Solubility Products. Solubility Products. Solubility Products. Solubility Products. Slide 2 / 57. Slide 1 / 57. Slide 3 / 57.

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1 Slide 1 / 57 Slide 2 / 57 Products queous equilibria II Products onsider the equilibrium that exists in a saturated solution of aso 4 in water: aso 4 (s) a 2+ (aq) + SO 4 2- (aq) Slide 3 / 57 Products The equilibrium constant expression for this equilibrium is K sp = [a 2+ ] [SO 4 2 ] where the equilibrium constant, K sp, is called the solubility product. Slide 4 / 57 1 Which Ksp expression is correct for gl? [g + ]/[l - ] [g + ][l - ] [g 2+ ] 2 [l 2- ] 2 [g + ] 2 [l - ] 2 None of the above. There is never any denominator in K sp expressions because pure solids are not included in any equilibrium expressions. Slide 5 / 57 2 Given the reaction at equilibrium: Zn(OH) 2 (s) <--> Zn 2+ (aq) + 2OH - (aq) what is the expression for the solubility product constant, K sp, for this reaction? K sp = [Zn 2+ ][OH - ] 2 / [Zn(OH) 2 ] K sp = [Zn(OH) 2 ] / [Zn 2+ ][2OH - ] Slide 6 / 57 Products K sp is not the same as solubility. is generally expressed as the mass of solute dissolved in 1 L (g/l) or 100 ml (g/ml) of solution, or in mol/l (M). K sp = [Zn 2+ ][2OH - ] K sp = [Zn 2+ ][OH - ] 2

2 Slide 7 / 57 The term "solubility" represents the maximum amount of solute that can be dissolved in a certain volume before any precipitate is observed. The solubility of a substance can be given in terms of or in terms of grams per liter moles per liter g/l mol/l The latter is sometimes referred to as "molar solubility." 2.5 L 1 Slide 9 / 57 What is the maximum amount (in grams) of a2o4 that could dissolve in 2.5 L (before a precipitate occurs)? x The solubility of a2o4 is 1.3 x 10-3 mol/l. a2o4 --> a O4 2-1 mol a 2O g = x 10-3 mol a 2O 4 x g = 0.293g a 2O 4 x 2.5L = 0.73g 1L This is the maximum amount that could dissolve in 2.5 L before a precipitate occurs. 1.3 x 10-3 mol a2o4 1 L g a2o4 x 1 mol a2o4 Slide 8 / 57 xample #1 onsider the slightly soluble compound barium oxalate, a 2 O 4. The solubility of a 2 O 4 is 1.3 x 10-3 mol/l. The ratio of cations to anions is 1:1. This means that 1.3 x 10-3 moles of a 2+ can dissolve in one liter. lso, 1.3 x 10-3 moles of 2 O 4 2- can dissolve in one liter. What is the maximum amount (in grams) of a 2 O 4 that could dissolve in 2.5 L (before a solid precipitate or solid settlement occurs)? Slide 10 / 57 xample #2 onsider the slightly soluble compound lead (II) chloride, Pbl 2. The solubility of Pbl 2 is mol/l. The ratio of cations to anions is 1:2. This means that moles of Pb 2+ can dissolve in one liter. Molar solubility always refers to the ion with the lower molar ratio. Twice as much, or 2(0.016) = moles of l - can dissolve in one liter. Slide 11 / 57 xample #3 onsider the slightly soluble compound silver sulfate, g 2 SO 4. The solubility of g 2 SO 4 is mol/l. The ratio of cations to anions is 2:1. This means that moles of SO 4 2- can dissolve in one liter. Twice as much, or 2(0.015) = moles of g + can dissolve in one liter. Slide 12 / 57 Remember that molar solubility refers to the ion with the lower mole ratio. It does not always refer to the cation, although in most cases it does. ompound a 2 O 4 Pbl 2 g 2 SO 4 Molar of [ation] ompound 1.3 x 10-3 mol 1.3 x 10-3 mol [nion] 1.3 x 10-3 mol mol/l mol/l mol/l mol/l mol/l mol/l

