Solubility Equilibrium

Chemical Equilibrium – Solubility The solubility of a substance is dependent on the forces holding the crystal together (the lattice energy) and the solvent acting on these forces. For now, we will consider only water as the solvent. As the solid dissolves, water molecules surround the ions in the solution by a process called hydration. During hydration, energy is released. The extent to which the energy of hydration is greater than the lattice energy determines the solubility. The equilibrium involved in the solubility of a substance is: MaXb (s) a Mn+ (aq) + b Xn- (aq) ( a general equation) where M is the cation of change n+ and X is the anion of charge n-. The subscripts of the salt become the coefficients for the ions. The equilibrium mass action expression would be: Ksp = [Mn+]a [Xn-]b Notice that only the aqueous ions are part of the mass action expression. Solids are never included. Also notice that the coefficient for each ion becomes the exponent in the mass action expression. There are four common types of solubility equilibria problems: solubility in pure water (will a precipitate form or finding the molar solubility), solubility in the presence of a common ion and selective precipitation. We will look at each of these in turn. Solubility in pure water Numerical problems involving solubility of a compound in pure water may fall into two categories: (1) determining whether a precipitate will form when two aqueous solutions are mixed or (2) determining the actual concentration of ions in a saturated solution. Let’s examine the mixing of two solutions. The problem might read as follows: Suppose 100.0 mL of a 0.0100 M CaCl2 solution and 200.0 mL of a 0.0200 M Na3PO4 solution were mixed. Would a precipitate form? To tackle this problem, you must do a stoichiometry problem first. Start by writing the balanced chemical equation: 3 CaCl2 (aq) + 2 Na3PO4 (aq) Ca3(PO4)2 (s) + 6 NaCl (aq) Convert this equation into net ionic form by crossing out the spectator ions from the total ionic form of the equation: 3 Ca2+ (aq) + 6 Cl¯(aq) + 6 Na+ (aq) + 2 PO43- (aq) Ca3(PO4)2 (s) + 6 Na+ (aq) + 6 Cl¯(aq) 3 Ca2+ (aq) + 2 PO43- (aq) Ca3(PO4)2 (s) Find the initial concentration of Ca2+ ions and PO43- ions: Next, determine the new, diluted concentration of calcium and phosphate ions resulting from the mixing of the two solutions: Calculate the solubility product quotient, Q, based on the mass action expression of the solubility equilibrium: Q = [Ca2+]3[PO43-]2 Q = [0.00333 M]3[0.0133 M]2 = 6.53 x 10-12 Compare the value of Q to the value of Ksp. If the Q Ksp , then the solubility has not been exceeded and a precipitate will not form. If the Q > Ksp, then the solubility has been exceeded and a precipitate will form. The Ksp of Ca3(PO4)2 is 1.0 x 10-26. Since Q, 6.53 x 10-12, exceeds the Ksp of 1.0 x 10-26, a precipitate of Ca3(PO4)2 will form in this case. A second type of solubility problem would ask to find the solubility of a pure substance in water. Suppose the question asks for the solubility of calcium phosphate in water. The equilibrium taking place is: Ca3(PO4)2 (s) 3 Ca2+ (aq) + 2 PO43- (aq) The Ksp of calcium phosphate is 1.0 x 10-26, and is either given in the problem or can be found on a table of solubility product constants. Below the equilibrium, show what is occurring initially, the change needed to establish equilibrium and the conditions at equilibrium: Ca3(PO4)2 (s) 3 Ca2+ (aq) + 2 PO43- (aq) Initial 0 0 Change + 3x + 2x Equilibrium 3x 2x Initially, calcium phosphate has not dissociated, thus the concentrations of calcium ions and phosphate ions is considered to be 0. Think of it as a reference point. To establish equilibrium, some calcium phosphate must dissociate as a 3x amount of calcium ions and 2x amount of phosphate ions. Thus, at equilibrium, the amount of calcium ions is 3x and the amount of phosphate ions is 2x. These values are substituted into the mass action expression: Ksp = [Ca2+]3[PO43-]2 = 