6.1.3 Relationship between solubility and the solubility product The conversion between K and solubility Because concentrations in theKeexpression must in molarity and the unit of solubility is g solute /100g water,So we need convert the solubility data to molarity(mol-L). A B (s)--nA*(aq)+mB"-(ag) Equilibrim /mol.L nS mS K=(nS)"(mS)"ABS=
6.1.3 Relationship between solubility and the solubility product 1 L /molmEquilibriu − ⋅ n S m S m + n − BA mn (s) nA (aq) + mB (aq) n m K = SS )m()n( sp AB 型 S = Ksp The conversion between and solubility Because concentrations in the expression must in molarity and the unit of solubility is g solute /100g water, So we need convert the solubility data to molarity(mol·L － 1). Ksp Ksp
Example The solubility of AgCl is found experimentally to be 1.92X10-3 gL at 25C. Calculate the value of K for AgCl. Answer:We know M(AgCl)=143.3 S=1.92x103 mol.L=1.34X10mol.L 143.3 AgCl(s)=Ag"(aq)+Cl (ag) Equilibrium mol.L-1 S S K9(AgC)={c(Ag)}{c(CI)}=S2=1.80X10-10
Example ：The solubility of AgCl is found experimentally to be 1.92×10-3 g·L-1 at 25oC. Calculate the value of for AgCl. Equilibrium / mol.L-1 SS M 3.143 r Answer：We know (AgCl)= 1 5 1 3 mol .1L 1034 mol L 3.143 .1 1092 − − − − S = × =⋅ × ⋅ AgCl(s) Ag (aq) Cl (aq) + − + 2 10 (AgCl ({) Ag )}{ (Cl )} .1 1080 + − − Ksp = Scc == × Ksp
Example:The Ke for AgCrO is 1.1X10-12at 25C.Calculate the solubility of Ag,CrO in gL-1. Answer:Ag2 CrO,(s)-2Ag"(aq)+CrO(aq) Equi. /(mol.L) 2x X Ke(Ag2CrO)=(c(Ag)(c(CrO 1.1X1012=4x3,x=6.5×10-5 M(Ag2Cr04)=331.7 S=6.5X10-5×331.7gL1=2.2X102gL
/(mol )L 2 1 xx − Equi. ⋅ Mr(Ag 42 )CrO = 331.7 5 1 12 5.6 10 331.7 2.2Lg 10 Lg − − −− S = ×× =⋅ × ⋅ 12 3 5 1.1 10 5.6,4 10 − − × xx == × Ag CrO 2Ag(s) (aq) (CrO aq) 42 4 + 2− + + 2− 4 2 42 K (Ag = ({)CrO Ag )} cc CrO({ )} sp Answer： Example：The for Ag2CrO4 is 1.1×10-12 at 25oC. Calculate the solubility of Ag2CrO4 in g·L-1. Ksp
Question:Determine the relationship between molar solubility (S,unit:mol.L)and Ke for Ca(PO)2. 5 108
Question：Determine the relationship between molar solubility (S, unit: mol.L-1) and for Ca3(PO4)2. Ksp 5 108 S = Ksp
Comparasion of solubility and solubility product of AgCl,AgBr,Agl,Ag,CrO formula Kype Solubility/mol.L-1 AgCI 1.8×10-10 1.3×105 AgBr 5.0×1013 7.1×107 AgI 8.3×10-17 9.1×10-10 Ag2CrO 1.1×1012 6.5×10-5 Sparingly soluble electrolytes of same type with bigger Kphave bigger solubility. We can't compare straightly the solubility of solutes of different types according to their solubility product.Calculaton!! K(AgCI)>K(Ag2CrO),but S(AgCI)<S(Ag2CrO)
(AgCl < SS () Ag 42 )CrO We can’t compare straightly the solubility of solutes of different types according to their solubility product. Calculaton!! * Sparingly soluble electrolytes of same type with bigger KspΘ have bigger solubility. Ksp (AgCl > Ksp () Ag 42 )CrO Comparasion of solubility and solubility product of AgCl, AgBr, AgI, Ag2CrO4 , but 分子式 溶度积 溶解度/ AgBr AgI AgCl 5 5.6 10− × 1 mol L− ⋅ 10 8.1 10 − × 13 0.5 10 − × 17 3.8 10− × 12 1.1 10− × 10 1.9 10 − × 7 1.7 10 − × 5 3.1 10− × Ag2CrO4 formula Ksp Solubility/ Θ