Chapter20 Solubility Product Constant

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<ul><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 1/32</p><p>2</p><p>20Ionic Equilibria III:</p><p>The SolubilityProduct Principle</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 2/32</p><p>3</p><p>Chapter Goals</p><p>1. Solubility Product Constants</p><p>2. Determination of Solubility ProductConstants</p><p>3. Uses of Solubility Product Constants</p><p>4. Fractional Precipitation</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 3/32</p><p>4</p><p>Solubility Product Constants</p><p> Silver chloride, AgCl,is rather insoluble in water.</p><p> Careful experiments show that if solid AgCl is</p><p>placed in pure water and vigorously stirred, a</p><p>small amount of the AgCl dissolves in the water.</p><p>aqaq ClAgClAg</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 4/32</p><p>5</p><p>Solubility Product Constants</p><p> The equilibrium constant expression for this</p><p>dissolution is called a solubility product constant.</p><p> Ksp = solubility product constant</p><p>-10-</p><p>sp 101.8]][Cl[AgK </p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 5/32</p><p>6</p><p>Solubility Product Constants</p><p> The solubility product constant, Ksp, for a</p><p>compound is the product of the concentrations of</p><p>the constituent ions, each raised to the power</p><p>that corresponds to the number of ions in oneformula unit of the compound.</p><p> Consider the dissolution of silver sulfide in water.</p><p>Ag S 2 Ag + S2+ 2- H O + 2-2 </p><p> 100%</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 6/32</p><p>7</p><p>Solubility Product Constants</p><p> The solubility product expression for Ag2S is:</p><p> K Ag Ssp 2 2 10 10 49.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 7/32</p><p>8</p><p>Solubility Product Constants</p><p> The dissolution of solid calcium phosphate in</p><p>water is represented as:</p><p> 2</p><p>2+ 3 2 3</p><p>3 4 ( ) 4 ( )2 s100Ca PO 3 Ca 2 POo</p><p>H O</p><p>aq aqC</p><p>The solubility product constant expression is:</p><p>You do it!</p><p> K Ca POsp 2 3</p><p>43 2</p><p>1 0 10 25</p><p>.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 8/32</p><p>9</p><p>Solubility Product Constants</p><p> In general, the dissolution of a slightly solublecompound and its solubility product expressionare represented as:</p><p>2s r</p><p>r s s100</p><p>r ss r</p><p>sp</p><p>M Y r M s Y</p><p>K M Y</p><p>o</p><p>H O</p><p>C</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 9/32</p><p>10</p><p>Solubility Product Constants</p><p> The same rules apply for compounds that have</p><p>more than two kinds of ions.</p><p> One example of a compound that has more than</p><p>two kinds of ions is calcium ammoniumphosphate.</p><p>3</p><p>44</p><p>2</p><p>sp</p><p>3</p><p>aq4</p><p>1</p><p>aq4</p><p>2</p><p>aqs44</p><p>PONHCaK</p><p>PONHCaPOCaNH</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 10/32</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 11/32</p><p>12</p><p>Determination of Solubility</p><p>Product Constants The equation for the dissociation of silver</p><p>chloride, the appropriate molar concentrations,</p><p>and the solubility product expression are:</p><p>AgCl Ag Cl</p><p> 1.34 10 1.34 10K Ag Cl</p><p>s</p><p>-5 -5</p><p>sp</p><p>M M</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 12/32</p><p>13</p><p>Determination of Solubility</p><p>Product Constants Substitution of the molar concentrations into the</p><p>solubility product expression gives:</p><p>K Ag Clsp </p><p>134 10 134 10</p><p>18 10</p><p>5 5</p><p>10</p><p>. .</p><p>.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 13/32</p><p>14</p><p>Determination of Solubility</p><p>Product ConstantsExample 20-2: One liter of saturated calcium</p><p>fluoride solution contains 0.0167 gram of CaF2at</p><p>25oC. Calculate the molar solubility of, and Kspfor,</p><p>CaF2.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 14/32</p><p>15</p><p>Uses of Solubility</p><p>Product Constants The solubility product constant can be used to</p><p>calculate the solubility of a compound at 25oC.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 15/32</p><p>16</p><p>Uses of Solubility</p><p>Product ConstantsExample 20-3: Calculate the molar solubility of</p><p>barium sulfate, BaSO4, in pure water and the</p><p>concentration of barium and sulfate ions in</p><p>saturated barium sulfate at 25o</p><p>C. For bariumsulfate, Ksp= 1.1 x 10</p><p>-10.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 16/32</p><p>17</p><p>Uses of Solubility</p><p>Product ConstantsExample 20-4: The solubility product constant for</p><p>magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11.</p><p>Calculate the molar solubility of magnesium</p><p>hydroxide and the pH of a saturated magnesiumhydroxide solution at 25oC.</p><p>You do i t !</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 17/32</p><p>18</p><p>The Common Ion Effect in</p><p>Solubility CalculationsExample 20-5: Calculate the molar solubility of</p><p>barium sulfate, BaSO4, in 0.010 Msodium sulfate,</p><p>Na2SO4, solution at 25oC. Compare this to the</p><p>solubility of BaSO4 in pure water. (Example 20-3).