I need help calculating solubility for this question. thank you C. Calcium Fluoride, CaF2 (aka, Fluorite), in a solution containing 19 mg/L of LiF (aka, Griceite), with all of the LiF being dissolved into solution. Please assume that LiF dissolves such that [Li] = [F] = 7.33 x 10-4 m. After adding CaF2 to the solution and allowing it to dissolve, what are final values of [Ca], [Li] and [F]? Note: you are given the concentrations of [Li] and [F] that result from dissolving LiF. No solubility calculation is required for dissolving LiF; just use the concentration given above. Steps: - First, write the stoichiometric equation for dissolution of CaF2, with the solid on the left and the two ions in solution on the right, separated by a "two-way" arrow. Hint: when CaF2 dissolves, you get 2 moles of F for every 1 mole of Ca. - Second, set up a law of mass action equation for dissolution of CaF2. There is a common ion effect for F, so if [Ca] = x, then [F]= 2x + [FLiF], where [FLiF] is from dissolution of the LiF. Rearrange the equation for dissolution of CaF2 so that x appears on both sides of the equation, label them x0 on LHS and x1 on RHS. Now iterate to find the value of x, as shown in my notes - Finally, report total concentrations for all three of the main ions. For Ca and Li, there is no common ion effect, but for F, you have to account for [FLiF] and [FCaF2].

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter15: Equilibria Of Other Reaction Classes

Section: Chapter Questions

Problem 11E: The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities...

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I need help calculating solubility for this question. thank you C. Calcium Fluoride, CaF2 (aka, Fluorite), in a solution containing 19 mg/L of LiF (aka, Griceite), with all of the LiF being dissolved into solution. Please assume that LiF dissolves such that [Li] = [F] = 7.33 x 10-4 m. After adding CaF2 to the solution and allowing it to dissolve, what are final values of [Ca], [Li] and [F]? Note: you are given the concentrations of [Li] and [F] that result from dissolving LiF. No solubility calculation is required for dissolving LiF; just use the concentration given above. Steps: - First, write the stoichiometric equation for dissolution of CaF2, with the solid on the left and the two ions in solution on the right, separated by a "two-way" arrow. Hint: when CaF2 dissolves, you get 2 moles of F for every 1 mole of Ca. - Second, set up a law of mass action equation for dissolution of CaF2. There is a common ion effect for F, so if [Ca] = x, then [F]= 2x + [FLiF], where [FLiF] is from dissolution of the LiF. Rearrange the equation for dissolution of CaF2 so that x appears on both sides of the equation, label them x0 on LHS and x1 on RHS. Now iterate to find the value of x, as shown in my notes - Finally, report total concentrations for all three of the main ions. For Ca and Li, there is no common ion effect, but for F, you have to account for [FLiF] and [FCaF2].

2. For each of the following compounds, calculate the solubilities (concentrations at equilibrium,
in molality) for ions involved using the law of mass action, where reversible reaction:
CC
dD+eE
reaches equilibrium when:
K =
sp
e
ad α [D]¹[E]ª
D
[c]
C
αc

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