! ! Name:____________________''Revised'CR'12/10/13' ' ©'LaBrake'&'Vanden'Bout'2013'Department of Chemistry University of Texas at Austin Solubility)Product)Constant)(Ksp))–)Supplemental)Worksheet)))))KEY)1. What'is'the'solubility,'in'moles/liter,'of'AgI'if'the'Ksp'='8.5•10O12?'AgI$(s)$à$Ag+$(aq)$+$I–$(aq)$Ksp$=$[Ag+]1[$I–]1$We'know'our'Ksp'='8.5•10O12'and'we'can'insert'this'into'our'Ksp'equation'8.5•10–12$$=$[Ag+]1[$I–]1$Since'we'have'a'1:1'ionOion'dissociation'of'AgI,'we'can'set'x'to'the'following'then'plug'into'our'Ksp'equation:''x$=$[Ag+]$=$[$I–]$8.5•10–12$$=$x$.$x$8.5•10–12$=$x2$2.93•10–6$=$x$x$=$2.93•10–6$mol$L@1$=$[Ag+]$=$[$I–]$SAgI$=$$2.93•10–6$mol$L@1$2. If'the'solubility'of'Li2CO3'is'1.32'g/100'mL,'what'is'its'Ksp'at'room'temperature?'Li2CO3$(s)$à$2Li+$(aq)$+$CO32–$Ksp$=$[Li+]2[CO32–]1$2Li+$=$2x$;$CO32–$=$x$Ksp$=$(2x)2(x)$=$4x3$Next,$we$need$to$determine$the$solubility$in$terms$of$mol/L.$Since$we$have$been$given$the$solubility$in$terms$of$g/mL,$we$need$to$convert$this.$$SLi2CO3$=$1.32$g$Li2CO3$1$mol$Li2CO3$1000$mL$100$mL$73.891g$Li2CO3$1$L$=$0.1786$mol$L@1$Li2CO3$! ! Name:____________________''Revised'CR'12/10/13' ' ©'LaBrake'&'Vanden'Bout'2013'Department of Chemistry University of Texas at Austin Since$we$have$converted$the$solubility$into$mol/L$we$can$insert$this$into$our$Ksp.$The$solubility$is$for$one$mole$of$Li2CO3$dissolving.$For$every$one$mole$of$the$substance$we$observe$one$mole$of$CO32@$which$we$earlier$called$“x”.$So$the$solubility$of$Li2CO3$is$equal$to$x.$Ksp$=$4(0.1786$mol$L@1)3$=$2.3$x$10–2$3. What'is'the'solubility,'in'moles/liter,'of'HgBr2'if'the'Ksp'='6.2•10O9?'HgBr2$(s)$à$Hg2+$(aq)$+$2Br$–$(aq)$Ksp=$[Hg2+]1[Br$–]1$2Br$–$=$2x$;$Hg2+$=$x$Ksp=$(2x)2(x)$=$4x3$6.2•10–9$=$4x3$In$order$to$find$the$solubility$at$this$step,$we$must$solve$for$x!$6.2•10–9$=$4x3$6.2•10–9$÷$4$=$x3$€ 6.2 • 10−943= x!x$=$0.00157$=$1.157•10–3$mol$L@1$SHgBr2$=$1.157•10–3$mol$L@1$4. If'Cu3(AsO4)2'has'a'Ksp'='8.0•10O36,'then'what'is'the'concentration'of'[Cu2+]'in'a'saturated'solution?'Cu3(AsO4)2$(s)$à$3Cu2+$(aq)$+$2AsO43–$Ksp$=$[Cu2+]3[AsO43–]2$3Cu2+$=$3x$;$2AsO43–$=$2x$Ksp$=$(3x)3(2x)2$=$(27x3)(4x2)$=$108x5$8.0•10–36$=$108x5$8.0•10–36$÷$108$=$x5$! ! Name:____________________''Revised'CR'12/10/13' ' ©'LaBrake'&'Vanden'Bout'2013'Department of Chemistry University of Texas at Austin € 8.0 • 10−361085$=$x$x$=$3.75•10–8$[Cu2+]$=$3x$=$3(3.75•10–8)$=$1.13•10–7$mol$L@1$5.
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