(What is the common ion?)</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 18/32</p><p>19</p><p>The Common Ion Effect in</p><p>Solubility Calculations</p><p> The molar solubility of BaSO4in 0.010 MNa2SO4</p><p>solution is 1.1 x 10-8M.</p><p> The molar solubility of BaSO4in pure water is 1.0 x</p><p>10-5M.</p><p> BaSO4is 900 times more soluble in pure water than in</p><p>0.010 Msodium sulfate!</p><p> Adding sodium sulfate to a solution is a fantastic</p><p>method to remove Ba2+ ions from solution!</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 19/32</p><p>20</p><p>The Reaction Quotient in</p><p>Precipitation Reactions The reaction quotient, Q, and the Ksp of a</p><p>compound are used to calculate the</p><p>concentration of ions in a solution and</p><p>whether or not a precipitate will form.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 20/32</p><p>The Reaction Quotient in</p><p>Precipitation Reactions</p><p>Qsp= Ksp saturated solution</p><p>Qsp&lt; Ksp unsaturated soln; precipitation cannot occur</p><p>Qsp&gt; Ksp supersaturated soln; precipitation should occur</p><p>21</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 21/32</p><p>22</p><p>The Reaction Quotient in</p><p>Precipitation ReactionsExample 20-6: We mix 100 mL of 0.010 M</p><p>potassium sulfate, K2SO4, and 100 mL of 0.10 M</p><p>lead (II) nitrate, Pb(NO3)2solutions. Will a</p><p>precipitate form?</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 22/32</p><p>23</p><p>The Reaction Quotient in</p><p>Precipitation ReactionsExample 20-7: Suppose we wish to remove</p><p>mercury from an aqueous solution that contains a</p><p>soluble mercury compound such as Hg(NO3)2. We</p><p>can do this by precipitating mercury (II) ions as theinsoluble compound HgS. What concentration of</p><p>sulfide ions, from a soluble compound such as</p><p>Na2S, is required to reduce the Hg2+concentration</p><p>to 1.0 x 10-8M? For HgS, Ksp=3.0 x 10-53.You do i t !</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 23/32</p><p>24</p><p>The Reaction Quotient in</p><p>Precipitation ReactionsExample 20-7: What concentration of sulfide ions,</p><p>from a soluble compound such as Na2S, is required</p><p>to reduce the Hg2+concentration to 1.0 x 10-8M?</p><p>For HgS, Ksp=3.0 x 10-53</p><p>.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 24/32</p><p>25</p><p>The Reaction Quotient in</p><p>Precipitation ReactionsExample 20-8: Refer to example 20-7. What volume</p><p>of the solution (1.0 x 10-8M Hg2+) contains 1.0 g of</p><p>mercury?</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 25/32</p><p>26</p><p>Fractional Precipitation</p><p> The method of precipitating some ions from a</p><p>solution while leaving others in solution is called</p><p>fractional precipitation.</p><p> If a solution contains Cu+</p><p>, Ag+</p><p>, and Au+</p><p>, each ion canbe precipitated as chlorides.</p><p> 13sp</p><p>10sp</p><p>7</p><p>sp</p><p>100.2ClAuKClAuAuCl</p><p>108.1ClAgKClAgAgCl</p><p>109.1ClCuKClCuCuCl</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 26/32</p><p>27</p><p>Fractional Precipitation</p><p>Example 20-9: If solid sodium chloride is slowly</p><p>added to a solution that is 0.010 Meach in Cu+, Ag+,</p><p>and Au+ions, which compound precipitates first?</p><p>Calculate the concentration of Cl-</p><p>required to initiateprecipitation of each of these metal chlorides.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 27/32</p><p>28</p><p>Fractional Precipitation</p><p> These three calculations give the [Cl-] required</p><p>to precipitate AuCl ([Cl-] &gt;2.0 x 10-11M), to</p><p>precipitate AgCl ([Cl-] &gt;1.8 x 10-8 M), and to</p><p>precipitate CuCl ([Cl-] &gt;1.9 x 10-5 M). It is also possible to calculate the amount of</p><p>Au+precipitated before the Ag+begins to</p><p>precipitate, as well as the amounts of Au+and</p><p>Ag+precipitated before the Cu+begins to</p><p>precipitate.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 28/32</p><p>29</p><p>Fractional Precipitation</p><p>Example 20-10: Calculate the percentage of Au+</p><p>ions that precipitate before AgCl begins to</p><p>precipitate.</p><p> Use the [Cl-</p><p>] from Example 20-9 to determine the [Au+</p><p>]remaining in solution just before AgCl begins to</p><p>precipitate.</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 29/32</p><p>30</p><p>Fractional Precipitation</p><p> The percent of Au+ions unprecipitated just</p><p>before AgCl precipitates is:</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 30/32</p><p>31</p><p>Fractional Precipitation</p><p> A similar calculation for the concentration</p><p>of Ag+ions unprecipitated beforeCuCl</p><p>begins to precipitate is:</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 31/32</p><p>32</p><p>Fractional Precipitation</p><p> The percent of Ag+ions unprecipitated justbefore AgCl precipitates is:</p></li><li><p>8/13/2019 Chapter20 Solubility Product Constant</p><p> 32/32</p><p>33</p><p>20</p><p>Ionic Equilibria III:</p><p>The SolubilityProduct Principle</p></li></ul>

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