Ksp: An Overview with objectives and examples Be sure to review the Solubility Rules when preparing for this topic.

Solubility Equilibria Will it all dissolve, and if not, how much? Will it precipitate? If not, what must be done to cause it to precipitate?

The Solubility of Salts An insoluble substance is defined as one that has a solubility less than about 0.01 mol per liter of water. General solubility rules should be memorized by all AP students. Many minerals are insoluble sulfides, oxides, and carbonates and methods had to be developed for extracting the various metals from these ores. Very often a sequence of solution and precipitation reactions is involved.

Ksp Expression and Molar Solubility Calculations Compounds that have low solubility in aqueous solutions are known as precipitates. For example, when a solution of silver nitrate is mixed with a solution of sodium chloride, the precipitate of silver chloride forms. The nitrate and the sodium ions are spectator ions, ions that do not participate in the reaction. The net ionic equation is: Ag+(aq) + Cl- (aq)  AgCl (s) Precipitates, compounds that have low solubility in solution, no matter how insoluble, are in equilibrium with the saturated solution.

The Solubility Product Constant Ksp When a salt dissolves to the greatest extent possible, the solution is said to be saturated. For a salt such as sodium bromide, which is only very sparingly soluble in water, the concentration of a saturated solution is very small. At 25ºC, the concentration of silver ions and bromide ions in a saturated solution is only 5.7 x 10-7 M. AgBr(s)  Ag+(aq) + Br-(aq) The equilibrium constant for the system is: Ksp = [Ag+][Br-] = (5.7 x 10-7)(5.7 x 10-7) = 3.3 x 10-13 The equilibrium constant is called the solubility product constant, or more simply, the solubility product, symbol Ksp. If the solubility product is known, then the concentrations of ions in solution can be calculated.

Ksp Expression and Molar Solubility Calculations The Ksp expression is the product of the soluble ions, expressed in molarity, raised to their coefficients from the balanced dissociation equation. For examples, the Ksp expression for Ba3(PO4)2 is: Ksp = [Ba 2+ ]3 [PO43-]2 The superscripts are obtained from the balanced dissociation equation: Ba3(PO4)2(s)  3 Ba2+(aq) + 2 PO43-(aq) The numerical value of Ksp is known as the solubility product constant. Notice as with other equilibrium expressions that the solid has been eliminated from the expression.

Molar Solubility Molar solubility refers to the number of moles of precipitate that dissolve per liter of solution. Molar solubility can be affected in two ways: Molar solubility can be reduced by adding a product ion; this is an example of the common ion effect. For example, adding NaCl solution to a saturated solution of AgCl will lower the solubility of AgCl.

Molar Solubility Two ways to affect solubility (continued) The molar solubility of some precipitates can be affected by the pH of the solution. If a product ion interacts with H+, the precipitate dissolves in acidic solution. For example, the precipitate AgF(s) dissolves in strong acid. The product ion, F-, reacts with H+ to form HF, a weak acid. As the F- is removed from solution in the form of HF, the precipitation equilibrium shifts to the right (precipitate dissolves) according to Le Chatelier’s Principle.

AP TESTING NOTE: Precipitation questions on the AP exam often involve the following precipitates: BaSO3, Ba3(PO4)2, CaCO3, MgS, Al(OH)3 and Zn(OH)2. The common ion effect and the effect that pH has on molar solubilities are common topics for lab questions. You will also need to know the colors of some common precipitates. Most precipitates are white, but some are very distinctive, such as Ag2CrO4, which is red-brown.

Determining Ksp from Experimental Measurements Values for the Ksp constants for various salts are determined experimentally. If the solubility can be determined, then the Ksp can be calculated.

Estimating Salt Solubility from Ksp Just as Ksp can be calculated from solubility of a salt, so the solubility can be calcualted form the Ksp. Comparison of solubility products does not necessarily indicate relative solubilities—it depends upon the stoichiometries of the salts. For example, compare silver chloride and silver chromate: AgCl: Ksp = 1.8 x 10-10 solubility = 1.3 x 10-5 mol/L Ag2CrO4: Ksp = 9.0 x 10-12 solubility = 1.3 x 10-4 mol/L Comparison of solubilities by examining the Ksp values is useful only when the stoichiometries of the salts are the same.

