Solubility Equilibria. Dissolving a salt... Chem 30S Review Solubility Rules. Solubility Equilibrium: Dissociation = Crystalization

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1 Chem 30S Review Solubility Rules Solubility Equilibria Salts are generally more soluble in HOT water(gases are more soluble in COLD water) Alkali Metal salts are very soluble in water. NaCl, KOH, Li 3 PO 4, Na 2 SO 4 etc... Ammonium salts are very soluble in water. NH 4 Br, (NH 4 ) 2 CO 3 etc Salts containing the nitrate ion, NO 3-, are very soluble in water. Most salts of Cl -, Br - and I - are very soluble in water - exceptions are salts containing Ag + and Pb 2+. soluble salts: FeCl 2, AlBr 3, MgI 2 etc... insoluble salts: AgCl, PbBr 2 etc Dissolving a salt... A salt is an ionic compound - usually a metal cation bonded to a non-metal anion. The dissolving of a salt is an example of equilibrium. The cations and anions are attracted to each other in the salt. They are also attracted to the water molecules. The water molecules will start to pull out some of the ions from the salt crystal. At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions. However, soon the ions floating in the water begin to collide with the salt crystal and are pulled back in to the salt. (crystalization) Eventually the rate of dissociation is equal to the rate of crystalization. The solution is now saturated. It has reached equilibrium. 3 4 Solubility Equilibrium: Dissociation = Crystalization NaCl Crystal Dissolving NaCl in water Na + and Cl - ions surrounded by water molecules In a saturated solution, there is no change in amount of solid precipitate at the bottom of the beaker. Concentration of the solution is constant. The rate at which the salt is dissolving into solution equals the rate of crystalization. Dissolving silver sulfate, Ag 2 SO 4, in H2O When silver sulfate dissolves it dissociates into ions. When the solution is saturated, the following equilibrium exists: Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 (aq) Since this is an equilibrium, we can write an equilibrium expression for the reaction: Ksp = [Ag + ] 2 [SO 4 ] Ksp (solubility product constant) Notice that the Ag 2 SO 4 is left out of the expression! Why? Ag 2 SO 4 is a solid and not included in the expression. 5 6

2 Writing solubility product expressions... Some Ksp Values For each salt below, write a balanced equation showing its dissociation in water. Then write the Ksp expression for the salt. Iron (III) hydroxide, Fe(OH) 3 Nickel sulfide, NiS Silver chromate, Ag 2 CrO 4 Zinc carbonate, ZnCO 3 Calcium fluoride, CaF 2 Note: These are experimentally determined, and may be slightly different on a different Ksp table. 7 8 Calculate Solubility from Ksp Ksp for AgCl = 1.8 x 10-10, what is the molar solubility in pure water? AgCl (s) Ag + (aq) + Cl - (aq) Ksp = [Ag + ] [Cl - ] At equilibrium the concentrations of Ag + and Cl - are equal let x = [Ag + ] and also [Cl - ] 1.8 x = (x) (x) 1.8 x = x 2 Calculate Solubility from Ksp 1.8 x = x 2 Square root both sides 1.3 x 10-5 = x = [Ag + ] and [Cl - ] Applying the concentration values to the balanced solubility equation (1:1:1 ratio) indicates the molar solubility is 1.3 x 10-5 mol/l 9 10 Calculate Solubility from Ksp 2 Lead iodide (PbI 2 ) has a Ksp = 7.9 x 10-9 What is the molar solubility of PbI 2 in water? PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) Ksp = [Pb 2+ ] [I - ] 2 let x = [Pb 2+ ] and let 2x = [I - ] 7.9 x 10-9 = (x) (2x) 2 Calculate Solubility from Ksp x 10-9 = (x) (4x 2 ) 7.9 x 10-9 = 4x x 10-3 = x = [Pb 2+ ] Applying the concentration values to the balanced solubility equation (1:1:2 ratio) indicates the molar solubility is 1.3 x 10-3 mol/l 11 12

