3 AsO4(s) 3Ag+(aq)+ AsO3 4(aq)moles dissolved of silver arsenate = 10 4g463 g mol 1= 10 6mol[AsO3 4] = 10 L= 10 3M[Ag+] = 3[AsO3 4] = 10 3 MKsp= ( 10 3)3( 10 3) = 10 104.

2 The solubility product of Ag2CO3isKsp= 10 12. Calculate the molar solubilityof silver carbonate in a) water and b) a M aqueous AgNO3solution (silver nitrateis 100% ionized).Answer:Ag2CO3(s) 2Ag+(aq)+ CO2 3(aq)(a)[Ag+][CO2 3]initial0 M0 Mchange2ssequilibrium2sMsM[Ag+]2[CO2 3] = 4s3= 10 12s= solubility = [CO2 3] = 10 4M(b)[Ag+][CO2 3] M0 Mchange2ssequilibrium( + 2s) MsMs( + 2s)2 s( )2= 10 12solubility =s= 10 10M25. The solubility product of AgCl isKsp= 10 10. Calculate the weight of silvernitrate that must be added to 10. mL of a M sodium chloride solution to initiatethe silver chloride precipitation :[Ag+][Cl ] = 10 10[Ag+] = 10 10 9M( 10 9mol L 1)( L) = 10 11mol( 10 11mol)( mol 1) = 10 9g6. The solubility product of Ag2SO4is 10 5. A laboratory student mixes 10. mL ofa M silver nitrate solution with 10. mL of a M sodium sulfate (Na2SO4)solution. Calculate a suitable reaction quotient to determine if silver sulfate precipitateshould form when the two Solutions are :Ag2SO4(s) 2Ag+(aq)+ SO2 4(aq)[Ag+] = M[SO2 4] = MQ= [Ag+]2[SO2 4] = ( )2( ) = 10 7< KspNo The solubility product of Cu(OH)2is 10 16.

3 Calculate(a) the pH of a saturated solution of copper (II) hydroxide in water;Answer:Cu(OH)2(s) Cu2+(aq)+ 2HO (aq)Ksp= [Cu2+][OH ]2[Cu2+][OH ]initial0 M0 Mchanges2sequilibriumsM2sM4s3= 10 16s= 10 6[OH ] = 2s= 10 6pOH = log10( 10 6) = = pOH = (b) the molar solubility of copper (II) hydroxide in a solution having pH= ;Answer:[OH ] = 10 10 1= 10 133[Cu2+] = 10 16( 10 13)2= completely soluble.(c) the molar solubility of copper (II) hydroxide in a solution having pH= :[OH ] = M[Cu2+] = 1) 16( )2= 10 14M8. Problem 54, page 814 :Cu(NH3)2+4(aq) Cu2+(aq)+ 4NH3(aq)K=1Kf= 1013= 10 14[Cu(NH3)2+2][Cu2+][NH3] M0 = Mchange-xx4xequilibrium( x) MxM( + 4x) M[Cu2+][NH3]4[Cu(NH3)2+4]=x( +x) x x( ) 10 14x= [Cu2+] = 10 18M9. Problem 56, page 814 :As in Problem 8Cu(NH3)2+4(aq) Cu2+(aq)+ 4NH3(aq)K=1Kf= 1013= 10 14[Cu2+][NH3]4[Cu(NH3)2+4]=[Cu2+]( ) 10 14[Cu2+] = 10 11 MNH3(aq)+ H2O(`) NH+4(aq)OH (aq)[NH+4][OH ][NH3]= 10 54[NH3][NH+4][OH ] M0 Mchange-xxxequilibrium( x) M( +x) MxM( +x) x x= 10 5= [OH ]Cu(OH)2(s) Cu2+(aq)+ 2OH (aq)Q= [Cu2+][OH ]2= ( 10 11)( 10 5)2= 10 21< KspNo Aqueous silver ions form a coordination complex with thiosulfate anions according tothe reactionAg+(aq)+ 2S2O2 3(aq) Ag(S2O3)3 2(aq),where the formation equilibrium constant for the complex isKf= 1013.

4 Giventhe solubility product of silver iodide, AgI, isKsp= 10 17, calculate the molarsolubility of silver iodide in a solution that is M in :AgI(s) Ag+(aq)+ I (aq)Ksp= 10 17Ag+(aq)+ 2S2O2 3(aq) Ag(S2O3)3 2(aq)Kf= 1013 AgI(s)+ 2S2O2 3(aq) Ag(S2O3)3 2(aq)+ I (aq)K=KspKf= 10 3[S2O2 3][Ag(S2O3)3 2][I ] M0 M0 Mchange-2yyyequilibrium( 2y) MyMyMy2( 2y)2= 10 2y= 10 3y= 10 3M511. Zinc ions form a complex in cyanide Solutions according to the reactionZn2+(aq)+ 4CN (aq) Zn(CN)2 4(aq)with a formation constantKf= 1018. It is found that the solubility of solid zincselenide (ZnSe) in a M cyanide solution is 10 5M. Calculate the solubilityof zinc selenide in :ZnSe(s)+ 4CN (aq) Zn(CN)2 4(aq)+ Se2 (aq)[CN ][Zn(CN)2 4][Se2 ] M0 M0 Mchange-4( 10 5) 10 10 10 10 5MK=( 10 5)2( )4= 10 5=KspKf= 1018 KspKsp= 10 23 ZnSe(s) Zn2+(aq)+ Se2 (aq)[Zn2+][Se2 ]initial0 M0 MchangesMsMequilibriumsMsMs2= 10 23s= 10 12M12.