Precipitation of Insoluble Salts You can predict whether a precipitate occurs by substituting the initial concentrations of the ions into the Ksp expression and calculating the reaction quotient, Q. A salt will precipitate from solution if the solubility product is exceeded. The quotient Qsp is the product of the actual concentrations of the ions in solution, for example: For silver chloride: Qsp = [Ag+][Cl-] For nickel sulfide: Qsp = [Ni2+][S2-] For silver chromate, Qsp = [Ag+]2[CrO42-] If Qsp > Ksp then the salt will precipitate If Qsp = Ksp then the solution is saturated If Qsp < Ksp then the solution is unsaturated and no precipitate will form.

Precipitation of Insoluble Salts Predicting whether a Precipitate will Form Example #1: If a solution contains 3.0 x 10-4 M Ag+ ions and sufficient potassium chromate is added to make the solution 5.0 x 10-3 M in chromate ions. Will precipitation occur? Silver chromate is Ag2CrO4 (Ksp= 9.0 x 10-12) Qsp = [Ag+]2 [CrO42-] = (3.0 x 10-4)2(5.0 x 10-3) = 4.5 x 10-10 This value does exceed the Ksp of 9.0 x 10-12 so precipitation will occur. Qsp= 4.5 x 10-10 > Ksp= 9.0 x 10-12; a precipitate forms

Precipitation of Insoluble Salts Predicting whether a precipitate will form Example #2: 10.0 mL of 0.010 M Pb(NO3)2 are mixed with 10.0 mL of 0.01 M NaI solution. Determine if a precipitate will occur. Ksp = 1.4 x 10-8. Justify your answer. Strategy: Identify the precipitate and formula Write the dissociation equations and Ksp expression Obtain the Ksp value from a solubility chart. Calculate Qsp (some just call this “Q”) Compare Qsp with Ksp

Precipitation of Insoluble Salts Example #2: 10.0 mL of 0.010 M Pb(NO3)2 are mixed with 10.0 mL of 0.01 M NaI solution. Ksp = 1.4 x 10-8 The precipitate is PbI2 (NO3- and Na+ are spectator ions The equation is: PbI2(s)  Pb2+(aq) + 2 I- (aq) Ksp = [Pb2+] [ I- ]2 = 1.4 x 10-8 Next do the stoichiometry calculations: Initial concentration of [Pb2+] = 0.010 L x 0.010 M Pb(NO3)2 x 1 mol Pb2+ = 0.0050 M Pb2+ 0.020 L 1 mol Pb(NO3)2 Initial concentration of [ I-] = 0.010 L x 0.010 M NaI x 1 mol I- = 0.0050 M I- 0.020 L 1 mol NaI Qsp = [0.0050] [0.0050]2 = 1.2 x 10-7 Qsp = 1.2 x 10-7 > K = 1.4 x 10-8 A precipitate will form until the ion concentrations are lowered to the point that Q = K.

Example #3: Calculating ion concentrations in a precipitation reaction Determine all ion concentrations at equilibrium when 10.0 mL of a 0.050 M Pb(NO3)2 are mixed with 20.0 mL of 0.010 M Na2CrO4. General strategy suggested: Identify the precipitate Write the dissociation equation Write the Ksp expression. Obtain the Ksp value from a solubility chart Determine IF a precipitate forms (Q versus K calculation) Remember to divide by the total volume for concentrations of spectator ions. For those ions involved in the precipitation, assume the maximum amount of precipitate forms for an instant. Determine how much solid re-dissolves using –s to represent the change. The ions that form are +s and follow coefficient ratios. Substitute into the Ksp expression, solve and validate assumptions.