3 Algebraic form of Ksp For salts the algebraic form of the Ksp are AB: (x) (x) = x 2 = Ksp e.g. AgCl AB 2 : (x) (2x) 2 = 4x 3 = Ksp e.g. PbI 2 AB 3 : (x) (3x) 3 = 27x 4 = Ksp e.g. AlBr 3 A2B3: e.g. Au2O3 The solubility of Ag 2 CrO 4 is 6.7 x 10-5 mol/l What is the Ksp? Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2 (aq) Ksp = [Ag + ] 2 [CrO 4 2 ] The equilibrium concentrations of the ions are determined from the solubility of Ag 2 CrO 4 The ratios of the coefficients from the balanced solubility equation are used e.g. 1 Ag 2 CrO 4 : 2 Ag + there is a 1:2 ratio, if 6.7 x 10-5 mol/l is the Ag 2 CrO 4 solubility the [Ag + ] eq = 2 ( 6.7 x 10-5 ) = 1.3 x 10-4 M e.g. 1 Ag 2 CrO 4 : 1 CrO 2 4 there is a 1:1 ratio, if 6.7 x 10-5 mol/l is the Ag 2 CrO 4 solubility the [CrO 4 ] eq = 1( 6.7 x 10-5 ) = 6.7 x10-5 M Apply the equilibrium values to the Ksp Ksp = [Ag + ] 2 [CrO 4 2 ] = (1.3 x 10-4 ) 2 (6.7 x 10-5 ) = 1.1 x More Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2 (aq) s 2s s solubility equilm conc. Ksp = [Ag + ] 2 [CrO 4 2 ] = (2s) 2 (s) = (4s 2 )(s) = 4s 3 Cu(OH) 2 (s) Solubility 3.42 x 10-7 Answer: Ksp = 1.6 x Fe(OH) 3 (s) Solubility 9.9 x Answer: Ksp = 2.6 x s = 6.7 x 10-5 mol/l Ksp = 4(6.7 x 10-5 ) 3 = 1.1 x

4 Ksp and Solubility Generally, salts with very small solubility product constants (Ksp) are only slightly soluble in water. But be careful... When comparing the solubilities of two salts, you can sometimes simply compare the relative sizes of their Ksp values. This works if the salts have the same ratio of ions! For example CuI has Ksp = 5.0 x and CaSO 4 has Ksp = 6.1 x Since the Ksp for calcium sulfate is larger than that for the copper (I) iodide, we can say that calcium sulfate is more soluble. Do you see the problem here?? can t compare Ksp to predict solubility for compounds with different ratios Mixing Solutions - Will a Precipitate Form? p. 581 If 15 ml of M lead (II) nitrate is mixed with 30 ml of M potassium chromate - will a precipitate form? Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) PbCrO 4 (s) + 2 KNO 3 (aq) Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) PbCrO 4 (s) + 2 KNO 3 (aq) Step 1: Is an insoluble salt formed? We can see that a double replacement reaction can occur and produce PbCrO 4. Since this salt has a low solubility (very small Ksp), it may precipitate from the mixture. The solubility equilibrium is: PbCrO 4 (s) Pb 2+ (aq) + CrO 4 (aq) Ksp = 2 x = [Pb 2+ ][CrO 4 ] If a precipitate forms, it means the amount of ions in solution is greater than the equilibrium concentrations This will happen only if product of the ion concentrations (Qsp) > Ksp in our mixture Step 2: Find the concentrations of the ions that form the insoluble salt. Since we are mixing two solutions in this example, the concentrations of the Pb 2+ and CrO 4 will be diluted. We have to do a dilution calculation! Dilution: M 1 V 1 = M 2 V 2 Pb(NO 3 ) 2 : Pb 2+ and the K 2 CrO 4 : CrO 4 ratios are 1:1 therefore: Step 3: Calculate IP (or Qsp) for the mixture. ion product (IP or Qsp) the product of the molar conc. in an ionic solution (raised to an appropriate power) PbCrO 4 (s) Pb 2+ (aq) + CrO 4 (aq) Qsp = [Pb 2+ ][CrO 4 ] = ( M)(0.020 M) Qsp = 1.6 x 10-4 [Pb 2+ ] = M1V1 = (0.024M)(15 ml) = Pb 2+ V2 (45 ml) [CrO 4 ] = M1V1 = (0.030M)(30 ml) = CrO 4 V2 (45 ml) 23 24

