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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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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
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
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
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
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
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
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
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
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
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
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.