1.0 x 10-26 = (3x)3(2x)2 = 1.0 x 10-26 As a reminder, Ca3(PO4)2 is a solid, and is therefore not a part of the mass action expression. Solve the above algebraic expression for x: 108 x5 = 1.0 x 10-26 x5 = 1.0 x 10-26/108 = 9.3 x 10-29 The value of x is the molar solubility of calcium phosphate. The actual concentrations of Ca2+ ions and PO43- ions in solution is given by: [Ca2+] = 3x = 3(2.5 x 10-6) = 7.5 x 10-6 M Ca2+ [PO43-] = 2x = 2(2.5 x 10-6) = 5.0 x 10-6 M PO43- Using the molar solubility, one may calculate the actual mass of calcium phosphate dissolved in one liter of water: Solubility in the presence of a common ion Suppose a salt of low solubility is found in the presence of a soluble salt in which the cation or anion of the soluble salt is common to the cation or anion of the insoluble salt. For instance, if a quantity of sodium phosphate is introduced to a saturated solution of calcium phosphate, what would happen to the solubility of calcium phosphate? Let’s look at this problem qualitatively, first. The common ion is the phosphate ion. According to LeChatelier’s Principle, if a stress is placed on an equilibrium, the equilibrium will shift in the direction which relieves the stress. The stress in this case is excess phosphate ion; the equilibrium will shift to the left, and one would see additional precipitate of calcium phosphate form. Ca3(PO4)2 (s) 3 Ca2+ (aq) + 2 PO43- (aq) 1143009906000 Equilibrium shifts to the left in the presence of excess phosphate ion. The lowered solubility of calcium phosphate becomes evident when the problem is approached quantitatively. A typical problem would read: What is the solubility of calcium phosphate in a 0.10 M solution of sodium phosphate? The equilibrium taking place is: Ca3(PO4)2 (s) 3 Ca2+ (aq) + 2 PO43- (aq) Keep in mind that while excess phosphate will be present from the sodium phosphate, sodium ion is just a spectator ion during the process. To start the problem, set up an equilibrium table: Ca3(PO4)2 (s) 3 Ca2+ (aq) + 2 PO43- (aq) Initial 0 0.10 M Change + 3x + 2x Equilibrium 3x 0.10 M + 2x Notice how the initial amount of phosphate is not zero. This initial amount of phosphate ion is contributed by the 0.10 M sodium phosphate solution. For equilibrium to become established, a 3x amount of calcium ion is formed and an additional 2x amount of phosphate ion from the dissociation of calcium phosphate is introduced. Thus, at equilibrium, the amount of calcium ion in solution is 3x and the amount of phosphate ion in solution is 0.10 + 2x. Substitute the algebraic terms into the mass action expression: Ksp = [Ca2+]3[PO43-]2 = 1.0 x 10-26 = (3x)3(0.10 + 2x)2 = 1.0 x 10-26 The above expression is a polynomial and its solution may require a serial approximation method. However, solving for x may be simplified if the 2x term in the second factor may be neglected. Recall that the solubility, x in pure water, had a magnitude of 10-6. Comparing a magnitude of 10-6 to 0.10, one can see that 2x will be negligibly small, thus can be neglected in the above expression. This simplifies the equation to: Ksp = (3x)3(0.10)2 = 0.27 x3 = 1.0 x 10-26 Solve for x by dividing through both sides by 0.27: x3 = 1.0 x 10-26/0.27 = 3.7 x 10-26 Next, take the cube root of both sides of the expression: As one can see, x, the solubility is quite small. Our assumption to neglect the 2x in the original algebraic expression was valid. Also, one can quantitatively see how the solubility has decreased, as predicted by LeChatelier’s Principle. In pure water, the solubility was 2.5 x 10-6; in the presence of a common ion, phosphate ion with a concentration of 0.10 M, the solubility of calcium phosphate decreased to 3.3 x 10-9. Selective precipitation Many compounds have variable solubility. This difference in solubility has a practical application when attempting to separate metal ions in solution. Water purification treatment plants use carbonates and sulfates to remove heavy metal cations from your drinking