5 Step 4: Compare Qsp to Ksp. Qsp = 1.6 x 10-4 > Ksp = 2 x Since Qsp > Ksp, a precipitate will form when the two solutions are mixed! Summary How do you know if a ppt forms? ion product > Ksp ppt. forms Note: If Qsp = Ksp, the solution is saturated If Qsp < Ksp, the solution is unsaturated Either way, no ppt will form! ion product = Ksp ion product < Ksp ppt. will not form Another ppt. problem with DILUTION! e.g. 50mL of M CaCl 2 and 50mL of M Na 2 SO 4 were mixed. Will a precipitate of CaSO 4 be formed? Ksp of CaSO 4 = 2.4 x 10-5 CaCl 2 (aq) + Na 2 SO 4 (aq) CaSO 4 (s) + 2 NaCl (aq) Step 1: Calc. The new ion concentrations using the new volume. In dilution problems we use: M i V i = M f V f The new volume is the combination of the two solutions i.e. 50 ml + 50 ml = 100 ml When calculating concentrations volume must be in litres (L) therefore we have L The new (final) concentration is calculated as follows: [CaCl 2 ] = M ivi = (0.0010)(0.050) = 5.0 x 10-4 M V f CaCl 2 Ca Cl - therefore [Ca 2+ ] = 5.0 x 10-4 M (1:1 ratio) M ivi (0.010)(0.050) [NaSO 4 ] = = = 5.0 x 10-3 M V f Na 2 SO 4 2 Na + + SO 4 therefore [SO 4 ] = 5.0 x 10-3 M (1:1 ratio) Step 2: calculate ion product We are looking for the formation of CaSO 4 precipitate therefore we use the following: CaSO 4 (s) Ca 2+ (aq) + SO 4 (aq) ion product = [Ca 2+ ] [SO 4 ] ion product = (5.0 x 10-4 ) (5.0 x 10-3 ) = 2.5 x 10-6 The Ksp for CaSO 4 is 2.4 x 10-5 < ion product Ksp 2.5 x x 10-5 therefore no precipitate! The Common Ion Effect The presence of a common ion in a solution will lower the solubility of a salt. shift left AgCl (s) Ag + (aq) + Cl - (aq) (adding NaCl) LeChatelier s Principle: The addition of the common ion will shift the solubility equilibrium to the left. This means that salt precipitates in the solution and the solubility is lowered. e.g. the addition of NaCl to a AgCl equilibrium lowers the solubility of the AgCl ( the common ion is Cl - ) 29 30

6 The Common Ion Effect on Solubility The solubility of MgF 2 in pure water is 2.6 x 10-4 mol/l. What happens to the solubility if we dissolve the MgF 2 in a solution of NaF, instead of pure water? Estimate the solubility of barium sulfate in a M sodium sulfate solution. The Ksp for barium sulfate is 1.1 x Write the equation and the equilibrium expression for the dissolving of barium sulfate. BaSO4(s) <=> Ba 2+ (aq) + SO4 (aq) Ksp = [Ba 2+ ][SO4 ] Make an "ICE" chart. Let "x" represent the barium sulfate that dissolves in the sodium sulfate solution expressed in moles per liter BaSO4(s) Ba 2+ (aq) SO4 (aq) I All solid M (from Na2SO4) C - x dissolves + x + x E Less solid x M + x Substitute into the equilibrium expression and solve for x. The assumption is that since x is going to be very small (the solubility is reduced in the presence of a common ion), the term " x" is the same as "0.020." (You can leave x in the term and use the quadratic equation or the method of successive approximations to solve for x, but it will not improve the significance of your answer.) 1.1 x = [x][ x] = [x][0.020] x = 5.5 x 10-9 mol/l Reduced Solubility BaSO Calculate molar solubility if Ca(IO3)2 in mol/l NaIO3. Ksp Ca(IO3)2 = 7.1 x 10-7 Ans: 2.0 x 10-4 mol/l Calculate the molar solubility of silver chloride in a 0.20 M sodium chloride solution. Ksp AgCl = 1.6 x Ans: 8.0 x mol/l Calc the molar solublity of silver chloride in a 1.5 x 10-3 mol/l silver nitrate solution. Ksp AgCl = 1.6 x Ans: 1.3 x 10-3 mol/l 35