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Aqueous Equilibria Entry Task: Feb 28 th Thursday Question: Provide the K sp expression for calcium phosphate, K sp = 2.0 x 10 -29. From this expression, will there be a great deal of dissolved substance? You have 5 minutes!

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Aqueous Equilibria I can… Define and explain the difference between: Solubility Solubility-product Molar solubility Write the Ksp expression for a given soluble ionic compound. Calculate the number of ions dissolved in solution.

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Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Sections 4

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Aqueous Equilibria Solubility Products Define saturated solution. Is a solution in which the solution is in contact with undissolved solute. It’s the maximum amount of solute dissolved in solution at a specific temperature.

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Aqueous Equilibria Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Equilibrium shift to the left, because not a lot of product. Slightly souble.

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Aqueous Equilibria Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2- ] where the equilibrium constant, K sp, is called the solubility product.

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Aqueous Equilibria Solubility Products Where can one find the Ksp values for a particular ionic solid? Appendix D on page 1023

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Aqueous Equilibria Solubility Products What does a small Ksp value mean? The smaller the Ksp value only a very small amount of solid would dissolve.

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Aqueous Equilibria Solubility Products Write the Ksp constant for CaF 2 including the # value. CaF 2 (s) Ca +2 (aq) + 2F - (aq) 3.9 x 10 -11 = [Ca +2 ] [F - ] 2

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Aqueous Equilibria Solubility Products Write the Ksp constant for barium carbonate include the # value. BaCO 3 (s) Ba +2 (aq) + CO 3 -2 (aq) 5.0 x 10 -9 = [Ba +2 ] [CO 3 -2 ]

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Aqueous Equilibria Solubility Products Write the Ksp constant for silver sulfate include the # value. Ag 2 SO 4 (s) 2Ag +1 (aq) + SO 4 -2 (aq) 1.5 x 10 -5 = [Ag +1 ] 2 [SO 4 -2 ]

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Aqueous Equilibria Solubility and Ksp It’s important to distinguish the difference between solubility and solubility-product constant, Ksp. What is solubility? Solubility of a substance is the quantity that dissolves to form a saturated solution

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Aqueous Equilibria Solubility and Ksp It’s important to distinguish the difference between solubility and solubility-product constant, Ksp. What is the molar solubility? The molar solubility is the number of moles of the solute that dissolves in forming a liter of saturated solution of the solute (mol/L)

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Aqueous Equilibria Solubility and Ksp It’s important to distinguish the difference between solubility and solubility-product constant, Ksp. What is the solubility-product constant (Ksp)? The solubility-product constant (Ksp) is the equilibrium constant for the equilibrium between ionic solid and its saturated solution.

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Aqueous Equilibria Solubility and Ksp How can solubility of a solid change? The solutions pH Concentration of other ions in solution Temperature

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Aqueous Equilibria Solubility Products K sp is not the same as solubility. 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).

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Aqueous Equilibria

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Aqueous Equilibria Calculating the Ksp

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Aqueous Equilibria Problem 1: Solid silver chromate is added to pure water at 25 C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 10 4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. First provide the equation: Ag 2 CrO 4 (s) 2Ag +1 (aq) + CrO 4 2- (aq)

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Aqueous Equilibria Problem 1: Solid silver chromate is added to pure water at 25 C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 10 4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. What do we know: Divide 1.3 x 10 -4 by 2 = 6.5 x 10 -5 We know that there is a 2 Ag 1+ ions to 1 of CrO 4 2- Ag 2 CrO 4 (s) 2Ag +1 (aq) + CrO 4 2- (aq) The Ag 1+ ions 1.3 10 4 M.

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Aqueous Equilibria Problem 1: Solid silver chromate is added to pure water at 25 C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 10 4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. Provide the Ksp expression: Ksp = [Ag +1 ] 2 [CrO 4 2- ]

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Aqueous Equilibria Problem 1: Solid silver chromate is added to pure water at 25 C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 10 4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. Plug in the numbers: Ksp = [1.3 x 10 -4 ] 2 [6.5 x 10 -5 ] Now solve: 1.1 x 10 -12

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Aqueous Equilibria Problem 2: A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25 C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound. First provide the equation: Mg(OH) 2 (s) Mg +2 (aq) + 2OH 1- (aq)