5 The silver cyanide coordination complex, Ag(CN) 2forms by the reactionAg+(aq)+ 2CN (aq) Ag(CN) 2(aq)with associated equilibrium constantKf= 1018. Given the solubility product ofsilver iodide (AgI) isKsp= 10 17, calculate the solubility of silver iodide in asolution having [CN ]= :AgI(s)+ 2CN (aq) Ag(CN) 2(aq)+ I (aq)K=KfKsp= 1018 10 17= 1026[CN ][Ag(CN) 2][I ] M0 M0 Mchange-2sssequilibrium( 2s) 102=[Ag(CN) 2][I ][CN ]2=s2( 2s) 2s= 22. s= 11. s= M13. Zinc ions form a coordination complex in aqueous ammoniaZn2+(aq)+ 4NH3(aq) Zn(NH3)2+4(aq)having formation equilibrium constantKf= 108. Zinc sulfide (ZnS) is only spar-ingly soluble in water with solubility product constantKsp= 10 25. Calculate themolar solubility of zinc sulfide in a M aqueous ammonia solution. Approximationswork for this :ZnS(s)+ 4NH3(aq) Zn(NH3)2+4(aq)+ S2 (aq)K=KfKsp= 10 17=[Zn(NH3)2+4][S2 ][NH3]4[NH3][Zn(NH3)2+4][S2 ] M0 M0 Mchange 4sssequilibrium( 4s) 10 17=s2( 4s)4 10 4s= 10 11M14.

6 Silver ions (Ag+) react with thiocyanate ions (SCN ) in aqueous solution to form thecoordination complex Ag(SCN)3 4with associated formation constantKf= a solution that is made by adding moles of silver ions to L ofa M thiocyanate solution. Calculate the concentration of free silver ions (Ag+)when equilibrium is reached. Approximations work for this (SCN)3 4(aq) Ag+(aq)+ 4 SCN (aq)7K=1Kf= 10 11 Initially[Ag(SCN)3 4] = L= M[Ag(SCN)3 4][Ag+][SCN ] M0 M( ( ))= Mchange-yy4yequilibrium( y) MyM( + 4y) 10 11=[Ag+][SCN ]4[Ag(SCN) 4]=y( + 4y)4( y) y( ) [Ag+] = 10 12M15. The formation equilibrium constant for the cobalt ammonia complex [Co(NH3)3+6] isKf= 1033. Calculate the molar concentration of free Co3+(aq)in a solution madeby mixing L of M Co3+to L of M aqueous ammonia. Approx-imations work for this :Co(NH3)3+6(aq) Co3+(aq)+ 6NH3(aq)K=1Kf= 10 34We assume initially all the cobalt ions are in the form of the complex.

7 The initialconcentrations are found usingnCo3+= ( mol L 1)( L) = 10 3molBefore complexationnNH3= ( mol L 1)( L) = molAfter complexationnNH3= mol 6( 10 3mol) = mol[Co(NH3)3+6] = 10 L= 10 3M[NH3] = L= M8[Co(NH3)3+6][Co3+][NH3] 10 3M0 Mchange-yy6yequilibrium( 10 3 y) MyM( + 6y) 10 34=[Co3+][NH3]6[Co(NH3)3+6]=y( + 6y) 10 3 y y( ) 10 3y= [Co3+] = 10 31M16. The solubility product constant for zinc oxalate (ZnC2O4) is 10 8and the for-mation constant for the zinc cyanide coordination complex [Zn(CN)2 4] is a mixture that is formed by combing L of a 10 4M CN solutionand L of a 10 6M Zn2+solution. After the Solutions are mixed molesof oxalate ions are added to the solution. Assuming the added oxalate does not changethe total volume of the solution, determine if a zinc oxalate precipitate will form. Ap-proximations work for this : Before complex formationnCN = ( 10 4mol L 1)( L) = 10 5molnZn2+= ( 10 6mol L 1)( L) = 10 6molAfter complex formationnCN = 10 5mol 4( 10 6mol) = 10 5mol[CN ] = 10 L= 10 5M [Zn(CN)2 4] = 10 10 6 MZn(CN)2 4(aq) Zn2+(aq)+ 4CN (aq)K=1Kf= 10 18=[Zn2+][CN ]4[Zn(CN)2 4][Zn(CN)2 4][Zn2+][CN ]] 10 6M0 10 5 Mchange-yy4yequilibrium( 10 6 y) MyM( 10 5+ 4y) 10 18=y( 10 5+ 4y) 10 6 y y( 10 5) 10 6y= [Zn2+] = 10 5M[C2O2 4] = L= MQ= [Zn2+][C2O2 4] = ( 10 5)( ) = 10 5> KspPrecipitate forms9