water. Qualitative analysis schemes in a general chemistry lab course often uses selective precipitation as the basis for identification of ions. A saturated hydrogen sulfide solution (0.10 M H2S) is often used. This solution can be adjusted to a desired sulfide concentration by changing the pH of the solution. By controlling the sulfide ion concentration, one can selectively precipitate out the less soluble metal sulfide. Recall in an earlier problem we examined how the reaction quotient, Q, compared to the Ksp. This concept is applied here. If the sulfide ion concentration in solution exceeds the solubility of one cation and not the other, the less soluble metal sulfide precipitates (Q > Ksp), and the more soluble metal sulfide remains in solution (Q Ksp). Suppose one wishes to separate nickel(II) cations with a concentration of 0.20 M from copper(II) cations with a concentration of 0.10 M. What pH must one adjust a saturated hydrogen sulfide solution to achieve the separation and what is the concentration of nickel(II) and copper(II) ions remaining in solution? First, consider all the equilibria involved: H2S(aq) H+ (aq) + HS¯(aq) Ka1 = 8.9 x 10-8 HS¯(aq) H+ (aq) + S2- (aq) Ka2 = 1.2 x 10-13 NiS (s) Ni2+ (aq) + S2- (aq) Ksp = 3 x 10-19 CuS (s) Cu2+ (aq) + S2- (aq) Ksp = 6 x 10-36 Determine the maximum sulfide concentration for each metal ion: The above calculation shows nickel(II) sulfide is many times more soluble than copper(II) sulfide. If the sulfide ion concentration is set to the more soluble [S2-]max of nickel(II) sulfide ([S2-]max = 1.5 x 10-18), copper(II) sulfide will precipitate and nickel(II) sulfide will remain in solution. The sulfide ion concentration is adjusted using a saturated hydrogen sulfide solution as a function of pH. Since hydrogen sulfide is a diprotic acid, there are two sources of hydrogen ions (H2S and HS¯), as shown by the first two equilibria, above. It will be easier to solve for hydrogen ion concentration if only one mass action expression were involved. With this in mind, combine the two equilibria into one overall equilibrium: H2S(aq) H+ (aq) + HS¯(aq) Ka1 = 8.9 x 10-8 HS¯(aq) H+ (aq) + S2- (aq) Ka2 = 1.2 x 10-13 H2S (aq) 2 H+ (aq) + S2- (aq) Koverall = Ka1Ka2 = 1.1 x 10-20 The mass action expression for the above equilibrium is: Rearrange the above equation to solve for the hydrogen ion concentration. Multiply both sides of the equation by hydrogen sulfide concentration and divide both sides by the sulfide ion concentration: Take the square root of both sides of the equation: The concentration of saturated solution of hydrogen sulfide is 0.10 M. For sulfide ion concentration, substitute the more soluble sulfide ion concentration calculated previously. Substitute for Koverall. Solve for hydrogen ion concentration: Finally, calculate the pH: pH = - log[H+] = - log(0.027) = 1.57 This pH represents the maximum pH which must be maintained in order to separate nickel(II) ions and copper(II) ions. Since the sulfide ion concentration is 1.5 x 10-18 M, the actual copper ions concentration still remaining in solution is: As one can see, the copper(II) ion concentration is indeed very negligible. The concentration of nickel(II) ions remains 0.20 M as long as the sulfide ion concentration is less than1.5 x 10-18 M. Problems 1. Calculate the molar solubility of the following compounds: a. PbCO3 Ksp = 7.4 x 10-14 b. Ag2CrO4 Ksp = 1.3 x 10-12 c. La(IO3)3 Ksp = 1.2 x 10-11 d. Fe3(PO4)2 Ksp = 1.0 x 10-36 2. Convert the molar solubility of each compound in problem 1 to the solubility expressed in grams /L. 3. Can one predict the relative solubility of a series of compounds such as the list of compounds in problem 1 by using only the Ksp value? Use BaCO3 and SrF2 as a test of your hypothesis. Ksp of BaCO3 = 5.0 x 10-9 Ksp of SrF2 = 3.2 x 10-9 4. If the carbonate ion concentration in a saturated lead(II) carbonate solution is adjusted to 0.020M, what is the concentration of lead(II) ion in solution? How does this quantitatively