6.1.3 Relationship between solubility and the solubility product The conversion betweenK and solubility Because concentrations in the Keexpression must be in molarity and the unit of solubility is g solute /100g water,So we need convert the solubility data to molarity(mol-L). A Bm(s)--nAT(aq)+mB"-(aq) Equilibrium /mol.L nS mS K8=(nS)"(mS)" AB type S=
6.1.3 Relationship between solubility and the solubility product 1 Equilibrium /mol L - nS mS m+ n- A Bm (s) nA (aq) + mB (aq) n n m K = (nS) (mS) sp AB type S = Ksp The conversion between and solubility Because concentrations in the expression must be 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 1.92X 10-3 gL-1 at 25C. Calculate the value of K for AgCl. Answer:We know M(AgCI)=143.3 S=1.92X103 143.3 mol-L=1.34X10mol-L AgCl(s)=Ag(aq)+Cl (aq) Equilibrium/mol.L-1 S S KP(AgC1)={c(Ag)}{c(CI)}=S2=1.80X1010
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 S S Answer：We know Mr (AgCl)= 143.3 1 5 1 3 mol L 1.34 10 mol L 143.3 1.92 10 - - - - S = × = × AgCl(s) Ag (aq) Cl (aq) + - + 2 10 (AgCl) { (Ag )}{ (Cl )} 1.80 10 + - - Ksp = c c = S = × Ksp
Example:The K for Ag,CrO 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 K(Ag2CrO)=(c(Ag))2(c(CrO ) 1.1×1012=4x3,x=6.5×10 M(Ag2Cr04)=331.7 S=6.5×105×331.7gL=2.2×102gL1
/(mol L ) 2 1 x x - Equi. Mr (Ag 2CrO4 ) = 331.7 5 1 2 1 6.5 10 331.7 g L 2.2 10 g L - - - - S = × × = × 12 3 5 1.1 10 4 , 6.5 10 - × - × = x x = Ag CrO (s) 2Ag (aq) CrO (aq) 2 4 4 + 2- + + 2- 4 2 2 4 K (Ag CrO ) ={c(Ag )} {c(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 Ca3(P04)2 5 K品 S= 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,AgI,Ag,CrO formula Kp° Solubility/mol L-1 AgCI 1.8 10-10 1.310-5 AgBr 5.0 10-13 7.110-7 AgI 8.310-17 9.11010 Ag2CrO4 1.11012 6.5105 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!! Ke(AgCI)>Ke(Ag2CrO),but S(AgCI)<S(Ag2CrO)
(AgCl) (Ag CrO ) S S 2 4 < 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 Ksp have bigger solubility. (AgCl) (Ag CrO ) 2 4 K > sp Ksp 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 6.5 10- 1 mol L - 10 1.8 10- 13 5.0 10- 17 8.3 10 - 12 1.1 10- 10 9.1 10- 7 7.1 10- 5 1.3 10- Ag2CrO4 formula Ksp Solubility/