K sp for Ca(OH) 2 is determined in two separate experiments, both at the same te
ID: 487045 • Letter: K
Question
Ksp for Ca(OH)2 is determined in two separate experiments, both at the same temperature.
- by titration of hydroxide in a simple saturated solution of Ca(OH)2.
- by titration of hydroxide in a 0.01 M CaCl2 solution saturated with Ca(OH)2.
Identify which of the following statements is either completely "True" or at least partially "False"
(assuming that the activity of each ion is equal to its molar concentration in each instance):
True False 1) In the second experiment, both [Ca2+] and [OH] are greater than in the the first experiment.
True False 2) You would expect more Ca(OH)2 to be able to dissolve in the simple solution, than in the CaCl2 solution.
True False 3) You would expect the value for Ksp to be the same from both experiments, since Ksp is a constant at a given temperature.
True False 4) You would expect the value for Ksp to be the same from both experiments, since the presence of CaCl2(aq) cannot affect the equilibrium position.
True False 5) You would expect the value for Ksp to be greater in the second experiment, since there is more Ca2+(aq) present in that solution.
Explanation / Answer
1) False: The equilibrium will shift with the addition of the Ca2+ common ion and therefore there will be a greater concentration of Ca2+ and a lower concentration of OH-
2) True: More Ca(OH)2 dissolves in the simple solution than the CaCl2 solutions because the added ions shift the equilibrium to the left.
3)True : Ksp is the same, the equilibrium just shifts to compensate.
4)False: Ksp is the same, but the addition of CaCl2 does affect equilibrium position due to the common ion effect.
5) False: More Ca2+ is present, but Ksp is still the same.
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