demonstrate LeChatelier’s Principle? Kxp for lead(II) carbonate is 7.4 x 10-14 5. If the phosphate ion concentration in a saturated iron(II) phosphate solution is adjusted to 0.50M, what is the concentration of iron(II) ion in solution? How does this quantitatively demonstrate LeChatelier’s Principle? Ksp for iron(II) phosphate is 1 x 10-36 6. Carbonate ion is added drop wise to a solution containing 0.30M Ca2+ and 0.30 M Ni2+. The Ksp for CaCO3 is 4.5 x 10-9 and the Ksp for NiCO3 is 1.3 x 10-7. a. Which cation will precipitate first? b. What will be the concentration of the less soluble cation remaining in solution when the more soluble cation begins to precipitate? 7. What is the concentration of Co(III) ions in solution if the pH of the solution is 9.00? Ksp for Co(OH)3 = 3.0 x 10-45 8. Will a precipitate form if 400 mL of 2.0 x 10-4 M Ca(NO3)2 is mixed with 600 mL of 1.5 x 10-5M KIO3? Ksp of Cd(IO3)2 = 2.3 x 10-8 9. Will a precipitate form if 300 mL of 5.0 x 10-3M Ca(NO3)2 is mixed with 400 mL of 3.5 x 10-4 M Na2CO3? Ksp of CaCO3 = 4.5 x 10-9 10. a. When additional Ba2+ ions are added to a saturated Ba3(PO4)2, one observes additional precipitate from in the container. Explain this observation in terms of LeChatelier’s Principle. b. When a saturated solution of Ba3(PO4)2 is acidified, the precipitate dissolves. Explain. c. When the solution from 10 b. is treated with base, the precipitate reappears. Explain. Answers 1. a.The equilibrium taking place is: PbCO3(s) Pb2+(aq) + CO32-(aq) The equilibrium constant expression is: Ksp = [Pb2+][CO32-] = 7.4 x 10-14. Notice that a denominator term is absent (actually equal to one), because the reactant in the chemical equilibrium is a solid. Equal amounts of lead(II) ions and carbonate ions are produced in solution for every lead(II) carbonate particle that dissociates. Let us use “x” for the amount of each of these ions in the equilibrium constant expression: Ksp = [Pb2+][CO32-] = 7.4 x 10-14 = (x)(x) = x2 The molar solubility will be “x” and represents the moles of solute soluble in one liter of solution: x = 7.4 x 10-14 = 2.7 x 10-7 M 1. b. The equilibrium taking place is: Ag2CrO4(s) 2 Ag+(aq) + CrO42-(aq) The equilibrium constant expression is: Ksp = [Ag+]2[CrO42-] = 1.3 x 10-12. Notice that a denominator term is absent (actually equal to one), because the reactant in the chemical equilibrium is a solid. The amount of silver(I) cation produced in solution is twice the amount of chromate ion. Let us use “2x” for the amount of silver(I) ion, and “x” for the amount of chromate ion in the equilibrium constant expression: Ksp = [Ag+]2[CO32-] = 1.3 x 10-12 = (2x)2(x) = 4x3 The molar solubility will be “x” and represents the moles of solute soluble in one liter of solution: 4x3 = 1.3 x 10-12 x3 = 1.3 x 10-124 = 3.25 x 10-13 x = 33.25 x 10-13 = 6.9 x 10-5 M 1. c. The equilibrium taking place is: La(IO3)3(s) La3+(aq) + 3 IO3¯(aq) The equilibrium constant expression is: Ksp = [La3+][ IO3¯]3 = 1.2 x 10-11. Notice that a denominator term is absent (actually equal to one), because the reactant in the chemical equilibrium is a solid. The amount of iodate ion produced in solution is three times as much as the amount of lanthanum(III) ions. Let us use “x” for the amount of lanthanum(II) ions and “3x” for the amount of iodate ions in the equilibrium constant expression: Ksp = [La3+][IO3¯] = 1.2 x 10-11 = (x) (3x)3 = 27x4 The molar solubility will be “x” and represents the moles of solute soluble in one liter of solution: 27x4 = 1.2 x 10-11 x4 = 1.2 x 10-1127 = 4.4 x 10-13 x = 44.4 x 10-13 = 8.2 x 10-4 M 1. d. The equilibrium taking place is: Fe3(PO4)2(s) 3 Fe2+(aq) + 2 PO43- (aq) The equilibrium constant expression is: Ksp = [Fe2+]3[ PO43-]2 = 1.0 x 10-36. Notice that a denominator term is absent (actually equal to one), because the reactant in the chemical equilibrium is a solid. Three particles of iron(II) ions and two particles of phosphate ions form in solution for every particle of