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SOLUBILITY Saturated Solution BaSO 4(s) Ba 2+ (aq) + SO 4 2- (aq) Equilibrium expresses the degree of solubility of solid in water. Equilibrium expresses the degree of solubility of solid in water. K sp = solubility product constant K sp = K eq [BaSO 4 ] (s) K sp = [Ba 2+ ] [SO 4 2- ] = 1.1 x 10 -10 K sp represents the amount of dissolution (how much solid dissolved into ions), the smaller the K sp value, the smaller the amount of ions in solution (more solid is present).

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Table 1 Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 0 C Name, FormulaK sp Aluminum hydroxide, Al(OH) 3 Cobalt ( II ) carbonate, CoCO 3 Iron ( II ) hydroxide, Fe(OH) 2 Lead ( II ) fluoride, PbF 2 Lead ( II ) sulfate, PbSO 4 Silver sulfide, Ag 2 S Zinc iodate, Zn( I O 3 ) 2 3 x 10 -34 1.0 x 10 -10 4.1 x 10 -15 3.6 x 10 -8 1.6 x 10 -8 4.7 x 10 -29 8 x 10 -48 Mercury ( I ) iodide, Hg 2 I 2 3.9 x 10 -6

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SOLUBILITY 1. Write the solubility product expression for each of the following: a) Ca 3 (PO 4 ) 2 b) Hg 2 Cl 2 c) HgCl 2. 2. In a particular sample, the concentration of silver ions was 1.2 x10 -6 M and the concentration of bromide was 1.7x10 -6 M. What is the value of K sp for AgBr?

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Solubility vs. Solubility Product Solubility: The quantity of solute that dissolves to form a saturated solution. ( g / L ) Molar Solubility: (n solute /L saturated solution ) K sp : The equilibrium between the ionic solid and the saturated solution.

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Interconverting solubility and Ksp SOLUBILITYOFCOMPOUND(g/L) MOLARSOLUBILITYOFCOMPOUND(mol/L) MOLARCONCENTRATIONOFIONS K sp

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Sample Problem 1 Determining K sp from Solubility PROBLEM:(a) Lead ( II ) sulfate is a key component in lead-acid car batteries. Its solubility in water at 25 0 C is 4.25x10 -3 g/100mL solution. What is the K sp of PbSO 4 ? (b) When lead ( II ) fluoride (PbF 2 ) is shaken with pure water at 25 0 C, the solubility is found to be 0.64g/L. Calculate the K sp of PbF 2. PLAN:Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. K sp = [Pb 2+ ][SO 4 2- ] = 1.40x10 -4 M PbSO 4 K sp = [Pb 2+ ][SO 4 2- ] = (1.40x10 -4 ) 2 = SOLUTION:PbSO 4 ( s ) Pb 2+ ( aq ) + SO 4 2- ( aq )(a) 1000mL L 4.25x10 -3 g 100mL soln 303.3g PbSO 4 mol PbSO 4 1.96x10 -8

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Sample Problem 1 Determining K sp from Solubility continued (b) PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = [Pb 2+ ][F - ] 2 = 2.6x10 -3 M K sp = (2.6x10 -3 )(5.2x10 -3 ) 2 = 0.64g L soln245.2g PbF 2 mol PbF 2 7.0x10 -8

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Sample Problem 2 Determining Solubility from K sp PROBLEM:Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH) 2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH) 2 in water if the K sp is 6.5x10 -6. PLAN:Write out a dissociation equation and K sp expression; Find the molar solubility (S) using a table. SOLUTION:Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq ) K sp = [Ca 2+ ][OH - ] 2 -Initial Change Equilibrium - - 00 +S+ 2S S2S K sp = (S)(2S) 2 S == 1.2x10x -2 M Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq )Concentration (M)

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Solubility vs. Solubility Product 1. A student finds that the solubility of BaF 2 is 1.1 g in l.00 L of water. What is the value of Ksp? 2. Exactly 0.133 mg of AgBr will dissolve in 1.00 L of water. What is the value of Ksp for AgBr? 3. Calomel (Hg 2 Cl 2 ) was once used in medicine. It has a Ksp = 1.3 x 10 -18. What is the solubility of Hg 2 Cl 2 in g/L?