3 Slide 13 / 57 3 If the solubility of barium carbonate, ao 3 is 7.1 x 10-5 M, this means that a maximum of barium ions, a 2+ ions can be dissolved per liter of solution. 4 Slide 14 / 57 If the solubility of barium carbonate, ao 3 is 7.1 x 10-5 M, this means that a maximum of carbonate ions, O 3 2- ions can be dissolved per liter of solution. 7.1 x 10-5 moles half of that twice as much 7.1 x 10-5 moles half of that twice as much one-third as much one-third as much one-fourth as much one-fourth as much Slide 15 / 57 Slide 16 / 57 5 If the solubility of g 2 ro 4 is 6.5 x 10-5 M, this means that a maximum of chromate ions, ro 4 2-, can be dissolved per liter of solution. 6 If the solubility of g 2 ro 4 is 6.5 x 10-5 M, this means that a maximum of g + ions can be dissolved per liter of solution. 6.5 x 10-5 moles 6.5 x 10-5 moles twice 6.5 x 10-5 moles half 6.5 x 10-5 moles one-fourth 6.5 x 10-5 moles four times 6.5 x 10-5 moles twice 6.5 x 10-5 moles half 6.5 x 10-5 moles one-fourth 6.5 x 10-5 moles four times 6.5 x 10-5 moles Slide 17 / 57 Slide 18 / 57 alculating K sp from the Sample Problem The molar solubility of lead (II) bromide, Pbr 2 is 1.0 x 10-2 at 25 o. alculate the solubility product, K sp, for this compound. [Pb 2+ ] = 1.0 x 10-2 mol/l [r - ] = 2.0 x 10-2 mol/l Substitute the molar concentrations into the K sp expression and solve. 7 For the slightly soluble compound,, the molar solubility is 3 x 10-8 moles per liter. alculate the K sp for this compound. No calculator. <--> x /2 (3 x 10-8 ) (3 x 10-8 )^1/2 2 (3 x 10-8 ) (3 x 10-8 )^2 K sp = [Pb 2+ ][r - ] 2 = (1.0 x 10-2 )(2.0 x 10-2 ) 2 = 4.0 x 10-6 gl < - - > g+ + l-

4 Slide 19 / 57 8 For the slightly soluble compound, XY, the molar solubility is 5 x 10-5 M. alculate the K sp for this compound. No calculator. XY <--> X + + Y - 5 x x x x x Slide 20 / 57 9 For the slightly soluble compound, MN, the molar solubility is 4 x 10-6 M. alculate the K sp for this compound. No calculator. MN <--> M + + N - 4 x x x x ao 3 < - - > a 2+ + O 3 2- Slide 21 / For the slightly soluble compound, 2, the molar solubility is 3 x 10-4 M. alculate the solubilityproduct constant for this compound. No calculator. 2 <--> x x x x x Pbl 2 < - - > Pb l - Slide 23 / 57 Slide 22 / For the slightly soluble compound, X 3 Y, the molar solubility is 1 x 10-4 M. alculate the solubility product for this compound. No calculator. X 3 Y <-> 3X + + Y 3-3 x x x x Fe(OH) 3 < - - > Fe (OH) - Na 3 P <--> 3Na+ + P 3- Slide 24 / 57 alculating from the K sp Sample Problem alculate the solubility of af 2 in grams per liter in a) pure water b) a 0.15 M KF solution c) a M a(no 3 ) 2 solution The solubility product for calcium fluoride, af 2 is 3.9 x alculating from the K sp a) pure water af 2 < - - > a F - If we assume x as the dissociation then, a 2+ ions = x and [F - ] = 2x K sp = [a 2+ ] [F - ] 2 = (x)(2x) 2 K sp = 3.9 x = 4x 3 So x = 2.13 x 10-4 mol/l x (78 g/mol af 2) is g/l 2.13 x10-4 a 2+ mol/l x 1mol/L af2 78g/L x mol/L a 2+ ions 1 mol af2