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Aqueous Equilibria Problem 2: A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25 C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound. What we know: pH is 10.17, what can we get out of this information? We can get OH- concentration. pH = 10.17 – 14 = 3.83 is pOH 10 -3.83 = 1.47 x 10 -4 is the OH -1 concentration AND there is twice the amount OH -1 to Mg +2 ions so divide OH by 2. 7.35 x 10 -5 Mg +2 ions

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Aqueous Equilibria Problem 2: A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25 C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound. Provide the Ksp expression: Ksp = [Mg +2 ] [OH -1 ] 2

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Aqueous Equilibria Problem 2: A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25 C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound. Plug in the numbers: Ksp = [7.35 x 10 -5 ] [1.47 x 10 -4 ] 2 Ksp = 1.6 x 10 -12

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Aqueous Equilibria Problem 3: Calculate the K sp for CaCl 2 if 200.0 g of CaCl 2 are required to saturate 100.0 mL of solution. First provide the equation: CaCl 2 (s) Ca +2 (aq) + 2Cl 1- (aq)

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Aqueous Equilibria Problem 3: Calculate the K sp for CaCl 2 if 200.0 g of CaCl 2 are required to saturate 100.0 mL of solution. What we know: CaCl 2 (s) Ca +2 (aq) + 2Cl 1- (aq) 200.0 g / 110.98 = 1.80 mol/0.1 L or 18.0M of CaCl 2 There’s 1 Ca +2 ions to 2Cl -1 ions. There’s 18.0M Ca +2 ions to 36.0M Cl -1 ions.

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Aqueous Equilibria Problem 3: Calculate the K sp for CaCl 2 if 200.0 g of CaCl 2 are required to saturate 100.0 mL of solution. Provide the Ksp expression: CaCl 2 (s) Ca +2 (aq) + 2Cl 1- (aq) Ksp = [Ca 2+ ][F - ] 2

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Aqueous Equilibria Problem 3: Calculate the K sp for CaCl 2 if 200.0 g of CaCl 2 are required to saturate 100.0 mL of solution. Plug in the numbers: CaCl 2 (s) Ca +2 (aq) + 2Cl 1- (aq) Ksp = [18.0][36.0] 2 There’s 18.0M Ca +2 ions to 36.0M Cl -1 ions. Ksp = 2.33 x10 4

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Aqueous Equilibria Calculating Molar Solubility

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Aqueous Equilibria Problem 4: If the solubility product of HgSO 4 is 6.4 x10 -5, then its molar solubility is: Provide equation: HgSO 4 (s) Hg +2 (aq) + SO 4 2- (aq)

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Aqueous Equilibria Problem 4: If the solubility product of HgSO 4 is 6.4 x10 -5, then its molar solubility is: What we know: Initial Change Equilibrium 0 0------- +x M x M HgSO 4 (s) Hg +2 (aq) + SO 4 2- (aq)

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Aqueous Equilibria Problem 4: If the solubility product of HgSO 4 is 6.4 x10 -5, then its molar solubility is: What we know: Notice that the stoichiometry of the equilibrium for every mole of HgSO 4 dissolves there is the same number of its ions. Provide the Ksp expression:Ksp = [Hg 2+ ][SO 4 -2 ] Plug in the equilibrium [ ]:6.3 x 10 -5 = [x][x]

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Aqueous Equilibria Problem 4: If the solubility product of HgSO 4 is 6.4 x10 -5, then its molar solubility is: Clean up and solve: Plug in the equilibrium [ ]: x 2 = 6.3 x 10 -5 x = 7.9 x 10 -3 What does this number represent? It’s the molar solubility of HgSO 4 is 7.9 x 10 -3 6.3 x 10 -5 = [x][x]

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Aqueous Equilibria Problem 5: The solubility product of NaCl is 1.5x10 -10 is at 298 K. its solubility in mol/L would be: Provide equation: NaCl(s) Na +1 (aq) + Cl 1- (aq)

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Aqueous Equilibria Problem 5: The solubility product of NaCl is 1.5x10 -10 is at 298 K. its solubility in mol/L would be: What we know: Initial Change Equilibrium 0 0------- +x M x M NaCl(s) Na +1 (aq) + Cl 1- (aq)

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Aqueous Equilibria Problem 5: The solubility product of NaCl is 1.5x10 -10 is at 298 K. its solubility in mol/L would be: What we know: Notice that the stoichiometry of the equilibrium for every mole of NaCl dissolves there is the same number of its ions. Provide the Ksp expression:Ksp = [Na 1+ ][Cl -1 ] Plug in the equilibrium [ ]:1.5 x 10 -10 = [x][x]

