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2. Calculate the Kop value, if [F]-4.10 x 104 M for a saturated solution of CaF2....
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+...
Consider a saturated AgBr solution that is also 0.050 M Ca(NO3)2. Ca(NO3)2(aq) + Ca2+ (aq) + 2NO3 (aq) AgBr(s) Ag+ (aq) + Br-(aq) The thermodynamic solubility product (Kop) of AgBr is 5.0 x 10-13. What must be the molarity of Ag+ in the solution?
Consider the dissolution of CaF2 in aqueous solution: CaF2 (s) ⇌ Ca(+2) (aq) + 2F(-) (aq). If we start with 1M of undissolved CaF2 in a solution that already contains 0.015M of F(-) ions, what is the solubility of CaF2 at 298.15K in terms of molarity? (Hint: Look at the final simulated concentration of Ca(+2).) A. 3.7E-7 M B. 2.5E-6 M C. 1.2E-5 M
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.
If 1.2 moles of CaF2 (s) are added to 2.7 L of water what will the concentration of F- be at equilibrium? The Kop - 3,9 x 10-11 CaF, (s) = Ca (aq) + 2F (aq) O(A) 5.38 x 10 M (B) 3. 12 x 10-6 M (C) 2.14 x 10- M (D) 7.91 x 10-M (E) 4.27 x 10-4 M
Calculate the concentration of ca2+, including activity coefficients, in a 0.08 M NaF solution saturated with CaF2. Ksp=3.2x10^-11 for CaF2, alphaCa2+=600 pm, alphaF-=350 pm.
2. The equilibrium constant for the following reaction is called the "solubility product" of calcium fluoride: CaF2(s) - Cal(aq) + 2F-(ay) K = Kp = 3.2 x 10-11 (a) Write an expression for the equilibrium constant of this reaction in terms of concentrations. Why do you suppose we call this a solubility product instead of a solubility quotient or ratio? (b) Calculate the equilibrium concentrations of Ca2+ and F if excess solid CaF is placed in water. (c) in which...
Learning Goal: To learn how to calculate the solubility from Kspand vice versa. Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution: CaF2(s)⇌Ca2+(aq)+2F−(aq) At equilibrium, the ion concentrations remain constant because the rate of dissolution of solid CaF2 equals the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is Ksp=[Ca2+][F−]2 Ksp is called the solubility product and can be determined experimentally by measuring thesolubility, which is the amount...
2. Solubility and Ksp of saturated CallO3)2 in o.0100 M KIO3 About 30-35 mL of this saturated solution was filtered. Two 10.00 mL samples of this solution were combined with -2 g KI, -50 mL of H20, and 10 mL 1.0 M HCL Each solution was titrated with standard thisolulfate solution (0.04913 M) exactly as in part 2 to a colorless end-point. The following data was collected Trial 1 Trial 2 Vr 23.79 mL 47.64 mL V- 0.00 mL 23.79...