Calcium fluoride, the main component of the mineral fluorite, dissolves to a slight extent in water.
CaF2 (s) ⇄ Ca2+ (aq) + 2F-(aq)
Ksp = [Ca2+][F-]2
Calculate the Ksp value for CaF2 if the calcium ion concentration has been found to be 2.3×10-4 M.
Ksp = [Ca2+][F-]2
[Ca+2] = 2.3 x 10-4 M
[F-] = 2 x 2.3 x 10-4 = 4.6 x 10-4 M
Ksp = [2.3 x 10-4][4.6 x 10-4]2
Ksp = 5.0 x 10-11
Calcium fluoride, the main component of the mineral fluorite, dissolves to a slight extent in water....
I need all the help possible on this....thank you so much. I attached two pics...its the same question. The first picture gives background on the actual questions 2. A geologist found a large deposit of the mineral Fluorite, CaF2, within a pond of stagnant water that is undisturbed by biological activities. This mineral had been leached into the water and had established equilibrium at 25°C for a considerable length of time. The equation for the solvation of Fluorite in water...
17. Calculate the solubility of calcium fluoride in a 0.035 M sodium fluoride solution. Some important information is given below: ▪ NaF will ionize completely, and the reaction is: NaF (aq) → Na+ (aq) + F− (aq) ▪ The calcium fluoride equilibrium reaction is: CaF2 (s) → Ca+2 (aq) + 2F−1 (aq) ▪ For CaF2 (s), Ksp is 4.0 × 10−11 ▪ Na+ (aq) is a spectator ion; The F− (aq) is the common ion.
10 L of a saturated solution of calcium fluoride, CaF2, was evaporated and found to contain 4.28 millimoles (mmol) fluoride. What is the value of Ksp? CaF2(s) -> <- ca2+ (aq) +2 F- (aq) How did Ksp= [Ca2+] [F-]2 become Ksp= [s] [2s]2= 4s3 Cause I thought it was only Ksp= [s] [s]2 = s3 and why 4.28x10-3 have to devide by 10 and why the [F-] = 4.28 x 10-4 have to eqautes to 2 to have s= 2.14x10-4...
1. Solid calcium fluoride (CaF2) establishes the following equilibrium in solution: CaF2(s) = Ca2+(aq) + 2F-(aq) Ke = 1.5 x 10-10 A solution initially contains 2.45 g of CaF2. a. In which direction will the reaction move to reach equilibrium? Explain your reasoning b. Calculate the equilibrium concentrations of Ca2+ and F. How much CaF2 (in mg) dissolves in solution? c. Another solution initially contains 2.45 g of CaF2 and 0.0025 M NaF. What are the equilibrium concentrations of Ca2+...
The mineral fluorite is calcium fluoride, CaF_2. The goal is to calculate the molar solubility of CaF_2 in water from its solubility product constant (k_sp=3.4 times 10^-11). Write down the ICE chart: Write down K_sp in terms of the solubility (s) and solve for s. The goal now is to calculate solubility after adding 0.15 MCaCl_2. Write down the new ICE chart and solve for the new s.
Fluorite, CaF2, is a slightly soluble salt in water. In a saturated solution, the concentration of [Ca2+] = 2.5 x 10^-4 and [F-] = 4.30 x 10^-4 a. write the dissolution equation for fluorite. b. Write Ksp expression for the dissolution of fluorite. c. Calculate the Ksp.
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The department of health and human services recently recommended the concentration of fluoride in drinking water be limited to 0.7mg/L (the current policy allows for a range of 0.7 to 1.2 mg/L). The concentration of fluoride in natural water depends on The presence of fluoride-containing minerals 9. (7 points) The Department of Health and Human Services recenty recommended oe concentration of fluoride in drinking water be limited to 0.7 mg (the current poley wws t a range of 0.7 to...
Groundwater at a temperature of 250C flows through strata containing abundant fluorite (CaF2) mineralization and becomes saturated with respect to that mineral. The equilibrium solubility product, Ksp, for this reaction is 10^-10.4. a) Calculate the concentrations of Ca2+ and F- once the water has reached saturation with respect to fluorite assuming that prior to entering the mineralized strata the water had an ionic strength of less than 10-5 mol/L and contained negligible Ca2+ and negligible F-. For this part of...