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Relationship Between K sp and Solubility at 25 0 C No. of IonsFormulaCation:AnionK sp Solubility (M) 2MgCO 3 1:13.5 x 10 -8 1.9 x 10 -4 2PbSO 4 1:11.6 x 10 -8 1.3 x 10 -4 2BaCrO 4 1:12.1 x 10 -10 1.4 x 10 -5 3Ca(OH) 2 1:25.5 x 10 -6 1.2 x 10 -2 3BaF 2 1:21.5 x 10 -6 7.2 x 10 -3 3CaF 2 1:23.2 x 10 -11 2.0 x 10 -4 3Ag 2 CrO 4 2:12.6 x 10 -12 8.7 x 10 -5

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Ion-Product Expression (Q sp ) & Solubility Product Constant (K sp ) At equilibrium Q sp = [M n+ ] p [X z- ] q = K sp For the hypothetical compound, M p X q

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Solubility and Common Ion effect CaF 2(s) Ca 2+ (aq) + 2F - (aq) The addition of Ca 2+ or F - shifts the equilibrium. According to Le Chatelier’s Principle, more solid will form thus reducing the solubility of the solid. The addition of Ca 2+ or F - shifts the equilibrium. According to Le Chatelier’s Principle, more solid will form thus reducing the solubility of the solid. Solubility of a salt decreases when the solute of a common ion is added.

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The effect of a common ion on solubility PbCrO 4 ( s ) Pb 2+ ( aq ) + CrO 4 2- ( aq ) CrO 4 2- added

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Sample Problem 3Calculating the Effect of a Common Ion on Solubility PROBLEM:In Sample Problem 19.6, we calculated the solubility of Ca(OH) 2 in water. What is its solubility in 0.10M Ca(NO 3 ) 2 ? K sp of Ca(OH) 2 is 6.5x10 -6. PLAN:Set up a reaction equation and table for the dissolution of Ca(OH) 2. The Ca(NO 3 ) 2 will supply extra [Ca 2+ ] and will relate to the molar solubility of the ions involved. SOLUTION:Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq )Concentration(M) Initial Change Equilibrium - - - 0.100 +S+2S 0.10 + S2S K sp = 6.5x10 -6 = (0.10 + S)(2S) 2 = (0.10)(2S) 2 S << 0.10 S = = 4.0x10 -3 Check the assumption: 4.0% 0.10M 4.0x10 -3 x 100 =

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Solubility and Common Ion effect CaF 2(s) Ca 2+ (aq) + 2F - (aq) 1. The K sp of the above equation is 3.2 x 10 -11. (a) Calculate the molar solubility in pure water. (b) Calculate the molar solubility in 3.5 x 10 -4 M Ca(NO 3 ) 2. 2. What is the molar solubility of silver chloride in 1.0 L of solution that contains 2.0 x 10 -2 mol of NaCl?

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CRITERIA FOR PRECIPITATION OF DISSOLUTION BaSO 4(s) Ba 2+ (aq) + SO 4 2- (aq) Equilibrium can be established from either direction. Q (the Ion Product) is used to determine whether or not precipitation will occur. Q < K solid dissolves Q = K equilibrium (saturated solution) Q > K ppt

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Sample Problem 3 Predicting Whether a Precipitate Will Form PROBLEM:A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO 3 ) 2 is mixed with 0.200L of 0.060M NaF? PLAN:Write out a reaction equation to see which salt would be formed. Look up the K sp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < K sp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION:CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = 3.2x10 -11 mol Ca 2+ = 0.100L(0.30mol/L) = 0.030mol[Ca 2+ ] = 0.030mol/0.300L = 0.10M mol F - = 0.200L(0.060mol/L) = 0.012mol[F - ] = 0.012mol/0.300L = 0.040M Q = [Ca 2+ ][F - ] 2 =(0.10)(0.040) 2 = 1.6x10 -4 Q is >> K sp and the CaF 2 WILL precipitate.

