First, find the solubility (i. e. equilibrium concentrations) of a Ca2+ and F ions in a solution of excess CaF2. Ksp = 4.0 x 10-11. Next, find the equilibrium concentrations of Ca2+, F", and Ht when you dissolve CaF2 in 0.1 M HNO3. Ka = 7.2 X 10-4 for HF.

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Chapter1: Chemical Foundations
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First, find the solubility (i. e. equilibrium
concentrations) of a Ca2+ and F ions in a
solution of excess CaF2. Ksp = 4.0 x 10-11.
%D
Next, find the equilibrium concentrations of
Ca2+, F", and H* when you dissolve CaF2 in
0.1 M HNO3. Ka = 7.2 X 10-4 for HF.
First, find the solubility (į. e. equilibrium concentrations) of a Ca²+ and F ions in a
solution of excess CaF2. Kse = 4.0 x 10-11.
Next, find the equilibrium concentrations of Ca2+. F, and H* when you dissolve CaF2 in
0.1 M HNO3. Ka = 7.2 X 104 for HF.
Hint: As always, start by writing out the relevant chemical equations and figuring out
what the net reaction will be. Also, you will need to remember what happens to Ka
when you reverse a reaction.
Transcribed Image Text:First, find the solubility (i. e. equilibrium concentrations) of a Ca2+ and F ions in a solution of excess CaF2. Ksp = 4.0 x 10-11. %D Next, find the equilibrium concentrations of Ca2+, F", and H* when you dissolve CaF2 in 0.1 M HNO3. Ka = 7.2 X 10-4 for HF. First, find the solubility (į. e. equilibrium concentrations) of a Ca²+ and F ions in a solution of excess CaF2. Kse = 4.0 x 10-11. Next, find the equilibrium concentrations of Ca2+. F, and H* when you dissolve CaF2 in 0.1 M HNO3. Ka = 7.2 X 104 for HF. Hint: As always, start by writing out the relevant chemical equations and figuring out what the net reaction will be. Also, you will need to remember what happens to Ka when you reverse a reaction.
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