Example #3: The precipitate is PbCrO4 with the equation: PbCrO4  Pb2+ + CrO42- Ksp = [Pb2+][CrO42-] = 2.0 x 10-14 Initial concentration of Pb2+ = 0.050 M x 0.010 L = 0.017 M [Pb2+] 0.030 L Initial concentration of CrO42- = 0.020 M x 0.010 L = 0.0067 M [CrO42-] 0.030 L Q = [0.017] [0.0067] = 1.1 x 10-4 which is greater than 2.0 x 10-14 A precipitate will continue to form until Q = K. Spectator Ion concentrations are as follows: The NO3- concentration = (0.050 mol/L Pb(NO3)2 x 0.010 L x 2 mol NO3- = 0.033 M NO3- 0.030 L 1 mol Pb(NO3)2 The Na+ concentration = (0.010 mol/L Na2CrO4 x 0.020 L x 2 mol Na+ = 0.013 M Na+ 0.030 L 1 mol Na2CrO4

Example #3 (continued): Determining what ions remain is similar to a limiting reactant problem: Pb2+ + CrO42-  PbCrO4(s) initially (using mol instead of M) 5.0 x 10-4 2.0 x 10-4 0 assume maximum ppt forms: -2.0 x 10-4 -2.0 x 10-4 + 2.0 x 10-4 Assume all CrO42- reacts and its concentration is 0 for an instant. Amount (mol) left: 3.0 x 10-4 0 2.0 x 10-4 Amount of solid that re-dissolves: + s + s - s We’re not looking for what remains of the solid only ions Concentrations at equilibrium (M): 3.0 x 10-4 + s s solid 0.030 L (not ions) Assume + s is negligible 1.0 x 10-2 M + s s solid Plug back in to Ksp expression: Ksp= 2.0 x 10-14 = [1.0 x 10-2] [ s ] s = 2.0 x 10-12 = [CrO42-] Verify the negligibility of “+s”: 1.0 x 10-2M + (2.0 x 10-12 ) = 1.0 x 10-2 M the value of “s” is negligible compared to 1.0 x 10-2 FINAL ANSWER: Pb2+= 1.0 x 10-2 M CrO42-= 2.0 x 10-12 NO3- = 0.033 M Na+= 0.013 M

Solubility and the Common Ion Effect The common ion effect is the effect on the ionization of a weak acid or base of the addition of a common ion (the conjugate partner of the weak electrolyte). The effect was the suppression of the ionization. The effect is exactly the same in the case of a sparingly soluble salt. Addition of a common ion suppresses the solubility of the salt. In calculations involving a common ion, you can usually make the assumption that the concentration of the common ion is derived entirely from the added soluble salt, and not at all from the small amount of the sparingly soluble salt actually in solution (it would be considered negligible).

Solubility, Ion Separations and Quantitative Analysis Qualitative analysis is the determination of which substances exist in a sample, not how much of the substance exists in the sample. In the qualitative analysis of aqueous solutions, the task is often to determine what cations and what anions are present. The solubility of various salts can be employed to good effect to establish which ions are present. For example, if a solution contains silver and copper salts, and sodium sulfide is added, both will precipitate. If sodium chloride is added, only silver will precipitate. Such experiments allow the identification of both metal cations and anions in a solution. It’s really like a puzzle game of determining what must be present versus what is not present based on the solubility rules.

Simultaneous Equilibria Sometimes more than one reaction occurs in solution at the same time; such systems are called simultaneous equilibria. Suppose that sodium bromide (a source of bromide ion) is added to a saturated solution of silver chloride. What happens?

Simultaneous Equilibria For silver chloride: AgCl(s)  Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] = 1.8 x 10-10 For silver bromide: AgBr(s)  Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 3.3 x 10-13 Reverse this equation and add to the first equation: AgCl(s) + Br-(aq)  AgBr(s) + Cl-(aq) K = = 545 So the reaction is product favored due to the large number; the bromide ions cause precipitation of the silver bromide and chloride ions replace the bromide ions in solution. If this equation was reversed K = 0.0018 What does this tell us? The smaller the Ksp value in competing precipitation reactions, the greater the favorability of this product being formed. Ksp1 x _1_ Ksp2

Solubility and pH Any salt that contains an anion that is the conjugate base of a weak acid dissolves in water to a greater extent than indicated by the value of Ksp. This is because these anions are hydrolyzed in aqueous solution: X-(aq) + H2O(l)  HX(aq) + OH-(aq) Addition of a strong acid to this equilibrium removes hydroxide OH- and shifts the equilibrium further to the right (to replace the OH-). As a result, more salt dissolves. In general, the solubility of a salt containing the conjugate base of a weak acid is increased by addition of a strong acid (a decrease in pH).

