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,
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,
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 - 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].
D. Barium Carbonate, BaCO3 (Witherite) in a solution containing 3.0 x 10-5 m dissolved Calcium Fluoride, CaF2 (Fluorite). What are [Ba] and [CO3] in solution?
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