5 Slide 25 / 57 alculating from the K sp Slide 26 / 57 alculating from the K sp b) a 0.15 M KF solution remember KF is a strong electrolyte, is completely ionized. the major source of F- ions, then [F - ] =0.15M The solubility product for calcium fluoride, af2 is 3.9 x [ F - ] = 0.15M Ksp = [a 2+ ] [F - ] 2 = (x)(0.15) 2 Ksp = 3.9 x = x So x = mol/l is = x (78 g/mol af2) = g/l 1.73 x10-9 a 2+ mol/l x 1mol/L af2 78g/L x mol/L a 2+ ions 1 mol af2 Slide 27 / 57 alculate the solubility of af 2 in grams per liter in c) a M a(no 3 ) 2 solution [a 2+ ] = 0.08M The solubility product for calcium fluoride,af 2 is 3.9 x af 2 (s) <---> a 2+ (aq) + 2 F - (aq) Ksp = [a 2+ ] [F-] 2 = (0.080)(x) 2 Ksp = 3.9 x = 0.080x 2 So x = 2.2 x 10-5 mol/l * (78 g/mol af 2 )/ 2 is g/l Slide 28 / 57 alculating from the K sp Recall from the ommon-ion ffect that adding a strong electrolyte to a weakly soluble solution with a common ion will decrease the solubility of the weak electrolyte. ompare the solubilities from the previous Sample Problem af 2 (s) <---> a 2+ (aq) + 2 F - (aq) 12 alculate the concentration of silver ion when the solubility product constant of gi is (1 x ) 2 (1 x ) (1 x ) 2 (1 x ) af 2 dissolved with: of af 2 pure water M KF g/l M a(no 3) g/l These results support Le hatelier's Principle that increasing a product concentration will shift equilibrium to the left. Slide 29 / The Ksp of a compound of formula 3 is 1.8 x The molar solubility of the compound is ---- Slide 30 / The Ksp of a compound of formula 3 is 1.8 x The solubility of the compound is ---- The molar mass is 210g/mol

6 Slide 31 / The concentration of hydroxide ions in a saturated solution of l(oh) 3 is 1.58 x The concentration of l 3+ is Slide 32 / The concentration of hydroxide ions in a saturated solution of l(oh) 3 is 1.58 x The ksp of l (OH) 3 is----- Slide 33 / 57 Factors ffecting Recall The ommon-ion ffect If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease. aso 4 (s) a 2+ (aq) + SO 4 2- (aq) So adding any soluble salt containing either a 2+ or SO 4 2- ions will decrease the solubility of barium sulfate. Slide 34 / 57 Factors ffecting The ommon-ion ffect applies when a basic anion is involved, such as conjugate base of a weak acid. onsider the dissociation of the salt calcium fluoride: af 2 <--> + What do you expect will happen to equilibrium point if the ph of this system is lowered by adding a strong acid? Slide 35 / 57 Factors ffecting Slide 36 / 57 Factors ffecting Write the dissociation equation for magnesium hydroxide: Suppose you have a saturated solution of magnesium hydroxide in a flask. What will happen if you add a small amount of strong acid to it? (Think Le hâtelier s Principle.) Main idea: dding an acid, or otherwise lowering the ph of a solution will the solubility of a salt containing a.

7 Slide 37 / 57 Factors ffecting ph If a substance has a basic anion, it will be more soluble in an acidic solution. Substances with acidic cations are more soluble in basic solutions. Slide 38 / Given the system at equilibrium gl (s) <--> g + (aq) + l - (aq) When 0.01M Hl is added to the sytem, the point of equilibrium will shift to the. right and the concentration of g + will decrese right and the concentration of g + will increase left and the concentration of g + will decrease left and the concentration of g + will increase Slide 39 / 57 Slide 40 / 57 Factors ffecting Factors ffecting omplex Ions Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. omplex Ions The formation of these complex ions increases the solubility of these salts. Slide 41 / 57 Factors ffecting Slide 42 / 57 Precipitation Problems mphoterism mphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. xamples of such cations are l 3+, Zn 2+, and Sn 2+. rules to memmorize by this time. Will help you to identify which will precipitate as you mix two solutions. ny salt made with a Group I metal is soluble. ll salts containing nitrate ion are soluble. ll salts containing ammonium ion are soluble.