iron(II) phosphate. Let us use “3x” for the amount of iron(II) ions in solution and “2x” for the amount of phosphate ions in solution: Ksp = [Fe2+]3[ PO43-]2 = 1.0x 10-36 = (3x)3(2x)2 = 108x5 The molar solubility will be “x” and represents the moles of solute soluble in one liter of solution: 108x5 = 1.0 x 10-36 x5 = 1.0 x 10-36108 = 9.3 x 10-39 x = 59.3 x 10-39 = 2.5 x 10-8 M 2. Since the molar solubility has units of moles solute per liter of solution, convert the moles of solute to grams using the molar mass of the compound. Using the molar solubilities calculated from problem 1: a. Molar mass of PbCO3 = AW Pb + AW C + 3 AW O where AW = atomic weight in g/mol MM PbCO3 = 207.2 g/mol + 12.01 g/mol + 3(16.00 g/mol) = 235.2 g/mol 2.7 x 10-7 mol PbCO31 L solution x 235.2 g PbCO31 mol PbCO3 = 6.4 x 10-5 g PbCO3/L soln b. Molar mass of Ag2CrO4 = 2 AW Ag + AW Cr + 4 AW O MM Ag2CrO4 = 2(107.9 g/mol) + 52.00 g/mol + 4(16.00 g/mol) = 331.8 g/mol 6.9 x 10-5 mol Ag2CrO4L solution x 331.8 g Ag2CrO41 mol Ag2CrO4 = 0.023 g Ag2CrO4/L soln 2. c. Molar mass of La(IO3)3 = AW La + 3 AW I + 9 AW O MM La(IO3)3 = 138.9 g/mol + 3(126.9 g/mol) + 9(16.00 g/mol) = 663.6 g/mol 8.2 x 10-4 mol La(IO3)31 L solution x 663.6 g La(IO3)31 mol La(IO3)3 = 0.54 g La(IO3)3/L soln d. Molar mass of Fe3(PO4)2 = 3 AW Fe + 2 AW P + 8 AW O MM Fe3(PO4)2 = 3(55.85 g/mol) + 2(30.97 g/mol) + 8(16.00 g/mol) = 357.5 g/mol 2.5 x 10-8 mol Fe3(PO4)21 L solution x 357.5 g Fe3(PO4)21 mol Fe3(PO4)2 = 8.9 x 10-6 g Fe3(PO4)2/L soln 3. One cannot easily predict the relative solubility of a series of compounds solely by looking at the Ksp value. For BaCO3, the Ksp = 5.0 x 10-9; for SrF2, the Ksp = 3.2 x 10-9. Initially one might think that the larger equilibrium constant for BaCO3 means the equilibrium lies further to the right, and thus BaCO3 would have a higher solubility. However, upon calculation of the actual molar solubility of each of these compounds, one discovers that the opposite is true. Notice that the numbers of particles produced by each compound in solution are not identical. BaCO3 produces 2 moles of ions per mole of substance in solution while SrF2 produces 3 moles of ions per mole of substance: BaCO3(s) Ba2+ (aq) + CO32? (aq) SrF2(s) Sr2+(aq) + 2 F¯(aq) An approximation could be done quickly by noting that one will have to take the square root of the Ksp of barium carbonate, while one will be taking the cube root of the Ksp of strontium fluoride. This will make the magnitude of the molar solubility of barium carbonate around 10-5 and the magnitude of the molar solubility of strontium fluoride around 10-3. Thus, even though it appears to have a smaller equilibrium constant, the strontium fluoride turns out to be more soluble. For those who are still skeptical, calculate the actual molar solubility of each compound: Ksp = [Ba2+][ CO32?] = 5.0 x 10-9 = x2 molar solubility = x = 5.0 x 10-9 = 7.1 x 10-5 M Ksp = [Sr2+][ F?]2 = 3.2 x 10-9 = (x)(2x)2 = 4x3 molar solubility = x = 33.2 x 10-94 = 9.3 x 10-4 M 4. This problem demonstrates LeChatelier’s Principle. Qualitatively, the stress is excess lead(II) ion which would drive the equilibrium towards products and lower the solubility. Let’s compare the solubility in pure water to the solubility in 0.020 M lead(II) ion. The equilibrium involved is: PbCO3(s) Pb2+(aq) + CO32? (aq) The equilibrium constant expression for this equilibrium would be: Ksp = [Pb2+][ CO32?] = 7.4 x 10-14 Equal amounts of lead(II) ions and carbonate ions are produced for every particle of lead(II) carbonate. Thus an “x” amount of both lead(II) and carbonate ions are produced in solution: Ksp = [Pb2+][ CO32?] = 7.4 x 10-14 = (x)(x) = x2 Solving for x gives the molar solubility in pure water: Molar solubility = x = 7.4 x 10-14 = 2.7 x 10-7 By adjusting lead(II) cations into solution to 0.020 M, the amount of lead(II) is no longer “x”. Adding extra lead(II) cations into solution totally disrupts the initial equilibrium. We can make the assumption that the intial concentration