Solubility and pH For example, consider the solution of magnesium hydroxide in water: Mg(OH)2(s)  Mg2+ + 2 OH- Ksp [Mg2+][OH-]2 = 1.5 x 10-11 This hydroxide is essentially insoluble. If, however a strong acid is added, the equilibrium is pulled to the right because the acid removes the hydroxide OH- and forms water: OH-(aq) + H3O+(aq)  2 H2O(l) K = 1/Kw = 1/1.0 x 10-14 Doubling this equation and adding it to the first equation produces the overall reaction. The equilibrium constant for the second equation must be squared and multiplied by the equilibrium constant for the first equation: [Mg2+][OH-]2 x = Mg(OH)2(s) + 2 H3O+(aq)  Mg2+ + 4 H2O(l) K = (1/Kw)2 x Ksp = ___1____ 2 x 1.5 x 10-11 = 1.5 x 1017 1.0 x 10-14 [H3O+]2 [Mg2+] [OH-]2 [H3O+]2 ____12____

Equilibria in the Environment The carbon dioxide cycle on this planet is chemistry on an immense scale. The atmosphere contains 700 billion tons of carbon dioxide (0.0325% of the atmosphere). This CO2 is in equilibrium with CO2 in the oceans (concentration about 10-5M). The oceans contain about 40 trillion tons of CO2 in the form of dissolved gas, as the bicarbonate or carbonate ions, as calcium or magnesium carbonates or as organic matter. Understanding the various equilibria involved in the carbon dioxide cycles allows a better understanding of the greenhouse effect, global warming and the capability of the ocean to absorb increasing amounts of carbon dioxide.

Example Problem: Calculating the molar solubility of a precipitate under varying conditions: The Ksp of zinc hydroxide, Zn(OH)2, at 25ºC is 1.8 x 10-14. a. Determine the molar solubility of Zn(OH)2 in pure water at 25ºC. b. Determine the molar solubility of Zn(OH)2 if 0.10 mol of Zn(NO3)2 are added to 1.0 L of a saturated solution of Zn(OH)2 at 25ºC (assume no volume change). c. Determine the molar solubility of Zn(OH)2 if the solution has an adjusted pH of 2.0 at 25ºC.

Example Problem: a. Determine the molar solubility of Zn(OH)2 in pure water at 25ºC First write the dissociation equation and the Ksp expression. Remember that solids do not appear in the expression. Place the data underneath the equation. The solid dissociates in a 1:2 ratio based on the coefficients in the equation. In solubility problems –x may be replaced with –s to refer to the amount that is soluble. Zn(OH)2(s)  Zn2+(aq) + 2 OH-(aq) Ksp = [Zn2+][OH-]2 Initial solid 0 0 Change -s +s +2s Equilibrium solid –s s 2s

Example Problem: a. Determine the molar solubility of Zn(OH)2 in pure water at 25ºC Substitute into the Ksp expression and solve: 1.8 x 10-14 = (s)(2s)2 = 4s3 because Zn2+ is 1:1 with Zn(OH)2 then: [Zn2+] = s = 1.6 x 10-5 = molar solubility of Zn(OH)2

Example Problem: b. Determine the molar solubility of Zn(OH)2 if 0 Example Problem: b. Determine the molar solubility of Zn(OH)2 if 0.10 mol of Zn(NO3)2 are added to 1.0 L of a saturated solution of Zn(OH)2 at 25ºC (assume no volume change). First write the dissociation equation and the Ksp expression. Remember that solids do not appear in the expression. Recall Stoichiometry: 0.10 mol Zn(NO3)2 = 0.10 mol Zn2+ Place the data underneath the equation. The solid dissociates in a 1:2 ratio. The amount of Zn2+ that results from the dissociation of Zn(OH)2 is insignificant when compared to the 0.10 M concentration. Ksp = [Zn2+][OH-]2 Zn(OH)2(s)  Zn2+(aq) + 2 OH-(aq) Zn(OH)2(s)  Zn2+(aq) + 2 OH-(aq) Initial solid 0.10 mol 0 1.0 L Equilibrium solid –s 0.10 M + s 2 s assume +s is negligible and [Zn2+] = 0.10 M