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Aqueous Equilibria Problem 5: The solubility product of NaCl is 1.5x10 -10 is at 298 K. its solubility in mol/L would be: Clean up and solve: Plug in the equilibrium [ ]: x 2 = 1.5 x 10 -10 x = 1.2 x 10 -5 What does this number represent? It’s the molar solubility of NaCl is 1.2 x 10 -5 1.5 x 10 -10 = [x][x]

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Aqueous Equilibria Calculating Solubility

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Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9 10 11 at 25 C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. First provide the equation: CaF 2 (s) Ca +2 (aq) + 2F 1- (aq)

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Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9 10 11 at 25 C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: **Remember solubility is the amount of solute that can dissolve in a solvent, where as the solubility-product constant is an equilibrium constant. **For this reason, we have to assume initially-that none of the salt has dissolved and allow x moles/liter of CaF 2 to dissociate completely when equilibrium is achieved

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Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9 10 11 at 25 C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: CaF 2 (s) Ca +2 (aq) + 2F 1- (aq) Initial Change Equilibrium 0 0------- +x M +2x M x M 2x M

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Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9 10 11 at 25 C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: Notice that the stoichiometry of the equilibrium dictates that 2x moles/liter of F- ions are produced for each x moles/liter of CaF 2 that dissolve. Provide the Ksp expression:Ksp = [Ca 2+ ][F - ] 2 Plug in the equilibrium [ ]:3.9 x 10 -11 = [x][2x] 2

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Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9 10 11 at 25 C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: Clean up and solve: Provide the Ksp expression:Ksp = [Ca 2+ ][F - ] 2 Plug in the equilibrium [ ]:3.9 x 10 -11 = [x][2x] 2 4x 3 = 3.9 x 10 -11 x = 3 3.9 x 10 -11 4 x = 2.1 x 10 -4 What does this number represent? It’s the molar solubility of CaF 2 is 2.1 x 10 -4

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Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9 10 11 at 25 C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: From the molarity solubility we can get to the number of grams dissolved in 1 liter of solvent Now what: 2.1 x 10 -4 mol CaF 2 1 liter X 1.6 x 10 -2 g CaF 2 /liter solution 78.1 g CaF 2 1 mol

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Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3 10 10 at 25 C. What is the solubility of Cu(N 3 ) 2 in grams per liter. Provide equation: Cu(N 3 ) 2 (s) Cu +2 (aq) + 2N 3 1- (aq)

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Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3 10 10 at 25 C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: **Remember solubility is the amount of solute that can dissolve in a solvent, where as the solubility-product constant is an equilibrium constant. **For this reason, we have to assume initially-that none of the salt has dissolved and allow x moles/liter of Cu(N 3 ) 2 to dissociate completely when equilibrium is achieved

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Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3 10 10 at 25 C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: Initial Change Equilibrium 0 0------- +x M +2x M x M 2x M Cu(N 3 ) 2 (s) Cu +2 (aq) + 2N 3 1- (aq)

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Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3 10 10 at 25 C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: Notice that the stoichiometry of the equilibrium dictates that 2x moles/liter of N 3 - ions are produced for each x moles/liter of Cu(N 3 ) 2 that dissolve. Provide the Ksp expression:Ksp = [Cu 2+ ][N 3 - ] 2 Plug in the equilibrium [ ]:6.3 x 10 -10 = [x][2x] 2

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Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3 10 10 at 25 C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: Clean up and solve: Provide the Ksp expression:Ksp = [Cu 2+ ][N 3 - ] 2 Plug in the equilibrium [ ]:6.3 x 10 -10 = [x][2x] 2 4x 3 = 6.3 x 10 -10 x = 3 6.3 x 10 -10 4 x = 5.4 x 10 -4 What does this number represent? It’s the molar solubility of Cu(N 3 ) 2 is 5.4 x 10 -4

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Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3 10 10 at 25 C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: From the molarity solubility we can get to the number of grams dissolved in 1 liter of solvent Now what: 5.4 x 10 -4 mol Cu(N 3 ) 2 1 liter X 7.97 x 10 -2 g Cu(N 3 ) 2 /liter solution 147.6 g Cu(N 3 ) 2 1 mol

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