Suppose, in the saturated solution of Ca(OH)2 prepared by the TA, some of the dissolved Ca2+ jons re-precipitate as Caco3. This removes Ca2+ ions from the solution. What will the solid Ca(OH)2 present in the bottom of the beaker do in order to minimize the effect of this shift away from equilibrium? What would the effect be on the [OH -] and the calculated Ksp value? O The removal of Ca2+ ions by the precipitation of CaCO3 from solution results in the dissolution of more Ca2+ from Ca(OH)2. As a result, the [OH-] will increase and the estimate of Kep will be too high. O Solid substances are not part of equilibrium equations, so their presence would leave the value of Ksp unaffected. O The removal of Ca2+ ions by the precipitation of CaCo3 from solution results in the co-precipitation of Ca(OH)2 as it attempts to maintain its equilibrium concentrations. As a result, the [OH ] will decrease and the estimate of Ksp will be too low.
Suppose, in the saturated solution of Ca(OH)2 prepared by the TA, some of the dissolved Ca2+ jons re-precipitate as Caco3. This removes Ca2+ ions from the solution. What will the solid Ca(OH)2 present in the bottom of the beaker do in order to minimize the effect of this shift away from equilibrium? What would the effect be on the [OH -] and the calculated Ksp value? O The removal of Ca2+ ions by the precipitation of CaCO3 from solution results in the dissolution of more Ca2+ from Ca(OH)2. As a result, the [OH-] will increase and the estimate of Kep will be too high. O Solid substances are not part of equilibrium equations, so their presence would leave the value of Ksp unaffected. O The removal of Ca2+ ions by the precipitation of CaCo3 from solution results in the co-precipitation of Ca(OH)2 as it attempts to maintain its equilibrium concentrations. As a result, the [OH ] will decrease and the estimate of Ksp will be too low.
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Chapter12: Chemical Equilibrium
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Problem 12.97PAE
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Question
![Suppose, in the saturated solution of Ca(OH)2 prepared by the TA, some of the dissolved Ca2+ ions re-precipitate as CaCO3. This removes Ca2+ ions from the solution. What will the solid Ca(OH)2 present in the bottom of the
beaker do in order to minimize the effect of this shift away from equilibrium? What would the effect be on the [OH-] and the calculated Ksp value?
O The removal of Ca2+ ions by the precipitation of CaCO3 from solution results in the dissolution of more Ca2+ from Ca(OH)2. As a result, the [OH-] will increase and the estimate of Ksp will be too high.
Solid substances are not part of equilibrium equations, so their presence would leave the value of Ksp unaffected.
The removal of Ca2+ ions by the precipitation of CaCO3 from solution results in the co-precipitation of Ca(OH)2 as it attempts to maintain its equilibrium concentrations. As a result, the [OH¯] will decrease and the
estimate of Ksp will be too low.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F668f7243-fecb-4dca-9846-a35dbaef67d8%2Ff990b56d-7a19-4c1c-97af-75a2412a6353%2Fqj6npva_processed.png&w=3840&q=75)
Transcribed Image Text:Suppose, in the saturated solution of Ca(OH)2 prepared by the TA, some of the dissolved Ca2+ ions re-precipitate as CaCO3. This removes Ca2+ ions from the solution. What will the solid Ca(OH)2 present in the bottom of the
beaker do in order to minimize the effect of this shift away from equilibrium? What would the effect be on the [OH-] and the calculated Ksp value?
O The removal of Ca2+ ions by the precipitation of CaCO3 from solution results in the dissolution of more Ca2+ from Ca(OH)2. As a result, the [OH-] will increase and the estimate of Ksp will be too high.
Solid substances are not part of equilibrium equations, so their presence would leave the value of Ksp unaffected.
The removal of Ca2+ ions by the precipitation of CaCO3 from solution results in the co-precipitation of Ca(OH)2 as it attempts to maintain its equilibrium concentrations. As a result, the [OH¯] will decrease and the
estimate of Ksp will be too low.
![A student does not filter his/her saturated solution before titrating. Will the calculated Ksp probably be too high, too low, or unaffected? Why?
Solid substances are not part of equilibrium equations, so their presence would leave the value of Ksp unaffected.
The remaining solids would absorb excess H30+ ions during the titration, making the calculated value of [OH ], and thus Ksp, too low.
O The remaining solids would dissolve during the titration to give more OH¯. The calculated Ksp would therefore be too high.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F668f7243-fecb-4dca-9846-a35dbaef67d8%2Ff990b56d-7a19-4c1c-97af-75a2412a6353%2Ffij1w1_processed.png&w=3840&q=75)
Transcribed Image Text:A student does not filter his/her saturated solution before titrating. Will the calculated Ksp probably be too high, too low, or unaffected? Why?
Solid substances are not part of equilibrium equations, so their presence would leave the value of Ksp unaffected.
The remaining solids would absorb excess H30+ ions during the titration, making the calculated value of [OH ], and thus Ksp, too low.
O The remaining solids would dissolve during the titration to give more OH¯. The calculated Ksp would therefore be too high.
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