Example Problem: b. Determine the molar solubility of Zn(OH)2 if 0 Example Problem: b. Determine the molar solubility of Zn(OH)2 if 0.10 mol of Zn(NO3)2 are added to 1.0 L of a saturated solution of Zn(OH)2 at 25ºC (assume no volume change). Substitute into the Ksp expression and solve: 1.8 x 10-14 = (0.10)(2s)2 = s = 2.1 x 10-7 = molar solubility of Zn(OH)2 Validate the assumption that “s” is negligible compared to 0.10M. 0.10 M + (2.1 x 10-7) = 0.10 M

Example Problem: c. Determine the molar solubility of Zn(OH)2 if the solution has an adjusted pH of 2.0 at 25ºC. Write the dissociation equation and the Ksp expression. Th pH is defined for this solution and thus the OH- value is known. pH = 2.0 so pOH = 12.0 Place the data underneath the equation. Ksp = [Zn2+][OH-]2 Zn(OH)2(s)  Zn2+(aq) + 2 OH-(aq) pH = 2.00 pH + pOH = 14.00 pOH = 14.00 – 2.00 = 12.00 [OH] = 10-pOH = 1.0 x 10-12 Initial solid 0 1.0 x 10-12 Equilibrium solid –s s 1.0 x 10-12 (Remember that this value of OH- is already doubled.)

Example Problem: c. Determine the molar solubility of Zn(OH)2 if the solution has an adjusted pH of 2.0 at 25ºC. Substitute into the Ksp expression and solve: 1.8 x 10-14 = (s)(1.0 x 10-12)2 = s = 1.8 x 10-10 = molar solubility of Zn(OH)2 Note the decrease of solubility of Zn(OH)2(s) as you go from dissolving it in water (part a of this Example) to dissolving it in the presence of the common ion, 0.10 M Zn(NO3)2 (part b in the Example). Note the tremendous increase in solbuility of Zn(OH)2 in the presence of acid (part c in this Example); this precipitate dissolves in strong acidic solution.

Predicting formation of a Precipitate and Ion Concentration Calculations When mixing two solutions of soluble salts, a precipitate may occur, depending on the identity of the ions and their concentrations. Take note: You need to know the solubility rules and be familiar with common precipitates (list follows) for the AP Exam. You do not need to know any Ksp values but you do need to have a sense of which precipitates are very insoluble and which ones are only moderately insoluble. Be SURE to review some of the slightly soluble compounds and their Ksp values in Appendix D on page 1126-1127 of your textbook.

Solubility and Complex Ions Metal ions are solvated in aqueous solution; each metal ion is surrounded by water molecules. The solvated ion is called a complex ion and is often written as [M(H2O)6]n+. The production of a complex ion is an equilibrium reaction and has a Kf, formation constant, associated with it. The formation of a complex ion can cause a precipitate of low solubility to dissolve as the complex ion is formed. The formation of the complex ion is a Lewis acid-base reaction. Other Lewis bases, such as ammonia, also complex with metal ions and such reactions can influence the solubility of metal ions. Lewis bases are easily identified by the existence of exposed unshared pairs of electrons ready to donate. (recall definition)

Solubility and Complex Ions For example, consider the solution of silver chloride: AgCl(s)  Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] = 1.8 x 10-10 Addition of ammonia complexes the metal ion: Ag+(aq) + 2 NH3(aq)  [Ag(NH3)2]+ Kf = 1.6 x 107 Adding the two equations, and multiplying the equilibrium constants: AgCl(s) + 2 NH3(aq)  [Ag(NH3)2]+ + Cl-(aq) K = 2.9 x 10-3 Because metal ions invariably form complex ions in solution, Ksp calculations can be quite inaccurate when these reactions forming complex ions are ignored.

Final Objective Review Writing solubility product expressions Calculating Ksp from solubility or vice versa Calculating the solubility of a slightly soluble salt in a solution of a common ion Predicting whether precipitation will occur Determining the completeness of precipitation Determining the qualitative effect of pH on solubility