Topic 14: Solubility AP Chemistry Mrs. Laura Peck, 2013 Goes with chapter 19: Silberberg’s Principles of General Chemistry AP Chemistry Mrs. Laura Peck, 2013

Objectives/Study Guide Write balanced equations for the dissolution of a salt and its corresponding solubility product expression. Predict the relative solubilities of salts which dissolve to give the same number of ions from their Ksp values Calculate the Ksp value from the solubility of a salt and also calculate the solubility of the salt in units of mol/L or g/L from the given Ksp value Predict the effect of a common ion on the solubility of a salt and perform calculations. Perform calculations to predict if a precipitate will form when two solutions are mixed Do problems involving selective precipitation. Perform calculations involving complex ions and solubility Use qualitative analysis to separate a mixture of ions

AP tip: Solubility problems would appear in the first free-response question, since they would be considered an equilibrium problem. Awareness of the types of problems outlined in this Topic and the methods used to solve them will lead you to succeed in this topic on the AP exam.

Solubility Product You should have already memorized the solubility rules and know which salts are soluble. For slightly soluble or insoluble salts, an equilibrium exists between the solid and its aqueous ions. Ex: PbCl2(s)   Pb2+(aq) + 2Cl-(aq) At first, when the salt is added to the water, there are no ions present. As the solid dissolves, the concentration of the ions increases. A simultaneous competing process is the reverse of the dissolution, that is, the reforming of the solid called crystallization. At some point, the maximum amount of dissolution is achieved, which is called the saturation point. However, remember that on a molecular level, a dynamic equilibrium exists between dissolved solute and undissolved solid (Rdissolution=Rcrystallization) The solution is saturated when no more solid dissolves and equilibrium is reached. Keq = Ksp = [Pb2+][Cl-]2 The constant, Ksp, is the solubility product constant. For salts producing the same number of ions, the Ksp value can be used to measure the extent to which the solid dissolves. The larger the Ksp value, the more soluble the salt.

Example #1 Given the following salts and their Ksp values, which salt is the most soluble? Which salt is the least soluble? Formula Ksp NiCO3 1.4x10-7 MnS 2.3x10-13 CaSO4 6.1x10-5 The most soluble salt is the salt with the largest Ksp value, CaSO4 The least soluble salt is the salt with the lowest value of Ksp, MnS. You are able to compare the Ksp values to determine the relative solubilities of the salts because they all produce the same number of ions.

Calculations involving solubility The solubility of a salt that will dissolve in 1 L of water. The solubility of a salt can be given in units of mol/L or g/L. The solubility of a salt can be used to determine the Ksp value for the salt.

Example #2: Calculating Ksp from solubility. The solubility of Pb3(PO4)2 is 6.2x10-12M. Calculate the Ksp value for the solid. Step 1: write the reaction for the dissolution of the solid. Step 2: Underneath the reaction, make an ICE chart. Pb3(PO4)2(s)   3Pb2+(aq) + 2PO43-(aq) 0 0 +3x +3x E. 3x 3x Or you can use stoich to solve this problem: 6.2x10-12 x molPb3(PO4)2 x 3mol Pb2+ = 1.9x10-11M Pb2+ 1 L 1mol Pb3(PO4)2 6.2x10-12 x mol Pb3(PO4)2 x 2 mol Pb43- = 1.2x10-11 M PO43- 1 L 1mol Pb3(PO4)2 Plug these values into the Ksp expression and solve for Ksp: Ksp = [Pb2+]3[PO43-]2 = (1.9x10-11)3(1.2x10-11)2 = 9.9x10-55 In the ICE chart, x represents x mol/L Of Pb3(PO4)2(s) dissolving to reach Keq which equals 6.2x10-12 M For every 1mol/L of Pb3(PO4)2, which Dissolves, 3 mols/L of Pb2+ and 2 mol/L Of PO43- form 3x is the mol/L of Pb2+ produced when The solid Pb3(PO4)2 dissolves 2x is the mol/L of PO43- produced when The solid, Pb3(PO4)2 dissolves Step 3: Write the Keq expression for the reaction And plug in the values from the E line of the ICE Keq = Ksp = [Pb2+]3[PO43-]2 = (3x)3(2x)2 = 108x5 The value of x is the solubility of Pb3(PO4)2, which Equals 6.2x10-12M Ksp = 108(6.2x10-12)5 = 9.9x10-55

Calculating solubility from Ksp If you are given a group of salts which do not all have the same cation to anion ratio and asked which is more soluble, you must perform a calculation to determine the solubility of each salt. If the salts each provide a number of ions in solution, so you cannot directly compare the Ksp values to predict which is more soluble.

