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I f you add additional solid calcium sulfate to the saturated solution, what will happen?

Assume you have a saturated solution of calcium sulfate at a temperature of 25  C represented by the following equation: CaSO 4 ( s )  Ca 2+ ( aq ) + SO 4 2- ( aq ). I f you add additional solid calcium sulfate to the saturated solution, what will happen?.

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I f you add additional solid calcium sulfate to the saturated solution, what will happen?

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  1. Assume you have a saturated solution of calcium sulfate at a temperature of 25 C represented by the following equation: CaSO4 (s)  Ca 2+ (aq) + SO42- (aq) If you add additional solid calcium sulfate to the saturated solution, what will happen? Nothing – Solid CaSO4 does not affect the equilibrium concentrations of Ca 2+ and SO4 2- If you allow a saturated solution of calcium sulfate to evaporate what will happen? Evaporation will reduce the volume of water and dissolved calcium sulfate must precipitate out since the concentrations of Ca 2+ and SO4 2- must remain constant for a system at equilibrium

  2. Assume you have a saturated solution of calcium sulfate at a temperature of 25 C represented by the following equation: CaSO4 (s)  Ca 2+ (aq) + SO42- (aq) If you add additional solid potassium nitrate to the saturated solution, what will happen? Solid KNO3 will dissolve into K + and NO3 – but the equilibrium concentrations of Ca 2+ and SO4 2- will not change since K + and NO3 – are not part of the equilibrium

  3. Assume you have a saturated solution of calcium sulfate at a temperature of 25 C represented by the following equation: CaSO4 (s)  Ca 2+ (aq) + SO42- (aq) If you add additional solid potassium sulfate to the saturated solution, what will happen? Solid K2SO4 will dissolve into K + and SO4 2- , tending to increase the concentration of SO4 2- . However, since the concentration SO4 2- cannot increase due to the equilibrium of Ca 2+ and SO42-, some dissolved Ca 2+ and SO42- must precipitate out as solid CaSO4

  4. Common Ion EffectThe equilibrium expression includes the total concentration of an ion regardless of the source of the ion. For example, if you have a mixture of 0.005 M calcium sulfate and 0.02 M sodium sulfate. Determine the concentration of each ion in the solution. Na + Na2SO4  2 Na+ + SO4 2- 2 (0.02) = 0.04 M Ca 2+ CaSO4 Ca 2+ + SO4 2- 1 (0.005) = 0.005 M SO4 2- Na2SO4  2 Na+ + SO4 2- 1 (0.02) = 0.02 M SO4 2-CaSO4 Ca 2+ + SO4 2- 1 (0.005) = 0.005 M 0.025 M

  5. Determine the solubility of calcium fluoride in a solution containing 0.010 M calcium nitrate if the solubility constant (Ksp) of calcium fluoride is 3.9 x 10 -11 . CaF2 (s)  Ca 2+ (aq) + 2 F - (aq) K = [Ca2+] [F -] 2 = 3.9 x 10 -11

  6. Determine the solubility of calcium fluoride in a solution containing 0.010 M calcium nitrate if the solubility constant (Ksp) of calcium fluoride is 3.9 x 10 -11 . Set up an ICE table for analysis of the information Table is organized by the balanced chemical equation CaF2 (s)  Ca 2+ (aq) + 2 F - (aq) Initial (from Ca(NO3)2) 0.010 0 Change x 2x Equil 0.010 + x 2x

  7. Determine the solubility of calcium fluoride in a solution containing 0.01 M calcium nitrate if the solubility constant (Ksp) of calcium fluoride is 3.9 x 10 -11 . CaF2 (s)  Ca 2+ (aq) + 2 F - (aq) Initial 0.010 0 Change x 2x Equil 0.010 + x 2x K = [Ca2+] [F -] 2 3.9 x 10 -11 = [0.010 + x] [2x]2 This would result in a quadratic equation and a complicated calculation. However, since the Ksp is very small, the concentration of Ca 2+,, x, must be very small – much smaller than 0.010. Therefore, (0.010 + x) is can be approximated to 0.010

  8. Determine the solubility of calcium fluoride in a solution containing 0.01 M calcium nitrate if the solubility constant (Ksp) of calcium fluoride is 3.9 x 10 -11 . CaF2 (s)  Ca 2+ (aq) + 2 F - (aq) Initial 0.010 0 Change x 2x Equil 0.010 + x  0.010 2x Ksp = [Ca2+] [F -] 2 3.9 x 10 -11 = [0.010] [2x]2 3.9 x 10 -11 = 0.04 x2 x = 3.12 x 10 –5 Molar Solubility of CaF2 = 3.12 x 10 –5 M

  9. CaF2 (s)  Ca 2+ (aq) + 2 F - (aq) Initial 0.010 0 Change x 2x Equil 0.010 + x  0.010 2x Ksp = [Ca2+] [F -] 2 3.9 x 10 -11 = [0.010] [2x]2 3.9 x 10 -11 = 0.04 x2 x = 3.12 x 10 –5 Molar Solubility of CaF2 = 3.12 x 10 –5 M

  10. The solubility constant (Ksp) of calcium fluoride is 3.9 x 10 -11 .If one liter of a 4.0 x 10 –6 M solution of calcium fluoride is mixedwith one liter of a 2.0 x 10 –5 M solution of calcium nitrate, would calcium fluoride precipitate from the final solution? Explain. CaF2 (s)  Ca 2+ (aq) + 2 F - (aq) Q = [Ca 2+] [F -] 2 CaF2 (s)  Ca 2+ (aq) + 2 F - (aq) Initial 2.0 x 10 -6 4.0 x 10 -6 Change 1.0 x 10 –5 0 Final 2.005 x 10 -6 4.0 x 10 -6

  11. The solubility constant (Ksp) of calcium fluoride is 3.9 x 10 -11 .If one liter of a 4.0 x 10 –6 M solution of calcium fluoride is mixedwith one liter of a 2.0 x 10 –5 M solution of calcium nitrate, would calcium fluoride precipitate from the final solution? Explain. CaF2 (s)  Ca 2+ (aq) + 2 F - (aq) Initial 2.0 x 10 -6 4.0 x 10 -6 Change + 1.0 x 10 –5 +0 Final 1.2 x 10 -5 4.0 x 10 -6 Q = (1.2 x 10 -5) (4.0 x 10 -6) 2 Q = 1.9 x 10 –16 Since Q is less than Ksp, all the calcium fluoride would remain in solution If Q were greater than Ksp, some calcium fluoride would precipitate.

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