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The Solubility Product Principle

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2 Silver chloride, AgCl,is rather insoluble in water. Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water. Solubility Product Constants AgCl(s) ⇌ Ag + (aq) + Cl - (aq)

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3 The equilibrium constant expression for this dissolution is called a solubility product constant. Ksp = solubility product constant Solubility Product Constants

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4 The solubility product constant, Ksp, for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound. Consider the dissolution of silver sulfide in water. Solubility Product Constants Ag 2 S(s) ⇌ 2Ag + (aq) + S 2- (aq) Ksp = [Ag + ] 2 [S -2 ]

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5 The dissolution of solid calcium phosphate in water is represented as: The solubility product constant expression is: Solubility Product Constants Ca 3 (PO 4 )(s) ⇌ 3Ca +2 (aq) + 2PO 4 3- (aq) Ksp = [Ca 2+ ] 3 [PO 4 -3 ] 2

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6 The same rules apply for compounds that have more than two kinds of ions. One example of a compound that has more than two kinds of ions is calcium ammonium phosphate. CaNH 4 PO 4 (s) ⇌ Ca +2 (aq) + NH 4 + aq) + PO 4 3- (aq) Solubility Product Constants

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7 All K sp values are small. All salts are only slightly soluble Ksp Table

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8 One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25°C. Calculate the molar solubility of, and Ksp for, AgCl. Molar solubility can be calculated from the data: Determination of Solubility Product Constants

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9 The equation for the dissociation of silver chloride and its solubility product expression are Determination of Solubility Product Constants

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10 Substitution into the solubility product expression gives Determination of Solubility Product Constants

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11 One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF 2. Calculate the molar solubility of, and Ksp for, CaF 2. Calculate the molar solubility of CaF 2. Determination of Solubility Product Constants

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12 From the molar solubility, we can find the ion concentrations in saturated CaF 2. Then use those values to calculate the Ksp. Determination of Solubility Product Constants

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13 Calculate the molar solubility of barium sulfate, BaSO 4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate. Ksp= 1.1 x 10 -10. Uses of Solubility Product Constants

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14 Now we can calculate the mass of BaSO 4 in 1.00 L of saturated solution. Uses of Solubility Product Constants

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15 The Ksp for magnesium hydroxide, Mg(OH) 2, is 1.5 x 10 - 11. Calculate the molar solubility of magnesium hydroxide and the pH of saturated magnesium hydroxide solution. Uses of Solubility Product Constants Ksp = [Mg 2+ ][OH - ] 2 Mg(OH) 2 (s) ⇌ Mg 2+ (aq) + 2OH - (aq)

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16 Substitute into the solubility product expression. Uses of Solubility Product Constants

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17 Calculate the concentration of calcium ion in a calcium phosphate, Ca 3 (PO 4 ) 2, in pure water. Ksp= 2.1 x 10 -33. Ca 3 (PO 4 ) 2 ⇌ 3Ca 2+ + 2PO 4 3- 3x 2x 2.1 x 10 -33 = [3x] 3 [2x] 2 = (27x 3 )(4x 2 ) 2.1 x 10 -33 = 108x 5 x 5 = 2.1 x 10 -33 /108 = 1.94 x 10 -35 x = x = 1.14 x 10 -7 ; [Ca 2+ ] = 3x = 3.42 x 10 -7 Uses of Solubility Product Constants

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18 Solubility is decreased when a common ion is added. This is an application of Le Châtelier’s principle: BaSO 4 ⇌ Ba 2+ + SO 4 2- as SO 4 2- (from Na 2 SO 4, say) is added, the equilibrium shifts away from the increase. Therefore, BaSO 4 (s) is formed and precipitation occurs. As Na 2 SO 4 is added to the system, the solubility of BaSO 4 decreases. Common Ion Effect

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19 Calculate the molar solubility of barium sulfate, BaSO 4, in 0.010 M sodium sulfate, Na 2 SO 4, solution. Compare this to the solubility of BaSO 4 in pure water. Write equations to represent the equilibria. The Common Ion Effect in Solubility

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20 Substitute the concentrations into the Ksp expression and solve for x. The molar solubility of BaSO 4 in 0.010 M Na 2 SO 4 solution is 1.1 x 10 -8 M. The molar solubility of BaSO 4 in pure water is 1.0 x 10 -5 M. 900 times greater The Common Ion Effect in Solubility

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21 We mix 100 mL of 0.010 M potassium sulfate, K 2 SO 4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO 3 ) 2 solutions. Will a precipitate form? K 2 SO 4 → 2K + + SO 4 2- Pb(NO 3 ) 2 → Pb 2+ + 2NO 3 - Will PbSO 4 precipitate? The Reaction Quotient in Precipitation

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22 Calculate the Qsp for PbSO 4. Solution volumes are additive. Concentrations of the important ions are: The Reaction Quotient in Precipitation Reactions

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23 Finally, we calculate Qsp for PbSO 4. Qsp = [Pb 2+ ][SO 4 2- ] = (0.050)(0.0050) = 2.5 x 10 -4 Ksp = 1.8 x 10 -8 Qsp > Ksp. therefore solid forms The Reaction Quotient in Precipitation Reactions

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24 Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO 3 ) 2. We can do this by precipitating mercury (II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na 2 S, is required to reduce the Hg 2+ concentration to 1.0 x 10 -8 M? For HgS, Ksp = 3.0 x 10 -53. The Reaction Quotient in Precipitation Reactions