Example #3: Given the two salts in the table below, which is more soluble? Show calculation to support your answer. Step 1: Write the reaction for the dissolution of the solid Salt #1 Step 2: Underneath the reaction, make an ICE chart Step 3: Write the equilibrium expression for Ksp, plug in The equilibrium line, and solve for x Step 4: Repeat process for salt #2 Salt Ksp FeC2O4 2.1x10-7 Cu(IO4)2 1.4x10-7 FeC2O4(s)   Fe2+(aq) + C2O42-(aq) 0 0 +x +x E. X x Ksp = [Fe2+][C2O42-]  2.1x10-7 = x2  x = 4.6x10-4 mol Fe2+/L 4.6x10-4 mol Fe2+ x 1mol FeC2O4/1L = 4.6x10-4 mol FeC2O4/L Cu(IO4)2(s)   Cu2+(aq) + IO4-(aq) 0 0 +x +2x E. X 2x Ksp = [Cu2+][IO4-]2  1.4x10-7 = x(2x)2  4x3 = 1.4x10-7  x = 3.3x10-3 mol/L 3.3x10-3mol Cu2+ x 1mol Cu(IO4)2/1L = 3.3x10-3mol Cu(IO4)2/L Turns out that Cu(IO4)2/L is greater than FeC2O4/L so its more soluble!

Common Ion Effect When a salt is dissolved in water containing a common ion, its solubility is decreased. Consider the solubility equilibrium of silver sulfate: Ag2SO4(s)   2Ag+(aq) + SO42-(aq) When silver sulfate is dissolved in 0.100M AgNO3, the Ag+ ion from silver nitrate causes the equilibrium to shift to the left, decreasing the solubility of silver sulfate.

Example #4 Calculate the molar solubility of Ag2SO4 in 0.10M AgNO3. Ksp for Ag2SO4(s) is 1.2x10-5 First, fill out the ICE chart. Be sure To include the initial [ ] of the ions From the soluble salt, AgNO3. Second, plug in the equilibrium line Into the Ksp expression. Ag2SO4(s)   2Ag+(aq) + SO42-(aq) I… 0.10 0 C… +2x +x E… 0.10+2x x Ksp = [Ag+]2[SO42-] = (0.10 + 2x)2(x) Assume that 0.10 x 2x is about 0.10, since Ksp is small you can assume that the change (2x) from the initial concentration (0.10) is negligible. Ksp = 1.2x10-5 = (0.10)2(x)  x = 1.2x10-3 mol SO42-/L 1.2x10-3mol/L SO42- x (1mol Ag2SO4)/(1mol SO42-) = 1.2x10-3mol/L Ag2SO4 The solubility of silver sulfate in 0.100M silver nitrate, 1.2x 10-3 mol/L is less Than the solubility of silver sulfate in pure water, 1.4x10-2 mol/L

pH and Solubility Chromium (III) hydroxide dissolves according to the equilibrium: Cr(OH)3(s)   Cr3+(aq) + 3OH-(aq) An increase in pH, caused by the addition of OH- ions, will shift the equilibrium to the left, decreasing the solubility of Cr(OH)3 A decrease in pH, caused by the addition of H+ ions, will shift the equilibrium to the right, increasing the solubility of Cr(OH)3. The H+ ions remove the OH- ions from the solution. A salt with the general formula, MX, will show increased solubility in acidic solution if the anion, X-, is an effective base (if HX is a weak acid). Common anions that make effective bases include S2-, OH-, and CO32-

Example #5 Calculate the solubility of Fe(OH)3 in a solution with a pH equal to 5.0. Ksp = 4.0x10-38 pOH = 14 – pH = 14.0-5.0 = 9.0 [OH-] = 1.0x10-9 M First steps are to write out the reaction Then fill out an ICE chart. Calculate The initial hydroxide concentration From the pH. Second, plug E values into Ksp equation Fe(OH)3(s)   Fe3+(aq) + 3OH-(aq) I.. 0 1.0x10-9 C.. X 3x E.. X 1.0x10-9 Because pH is 5.0, the equilibrium [OH-] must be equal to 1.0x10-9 Ksp = 4.0x10-38 = [Fe3+][OH-]3 = x(1.0x10-9)3 X = 4.0x10-11 M

Precipitate formation A precipitate may or may not form when two solutions are mixed, depending on the concentration of the ions involved in the formation of the solid. Ion Product: The ion product, Q, is written in the same way as the Ksp expression. F0r Lead(II) chloride, Q = [Pb2+][Cl-]2 Calculation of the value, Q, involves the use of the initial concentrations of the solutions mixed, [Pb2+]0 and [Cl-]0, instead of the equilibrium concentrations. A comparison of the value of Q to Ksp determines if a precipitate is formed. Q>Ksp – precipitation occurs Q<Ksp – no precipitation occurs Q=Ksp – the solution is saturated.