of lead(II) is 0.020 M, and that the initial concentration of carbonate after the equilibrium was stressed is zero. To re-establish the equilibrium, an x amount of additional lead(II) cations must form for an x amount of carbonate ions: Ksp = [Pb2+][ CO32?] = 7.4 x 10-14 = (0.020 + x)(x) Using a quadratic equation will not be necessary here. One can assume the “x” amount of additional lead(II) cations will be negligibly small. The approximation can be tested once the solution to the problem is calculated. This simplifies the above expression to: Ksp = [Pb2+][ CO32?] = 7.4 x 10-14 = (0.020)(x) Solving for x: x = 7.4 x 10-140.020 = 3.7 x 10-12 M Notice that the solubility dropped from 2.7 x 10-7 in a pure solution of lead(II) carbonate to 3.7 x 10-12 in a solution to which additional carbonate ions was introduced. The solubility is approximately 105 times lower! Also notice that our assumption that the additional “x” amount of lead(II) ions was negligibly small was valid: 0.020 + 3.7 x 10-12 = 0.020M. Thus, these results show quantitatively LeChatelier’s Principle in action. The stress of excess carbonate shifted the equilibrium towards reactants as indicated by the lower solubility. 5. The equilibrium constant expression for lead(II) phosphate is shown below. There will be a 3x amount of lead(II) ions in solution from the lead(II) precipitate. The phosphate ion concentration will be 0.50 M from whatever source of phosphate was used to adjust the concentration plus an additional 2x amount from the dissociation of lead(II) phosphate precipitate. Pb3(PO4)2 (s) 3 Pb2+(aq) + 2 PO43-(aq) The equilibrium constant expression is: Ksp = [Pb2+]3 [PO43-]2 = 1 x 10-36 Substitute the algebraic amounts into the above expression: Ksp = [Pb2+]3 [PO43-]2 = 1 x 10-36 = (3x)3(0.50 + 2x)2 Luckily, the equilibrium constant is very small (1 x 10-6). This allows us to neglect the additional 2x amount of phosphate from the dissociation of lead(II) phosphate. Otherwise, we would have to solve a polynomial equation using a serial approximation method. With our approximation, the equation simplifies to: 1 x 10-36 = (3x)3(0.50)2 = (27x3)(0.252) = 6.75x3 Solve for x: x3 = 1 x 10-366.75 = 1.5 x 10-37 x = 31.5 x 10-37 = 5.3 x 10-13 3x = [Pb2+] = 3(5.3 x 10-13) = 1.6 x 10-12 M To demonstrate how this quantitatively demonstrates LeChatelier’s Principle, we will compare this concentration of lead(II) ions in solution that the lead(II) ion concentration that would be present if the stress of additional phosphate were not added. To do this we will find the lead(II) ion concentration in a pure solution of saturated lead(II) phosphate. Using the same equilibrium constant expression, there will be a 3x amount of lead(II) ions in solution and a 2x amount of phosphate ions in solution at equilibrium. Substituting these algebraic amounts into the equilibrium constant expression gives: Ksp = [Pb2+]3 [PO43-]2 = 1 x 10-36 = (3x)3(2x)2 Solve for x: 1 x 10-36 = (3x)3(2x)2 = 108x5 x5 = 1 x 10-36108 = 9.3 x 10-39 x = 59.3 x 10-39 = 2.5 x 10-8 M 3x = [Pb2+] = 3(2.5 x 10-8) = 7.5 x 10-8 M A comparison of the results shows that the lead(II) concentration has been significantly reduced upon introducing additional phosphate into solution: 1.6 x 10-12 M Pb2+ in 0.50 M PO43- << 7.5 x 10-8 M Pb2+ in pure, saturated lead(II) phosphate solution. The stress placed on the equilibrium was the additional phosphate, hence the equilibrium shifted towards solid lead(II) phosphate to reduce the stress and the lead(II) ion concentration in solution has been decreased. 