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25 The Reaction Quotient in Precipitation Reactions

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26 A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? Ksp for Ag 2 CrO 4 = 9.0 x 10 -12 Ksp for PbCrO 4 = 1.8 x 10 -14 Solution The substance whose Ksp is first exceeded precipitates first. The ion requiring the lesser amount of CrO 4 2- ppts. first. Separating Salts

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27 A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? Ksp for Ag 2 CrO 4 = 9.0 x 10 -12 Ksp for PbCrO 4 = 1.8 x 10 -14 Solution Calculate [CrO 4 2- ] required by each ion. Separating Salts

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28 [CrO 4 2- ] to ppt. PbCrO 4 = Ksp / [Pb 2+ ] = 1.8 x 10 -14 / 0.020 = 9.0 x 10- 13 M [CrO 4 2- ] to ppt. Ag 2 CrO 4 = Ksp / [Ag + ] 2 = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M PbCrO 4 [CrO 4 2- ] = 9.0 x 10 -13 M Ag 2 CrO 4 [CrO 4 2- ] = 2.3 x 10 -8 M PbCrO 4 precipitates first. Separating Salts

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29 How much Pb 2+ remains in solution when Ag + begins to precipitate? Solution We know that [CrO 4 2- ] = 2.3 x 10 -8 M to begin to ppt. Ag 2 CrO 4. What is the Pb 2+ conc. at this point? Solution [Pb 2+ ] = Ksp / [CrO 4 2- ] = 1.8 x 10 -14 /2.3 x 10 -8 = 7.8 x 10 -7 M Lead ion has dropped from 0.020 M to < 10 -6 M Separating Salts

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30 99.9961% of Pb 2+ ions precipitates before AgCrO 4 begins to precipitate. Separating Salts

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31 Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution. Look at a solution that contains Cu+, Ag+, and Au+ We could precipitate them as chlorides CuCl ⇌ Cu + + Cl - Ksp = [Cu + ][Cl - ] = 1.9 x 10 -7 AgCl ⇌ Ag + + Cl - Ksp = [Ag + ][Cl - ] = 1.8 x 10 -10 AuCl ⇌ Au + + Cl - Ksp = [Au + ][Cl - ] = 2.0 x 10 -13 Fractional Precipitation

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32 If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu +, Ag +, and Au + ions, which compound precipitates first? Calculate the concentration of Cl - required to initiate precipitation of each of these metal(I) chlorides. Fractional Precipitation

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33 Repeat the calculation for silver chloride. Fractional Precipitation

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34 For copper(I) chloride to precipitate. Fractional Precipitation

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35 We have calculated the [Cl - ] required to precipitate AuCl, [Cl - ] > 2.0 x 10 -11 M to precipitate AgCl, [Cl - ] > 1.8 x 10 -8 M to precipitate CuCl, [Cl - ] > 1.9 x 10 -5 M We can calculate the amount of Au + precipitated before Ag + begins to precipitate, as well as the amounts of Au + and Ag + precipitated before Cu + begins to precipitate. Fractional Precipitation

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36 Calculate the percent of Au + ions that precipitate before AgCl begins to precipitate. Use the [Cl - ] = 1.8 x 10 -8 M to determine the [Au + ] remaining in solution just before AgCl begins to precipitate. Fractional Precipitation

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37 The percent of Au + ions unprecipitated just before AgCl precipitates is Therefore, 99.9% of the Au + ions precipitates before AgCl begins to precipitate. Fractional Precipitation

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38 Similar calculations for the concentration of Ag + ions unprecipitated before CuCl begins to precipitate gives Fractional Precipitation

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39 The percent of Ag + ions unprecipitated just before CuCl precipitates is Thus, 99.905% of the Ag + ions precipitates before CuCl begins to precipitate. Fractional Precipitation

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40 If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO 3 ) 2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? For Mg(OH) 2, Ksp = 1.5 x 10 -11 ; Kb for NH 3 = 1.8 x 10 -5. Calculate Qsp for Mg(OH) 2 and compare it to Ksp. Mg(NO 3 ) 2 is a soluble ionic compound so [Mg2+] = 0.010 M Aqueous ammonia is a weak base that we can calculate [OH - ] Simultaneous Equilibria

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41 Simultaneous Equilibria

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42 Since we now have the concentrations of both the magnesium and hydroxide ions, we can calculate the Qsp and compare it to the Ksp. Simultaneous Equilibria

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43 How many moles of solid ammonium chloride, NH 4 Cl, must be used to prevent precipitation of Mg(OH) 2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO 3 ) 2 ? Calculate the maximum [OH - ] that can exist in a solution that is 0.010 M in Mg 2+. Simultaneous Equilibria

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44 Calculate the maximum [OH - ] that can exist in a solution that is 0.010 M in Mg 2+. Simultaneous Equilibria

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45 Using the maximum [OH - ] that can exist in solution, we can calculate the number of moles of NH 4 Cl required to buffer 0.10 M aqueous ammonia so that [OH - ] does not exceed 3.9 x 10 -5 M. Simultaneous Equilibria

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46 NH 3 + H 2 O ⇌ NH 4 + + OH - (0.10- 3.9 x 10 -5 ) xM + 3.9 x 10 -5 3.9 x 10 -5 Simultaneous Equilibria

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47 Check our values by calculating Qsp for Mg(OH) 2. Simultaneous Equilibria

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48 Use the ion product for water to calculate the [H + ] and the pH of the solution. Simultaneous Equilibria

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