Will a precipitate for when 100. 0 mL of 4 Will a precipitate for when 100.0 mL of 4.0x10-4M Mg(NO3)2 is added to 100.0 mL of 2.0x10-4M NaOH? Example #6 Step 1: Determine the identity of the precipitate formed. Step 2: Determine the concentration of the ions after they are Mixed and before any reaction occurs. Determine the moles of concentration of each solute present. Be sure to divide by the total volume of the two solutions mixed. Step 3: Write the ion product expression, calculate its value, and Compare it to Ksp, which equals 8.8x10-12 Mg(OH)2 is the possible precipitate. NaNO3 is always soluble. [Mg2+]0 = (0.100L x (4.0x10-4mol/L))/0.200L = 2.0x10-4M [OH-]0 = (0.100L x(2.0x10-4mol/L))/0.200L = 1.0x10-4M Q = [Mg2+][OH-]2 = (2.0x10-4M)(1.0x10-4)2 = 2.0x10-12 Since Q<Ksp – no precipitate will form

Selective Precipitation A reagent is added to a mixture of metal ions, thus forming a precipitate. One metal ion will precipitate before the other, allowing the mixture to be separated. Example #7: a solution contains 0.25M Ni(NO3)2 and 0.25M Cu(NO3)2. A solution of Na2CO3 is slowly added to this solution. A) Will NiCO3 (Ksp = 1.4x10-7) or CuCO3 (Ksp = 2.5x10-10) precipitate first? B) Calculate the concentration of CO32- necessary to begin the precipitation of each salt. C) Determine the concentration of Cu2+ when NiCO3 begins to precipitate. CuCO3 will precipitate first because its Ksp is smaller For CuCO3 precip. begins: [CO32-]= Ksp/[Cu2+] = 2.5x10-10/0.25M = 1.0x10-9M For NiCO3 precip. Begins: [CO32-]=Ksp/[Ni2+]=1.4x10-7/0.25M = 5.6x10-7M [Cu2+] = Ksp/[CO32-] = 2.5x10-10/5.6x10-7M = 4.5x10-4M

Complex Ion Equilibria A complex ion consists of a metal ion surrounded by ligands which are Lewis bases such as H2O, OH-, NH3, Cl-, and CN- The coordination number is the number of ligands which attach to the transition metal ion. Common coordination numbers include 4 for Cu2+ and Co2+ and 2 for Ag+ The ligands attach one at a time to the metal ion; each step has a formation constant, Kf: Ag+ + S2O32-  Ag(S2O3)- Kf1 = 7.4x108 Ag(S2O3)- + S2O32-  Ag(S2O3)23- Kf2 = 3.9x104 Formation of a complex ion causes a precipitate to dissolve. The equilibrium, AgBr(s)  Ag+ + Br- is not affected by the addition of H+. But the concentration of Ag+ can be lowered by the addition of excess S2O32- forming the complex ion, Ag(S2O3)23-

Example #8 Calculate the mass of AgBr that can dissolve in 1.00L of 0.500M Na2S2O3. Ksp for AgBr = 5.0x10-13 Step 1: Determine the overall reaction by Adding up the reactions for the solubility Equilibrium and the stepwise formation Of the complex ion. Remember, to get the Equilibrium constant for the overall reaction, K, multiply together the K for each step. (from Preceding slide) Step 2: Complete an ICE chart with the overall reaction Step 3: Plug E line into Ksp expression AgBr(s)  Ag+(aq) + Br-(aq) Ksp = 5.0x10-13 Ag+ + S2O32-  Ag(S2O3)- Kf1 = 7.4x108 Ag(S2O3)- + S2O32-  Ag(S2O3)23- Kf2 = 3.9x104 AgBr(s) + 2S2O32-  Ag(S2O3)23- + Br- K’=14.4 I.. 0.500 0 0 C.. -2x +x +x E.. 0.500-2x x x K’ = 14.4 = [Ag(S2O3)23-][Br-]/[S2O32-]2 = x2/(0.500-2x)2 take square root of both sides 3.79 = x/(0.500-2x) X = 1.90-7.58x  x = 0.221MBr = 0.221M AgBr(s) 1.00L x (0.221mol AgBr/1L) x (187.8g/1mol) = 41.5g AgBr = 42g AgBr

Qualitative Analysis Qualitative analysis involves separating a mixture of cations or anions based on their solubilities. Cations can be separated into five major groups based on their solubilities. Group I: insoluble chlorides Group II: sulfides soluble in acidic solution Group II: sulfieds insoluble in basic solution Group IV: insoluble carbonates Group V: alkali metal and ammonium ions Each of these groups can be treated further to separate and identify the individual ions.

Example #9 Solution of Ag+, Hg22+, Pb2+ Precipitate of AgCl(s), Hg2, Cl2(s) Precipitate of Hg(l) (black), HgNH3Cl(s) (white) Solution of Ag(NH3)2, Cl- Precipitate of AgCl(s) (white) Solution of Pb2+ Precipitate of PbCrO4(s) (yellow) Precipitate of AgCl(s), Hg2, Cl2(s), PbCl3(s) Separate a mixture containing the Group I cations: Ag+, Hg22+ and Pb2+ Add cold HCl(aq) Heat Add CrO42- Add NH3(aq) Add H+

The End