6. a. This problem has competing equilibria. To determine which compound will precipitate first, find the molar solubility of each compound. The equilibrium for calcium carbonate is: CaCO3(s) Ca2+(aq) + CO32-(aq) The equilibrium constant expression for the above equilibrium is: Ksp = [Ca2+][CO32-] = 4.5 x 10-9 The initial concentration of calcium ion is 0.30 M. Once precipitation begins, every particle of calcium carbonate produces one calcium ion and one carbonate ion in solution, so we will use x as the unknown amount of additional calcium ions and an x amount of carbonate ions produced in solution. Substitute these algebraic amounts into the equilibrium constant expression: Ksp = [Ca2+][CO32-] = 4.5 x 10-9 = (0.30 + x)(x) Since the equilibrium constant of calcium carbonate is very small, we can neglect the additional x amount of calcium ions coming from the dissociation of calcium carbonate. This simplifies the problem of solving for x. The x amount represents the maximum amount of carbonate ion that will remain in solution. Any concentration exceeding this amount will cause calcium carbonate to precipitate. x = [CO32-] = Ksp[Ca2+] = 4.5 x 10-90.30 = 1.5 x 10-8 M Thus, any carbonate ion concentration greater than 1.5 x 10-8 M will cause calcium carbonate to precipitate under these conditions. We will do the analogous calculation for the nickel(II) carbonate. The equilibrium for nickel(II) carbonate is: NiCO3(s) Ni2+(aq) + CO32-(aq) The equilibrium constant expression for the above equilibrium is: Ksp = [Ni2+][CO32-] = 1.3 x 10-7 The initial concentration of nickel(II) ion is 0.30 M. Once precipitation begins, every particle of nickel(II) carbonate produces one nickel(II) ion and one carbonate ion in solution, so we will use x as the unknown amount of additional nickel(II) ions and an x amount of carbonate ions produced in solution. Substitute these algebraic amounts into the equilibrium constant expression: Ksp = [Ni2+][CO32-] = 1.3 x 10-7 = (0.30 + x)(x) Since the equilibrium constant of nickel(II) carbonate is very small, we can neglect the additional x amount of nickel(II) ions coming from the dissociation of nickel(II) carbonate. This simplifies the problem of solving for x. The x amount represents the maximum amount of carbonate ion that will remain in solution. Any concentration exceeding this amount will cause nickel(II) carbonate to precipitate. x = [CO32-] = Ksp[Ni2+] = 1.3 x 10-70.30 = 4.3 x 10-7 M Thus, any carbonate ion concentration greater than 4.3 x 10-7 M will cause nickel(II) carbonate to precipitate under these conditions. Based on these calculations, one can see that calcium carbonate is less soluble than nickel(II) carbonate, hence calcium carbonate will precipitate first. (1.5 x 10-8 M [CO32-] < 4.3 x 10-7 M [CO32-]) 6. b. One must find the concentration of calcium at the carbonate ion concentration at which nickel(II) carbonate begins to precipitate. The solubility of nickel(II) carbonate is 4.3 x 10-8 M as found in the previous problem. This means that any carbonate ion concentration greater than 4.3 x 10-7 M will exceed the solubility limit of nickel(II) carbonate and precipitation will begin. Using this concentration of carbonate, we can calculate what concentration of calcium ion remains in solution: Ksp = [Ca2+][CO32-] = 4.5 x 10-9 [Ca2+] = Ksp[CO32-] = 4.5 x 10-94.3 x 10-7 = 1.0 x 10-2 M Thus, at the point that nickel(II) carbonate begins to precipitate, 1.0 x 10-2 M Ca2+ ions remain in solution. 7. The equilibrium taking place is: Co(OH)3(s) Co2+(aq) + 3 OH¯(aq) The Ksp expression for this equilibrium is: Ksp = [Co2+][OH¯]3 = 3.0 x 10-45 In this equilibrium, the hydroxide ion concentration has been adjusted by controlling the pH by introducing hydroxide from a source other than cobalt(III) hydroxide . The initial concentration of hydroxide is calculated from the pH: pOH = 14 – pH = 14 – 9.00 = 5.00 [OH¯] = 10 ?pOH = 10?5.00 = 1.0 x 10-5 M At equilibrium, there will be an x amount of cobalt(II) ions present and 1.0 x 10-5 plus an additional 3x amount of hydroxide from the dissociation of cobalt(III) hydroxide. We can substitute these algebraic expressions into the Ksp expression: Ksp = [Co2+][OH¯]3 = 3.0 x 10-45 = (x)(1.0 x 10-5 + 3x)3 Due to the low solubility of this compound, we can simplify the above algebraic expression by neglecting the 3x amount of additional hydroxide. This simplifies the above polynomial expression to: Ksp = [Co2+][OH¯]3 = 3.0 x 10-45 = (x)(1.0 x 10-5)3 Solve for x which is equal to the [Co2+]: x = [Co2+] = Ksp(1.0 x 10-5)3 = 3.0 x 10-451.0 x 10-15 = 3.0 x 10-30 8. . From the solubility rules, one can write a chemical reaction that result from the mixing of these two solutions: Ca(NO3)2(aq) + Cd(IO3)2(aq) ? Cd(IO3)2(s) + NaNO3(aq) We need to find out whether or not the solubility limit for cadmium iodate has been exceeded from the mixing the two solutions. Calculating the initial concentration of cadmium ions and iodate ions is basically a dilution problem: C1V1 = C2V2 For Cd2+ , C1 = 2.0 x 10-4 M, V1 = 400 mL and V2 = 600 mL + 400 mL = 1000 mL = total volume Cd2+= 400 mL x 2.0 x 10-4 mol Cd(NO3)2L x 1mol Cd2+1 mol Cd(NO3)2 x 11000 mL = 8.0x 10-5 M For IO3¯, C1 = 1.5 x 10-3 M, V1 = 600 mL and V2 = 600 mL + 400 mL = 1000 mL = total volume IO3?= 600 mL x 1.5 x 10-3 mol KIO3L x 1mol IO3-1 mol KIO3 x 11000 mL = 9.0 x 10-4 M Calculate the solubility quotient, Qsp, and compare this to the Ksp for calcium carbonate: Qsp = [Cd2+][IO3-]2 Qsp = (8.0 x 10-5)(9.0 x 10-4)2 = 6.5 x 10-11 Compare the Qsp with the Ksp of cadmium iodate. The Ksp of cadmium iodate is 2.3 x 10-8. 6.5 x 10-11 < 2.3 x 10-8 Qsp < Ksp Since the Qsp < Ksp, the solubility limit for calcium carbonate has not been exceeded, and thus a precipitate will not form. 9. From the solubility rules, one can write a chemical reaction that result from the mixing of these two solutions: Ca(NO3)2(aq) + Na2CO3(aq) ? CaCO3(s) + 2 NaNO3(aq) We need to find out whether or not the solubility limit for calcium carbonate has been exceeded from the mixing the two solutions. Calculating the initial concentration of calcium ions and carbonate ions is basically a dilution problem: C1V1 = C2V2 For Ca2+ , C1 = 5.0 x 10-3M, V1 = 400 mL and V2 = 300 mL + 400 mL = 700 mL = total volume Ca2+= 300 mL x 5.0 x 10-3 mol Ca(NO3)2L x 1mol Ca2+1 mol Ca(NO3)2 x 1700 mL = 2.1x 10-3 M For CO32-, C1 = 3.5 x 10-4 M , V1 = 400 mL and V2 = 300 mL + 400 mL = 700 mL = total volume CO32-= 400 mL x 5.0 x 10-3 mol Ca(NO3)2L x 1mol CO32-1 mol Ca(NO3)2 x 1700 mL = 2.9 x 10-3 M Calculate the solubility quotient, Qsp, and compare this to the Ksp for calcium carbonate: Qsp = [Ca2+][CO32-] Qsp = (2.1 x 10-3)(2.9 x 10-3) = 6.1 x 10-6 Compare the Qsp with the Ksp of calcium carbonate. The Ksp of calcium carbonate is 4.5 x 10-9. 6.1 x 10-6 > 4.5 x 10-9 Qsp > Ksp Since the Qsp > Ksp, the solubility limit for calcium carbonate has been exceeded, and thus a precipitate will form. 10. The equilibrium involved is: Ba3(PO4)2(s) 3 Ba2+(aq) + 2 PO43-(aq) a. Addition of barium ions into the solution creates a stress on the products side of the equilibrium (too much barium ions), so the equilibrium will shift towards reactants to remove the excess barium ions. Thus, one observes additional precipitate form in the container. b. Addition of hydronium ions from an acid will consume phosphate ions by: H3O+(aq) + PO43-(aq) HPO42-(aq) + H2O(l) The stress created by this formation of hydrogen phosphate ions is the removal of phosphate ions from the solubility equilibrium (not enough phosphate), so the equilibrium will shift towards products to replenish the phosphate ions removed. Thus, one observes the precipitate dissolve. c. Addition of hydroxide ions will deplete hydronium ions from solution: H3O+(aq) + OH¯(aq) 2 H2O(l) This also removes hydronium ion from the previous equilibrium: H3O+(aq) + PO43-(aq) HPO42-(aq) + H2O(l) The stress on this equilibrium is now too little hydronium, so the equilibrium will shift towards reactants to replenish the hydronium removed by hydroxide ion. As a result, the phosphate ion concentration in solution will increase. This increase of phosphate ion will now impact the solubility equilibrium by creating a stress of too much phosphate: Ba3(PO4)2(s) 3 Ba2+(aq) + 2 PO43-(aq) The equilibrium will shift towards reactants to remove the excess phosphate, and one will observe the reappearance of the precipitate of barium phosphate.