An aqueous solution contains 2.1 x 10° mol/L Ca2*. Enough HF was then added to the solution to reach an equilibrium concentration of [HF] = 5.5 x 10“ mol/L. What must be the pH of the solution in order to observe a precipitate of CaF2? (Hint: Assume any change in pH will not change the value of [HF]). Ksp of CaF2 = 5.3 × 10-9 Ka of HF = 7.2 × 10-4 %3D
An aqueous solution contains 2.1 x 10° mol/L Ca2*. Enough HF was then added to the solution to reach an equilibrium concentration of [HF] = 5.5 x 10“ mol/L. What must be the pH of the solution in order to observe a precipitate of CaF2? (Hint: Assume any change in pH will not change the value of [HF]). Ksp of CaF2 = 5.3 × 10-9 Ka of HF = 7.2 × 10-4 %3D
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![5. a) An aqueous solution contains 2.1 x 10³ mol/L Ca²*. Enough HF was then added to the
solution to reach an equilibrium concentration of [HF] = 5.5 x 10“ mol/L. What must be
the pH of the solution in order to observe a precipitate of CaF2?
(Hint: Assume any change in pH will not change the value of [HF]).
Ksp of CaF2 = 5.3 × 10-9
Ka of HF = 7.2 × 10-4
%3D
5. b) Predict and briefly explain the sign of AS for each of the following processes.
(Don't write an essay, one short sentence is sufficient explanation).
Nis) + 4 CO(g) → Ni(CO)4(9)
i.
ii.
H20 (s) → H20(1)
H20 (1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F41e3df56-9703-4fb1-a607-01435b10fd2c%2Fcdfd59ea-c592-4654-bd3d-9a188aab8edf%2Fpxof0j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5. a) An aqueous solution contains 2.1 x 10³ mol/L Ca²*. Enough HF was then added to the
solution to reach an equilibrium concentration of [HF] = 5.5 x 10“ mol/L. What must be
the pH of the solution in order to observe a precipitate of CaF2?
(Hint: Assume any change in pH will not change the value of [HF]).
Ksp of CaF2 = 5.3 × 10-9
Ka of HF = 7.2 × 10-4
%3D
5. b) Predict and briefly explain the sign of AS for each of the following processes.
(Don't write an essay, one short sentence is sufficient explanation).
Nis) + 4 CO(g) → Ni(CO)4(9)
i.
ii.
H20 (s) → H20(1)
H20 (1)
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