CHM 170 1nd EditionExam # 3 Study Guide Lectures: 17 - 20Lecture 17 (March 26)The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp.For example, the dissociation reaction for PbCl2 isPbCl2(s)  Pb2+(aq) + 2 Cl−(aq)The equilibrium constant would be:Ksp = [Pb2+][Cl−]2The answer is C. Note: the bigger the K value then it favors the product more and therefore it is more soluble. The smaller the K value then it favors the reactant. The reactant is just a salt.The answer is C. The smaller the number then the bigger the K value.What is the molarity of Pb2+ ions in a saturated solution of PbCl2? (Ksp = 1.6 x 10-6)First: write out the expression and balance it. Then make an ICE chart and write the Ksp equation.PbCl2 → −¿2+¿+2 Cl¿Pb¿I -- 0 0C -- +x +2xE -- x 2xKsp = [2+¿−¿¿2Cl¿Pb¿¿Second: substitute and solve.Ksp = [2 x ¿2x ¿¿Ksp = [x][4x2¿ (add the exponents)Ksp = 4x3 (don’t forget that the Ksp value is given, substitute it in as well and solve)1.6x10−6 = 4x3 (divide by 4)x3 = 4x10−7 (square root 3 to get x by itself)X = 7.4x10−3 Third: find the molarity of 2+¿Pb¿by plugging the value of X into Pb. Since Pb is just X then the molarity of Pb is 7.4x10−3.However, if the question was to find Cl then it would be 2 x 7.4x10−3 and the answer for that would be the molarity for Cl. Solubility is the amount of solute that will dissolve in a given amount of solution; at a particular temperature. The molar solubility is the number of moles of solute that will dissolve in a liter ofsolution.Practice Examples:Calculate the molar solubility of PbCl2 in pure water at 25C. (Ksp = 1.6 x 10−6)PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)Ksp = [Pb2+][Cl−]2Ksp = [s][2s]2Ksp = 4s31.6 x 10−6 = 4s3s3 = 4x 10−7s = 7.4 x 10−3 (this is equal to Pb which is also equal to PbCl2)PbCl2= 7.4 x 10−3MDetermine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M.PbBr2(s)  Pb2+(aq) + 2 Br−(aq)Ksp = [Pb2+][Br−]2Ksp = (1.05 x 10-2)(2.10 x 10-2)2How many grams of BaCO3 (formula weight = 197) will dissolve in 1.2 L of water? (Ksp = 5.1 x 10-9)BaCO3  Ba+2 + CO3-2I -- 0 0C -- +x +xE -- x xKsp = [Ba+2][CO3-2]Ksp = [x][x]Ksp = x25.1 x 10-9 = x2X = 7.14 x 10-57.14 x 10−5 molL x 1.2 L❑x 197 g1 mol = 1.7 x 10-2 g BaCO3A 1.0-L sample of a saturated calcium oxalate solution, CaC2O4, contains 0.0061-g of the salt at 25°C. Calculate the Ksp for this salt at 25°C.We must first convert the solubility of calcium oxalate from 0.0061 g/liter to moles per liter.CaC2O4 (s) ↔ Ca +2 (aq)+ C2O4-2(aq)I -- 0 0C -- +4.8 x 10-5+4.8 x 10-5E -- 4.8 x 10-54.8 x 10-5Lecture 18 (March 31)The Effect of Common Ion on Solubility:Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt.o for example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)addition of Cl− shifts the equilibrium to the leftExample:The answer is A. realize that A, C, and D all have a common ion, however A has more common ions and will decrease the solubility more.Examples:Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C. CaF2(s)  Ca2+(aq) + 2 F−(aq)Ksp = [Ca2+][F−]2Ksp = [Ca2+][F−]2Ksp = (S)(0.100 + 2S)2Ksp = (S)(0.100)2Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound.o if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occurQ = Ksp, the solution is saturated, no precipitationQ < Ksp, the solution is unsaturated, no precipitationQ > Ksp, the solution would be above saturation, the salt above saturation will precipitateSome solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions.Predicting Whether Precipitation Will OccurThe concentration of calcium ion in blood plasma is 0.0025 M. If the concentration of oxalate ionis 1.0 x 10-7 M, do you expect calcium oxalate to precipitate? Ksp for calcium oxalate is 2.3 x 10-9.This value is smaller than the Ksp, so you do not expect precipitation to occur.A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF?Plan: Write out a reaction equation to see which salt would be formed. Look up the Ksp values ina table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations.CaF2(s) ↔ Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11C1V1=C2V2 (use this to find Ca and F)[Ca2+] = 0.10M [F-] = = 0.040M Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6x10-4 Q is >> Ksp and the CaF2 WILL precipitate.The effect of pH on solubility:The pH can affect the solubility of the salt in which, the base will decrease solubility and the acid will increase the solubility. Complex Ion Formation:Ions that form by combining a cation with several anions or neutral molecules are called complex ions.o e.g., Ag(H2O)2+The attached ions or molecules are called ligands.o e.g., H2OoThermodynamicsThe first Law of Thermodynamics: energy cannot be created or destroyed, but it will spread itself. It is the conservation of energy. For a exothermic reaction head is lost from the system and goes to the surroundings. The system is the chemical reaction and the surroundings is everything around the chemical reaction. If head is absorbed then it is a endothermic reaction. If the ∆ H is negative then energy is given off. If the ∆ H is positive then energy is being absorbed. Energy conservation requires that the energy change in the system equal the heat released + work done.o∆ E= q+wo∆ E=∆ H +P ∆ Vo∆ E and ∆ His a state function.Process that will occur are called spontaneous. Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction. If the system after reaction has less potential energy than before the reaction, the reaction is thermodynamically favorable.Note: spontaneity is not the same thing as fast or slow. There are two factors that determine whether a reaction is spontaneous. They are