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CRITERIA FOR PRECIPITATION OF DISSOLUTION 1. Calcium phosphate has a K sp of 1 x 10 -26, if a sample contains 1.0 x 10 -3 M Ca 2+ & 1.0 x 10 -8 M PO 4 3- ions, calculate Q and predict whether Ca 3 (PO 4 ) 2 will precipitate? 2.Exactly 0.400 L of 0.50 M Pb 2+ & 1.60 L of 2.5 x 10 -8 M Cl - are mixed together to form 2.00L. Calculate Q and predict if a ppt will occur. K sp = 1.6 x 10 -5

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EFFECT OF pH ON SOLUBILITY CaF 2 Ca 2+ + 2F - 2F - + 2H + 2HF CaF 2 + 2H + Ca 2+ + 2HF Salts of weak acids are more soluble in acidic solutions. Thus shifting the solubility to the right. Salts with anions of strong acids are largely unaffected by pH.

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Sample Problem 4 Predicting the Effect on Solubility of Adding Strong Acid PROBLEM:Write balanced equations to explain whether addition of H 3 O + from a strong acid affects the solubility of these ionic compounds: (a) Lead ( II ) bromide(b) Copper ( II ) hydroxide(c) Iron ( II ) sulfide PLAN:Write dissolution equations and consider how strong acid would affect the anion component. Br - is the anion of a strong acid. No effect. SOLUTION:(a) PbBr 2 ( s ) Pb 2+ ( aq ) + 2Br - ( aq ) (b) Cu(OH) 2 ( s ) Cu 2+ ( aq ) + 2OH - ( aq ) OH - is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility. (c) FeS( s ) Fe 2+ ( aq ) + S 2- ( aq )S 2- is the anion of a weak acid and will react with water to produce OH -. Both weak acids serve to increase the solubility of FeS. FeS( s ) + H 2 O( l ) Fe 2+ ( aq ) + HS - ( aq ) + OH - ( aq )

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EFFECT OF pH ON SOLUBILITY 1. Consider the two slightly soluble salts BaF 2 and AgBr.Which of these two would have its solubility more affected by the addition of a strong acid? Would the solubility of that salt increase or decrease. 2. What is the molar solubility of silver chloride in 1.0 L of solution that contains 2.0 x 10 -2 mol of HCl?

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3 STEPS TO DETERMINING THE ION CONCENTRATION AT EQUILIBRIUM I. Calculate the [Ion] i that occurs after dilution but before the reaction starts. II. Calculate the [Ion] when the maximum amount of solid is formed. - we will determine the limiting reagent and assume all of that ion is used up to make the solid. - The [ ] of the other ion will be the stoichiometric equivalent. III. Calculate the [Ion] at equilibrium*. *Since we assume the reaction went to completion, yet by definition a slightly soluble can’t, we must account for some of the solid re-dissolving back into solution.

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1. When 50.0 mL of 0.100 M AgNO 3 and 30 mL of 0.060 M Na 2 CrO 4 are mixed, a precipitate of silver chromate is formed. The solubility product is 1.9 x 10 -12. Calculate the [Ag + ] and [CrO 4 2- ] remaining in solution at equilibrium. 2. Suppose 300 mL of 8 x 10 -6 M solution of KCl is added to 800 mL of 0.004 M solution of AgNO 3. Calculate [Ag + ] and [Cl - ] remaining in solution at equilibrium.

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Chapter 17 sections 4-7 Solubility Equilibria © 2012 Pearson Education, Inc.

Chapter 17 sections 4-7 Solubility Equilibria © 2012 Pearson Education, Inc.

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