8 Slide 43 / What is the name of the solid precipitate that is formed when a solution of sodium chloride is mixed with a solution of silver nitrate? Slide 44 / What is the name of the solid precipitate that is formed when a solution of potassium carbonate is mixed with a solution of calcium bromide? sodium silver potassium bromide sodium nitrate calcium carbonate chloride nitrate potassium calcium silver chloride carbonate bromide Not enough information Not enough information Slide 45 / 57 Slide 46 / What is the name of the solid precipitate that is formed when a solution of lead (IV) nitrate is mixed with a solution of magnesium sulfate? PbSO 4 Pb(SO 4 ) 2 Pb 2 SO 4 Mg(NO 3 ) 2 Not enough information Precipitation Problems Most of the problems in this section ask you whether a precipitate will form after mixing certain solutions together. General Problem-Solving Strategy Step 1 - etermine which of the products is the precipitate. Write the K sp expression for this compound. Step 2 - alculate the cation concentration of this slightly soluble compound. Step 3 - alculate the anion concentration of this slightly soluble compound. Step 4 - Substitute the values into the reaction quotient (Q) expression. Recall that this is the same expression as K. Slide 47 / 57 Precipitation Problems nalyzing quantitative results: Step 5 - ompare Q to K to determine whether a precipitate will form. Slide 48 / The K sp for Zn(OH) 2 is 5.0 x Will a precipitate form in a solution whose solubility is 8.0x10-2 mol/l Zn(OH) 2? If Q = Ksp If Q > Ksp If Q < Ksp then you have an exactly perfect saturated solution with not one speck of undissolved solid. In a solution, then YS you will observe a precipitate; the number of cations and anions exceeds the solubility then NO precipitate will form; there are so few cations and anions that they all remain dissolved If Q = K sp, the system is at equilibrium and the solution is saturated. If Q > K sp, the salt will precipitate until Q = K sp. If Q < K sp, more solid can dissolve until Q = K sp. yes, because Q sp < K sp yes, because Q sp > K sp no, because Q sp = K sp no, because Q sp < K sp no, because Q sp > K sp

9 Slide 49 / 57 Slide 50 / 57 Precipitation Problems Sample Problem Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, al 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Qualitatively, you know that will be the solid that would theoretically form. Quantitatively, however, the question is, are there enough ions and ions to produce a solid precipitate, or will there be so few of them that they will all dissolve? Slide 51 / 57 Sample Problem - nswers Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, al 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2 SO 4? Step 1 - etermine which of the products is the precipitate. Write the K sp expression for this compound. aso 4 (s) <--> a 2+ (aq) + SO 4 2- (aq) Step 2 - alculate the cation concentration of this slightly soluble compound. M 1 V 1 =M 2 V 2 M 2 = (M 1 V 1 ) / V 2 M 2 = (0.20M*50.0mL) / 100 ml M 2 = 0.10 M al 2 [a 2+ ] = 0.10 M Step 3 - alculate the anion concentration of this slightly soluble compound. M 1 V 1 =M 2 V 2 M 2 = (M 1 V 1 ) / V 2 M 2 = (0.30M*50.0mL) / 100 ml M 2 = 0.15 M Na 2 SO 4 [SO 4 2- ] = 0.15 M Slide 52 / The K sp for zinc carbonate is 1 x If equivalent amounts of 0.1M sodium carbonate and 0.1M zinc nitrate are mixed, what happens? zinc carbonate precipitate forms, since Q>K. zinc carbonate precipitate forms, since Q<K. sodium nitrate precipitate forms, since Q>K. No precipitate forms, since Q=K. Slide 53 / Which of the following factors affect solubility? ph oncentration ommon-ion ffect and,, and Sample Problem - nswers (con't) Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, al 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2 SO 4? Step 4 - Substitute the values into the reaction quotient (Q) expression. Recall that this is the same expression as K. K sp = [a 2+ ] [SO 4 2- ] = (0.10) (0.15) = Step 5 - ompare Q to K to determine whether a precipitate will form. The K sp for barium sulfate is 1 x Therefore, since Q > K, there will be a precipitate formed when you mix equal amounts of 0.20 M al 2, and 0.30 M Na 2 SO 4. Slide 54 / 57 Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.

10 Slide 55 / 57 Slide 56 / 57 HW question To a solution containing 2.0 x 10-5 M arium ions and 1.8 x 10-4 M lead ions, Na 2 ro 4 is added. Which would precipitate first from this solution? Slide 57 / 57