Determine the molar solubility of BaF2 in a solution containing 0.0750 M LiF. Ksp (BaF2) = 1.7 ? 10^-6.
Please do not leave x(2x+.0750)^2 = 3.0x10^-4.
I know that is the set up, I know why that is the set up, and I know that, that is the answer. Please show me the work to get that answer.
If you have to foil show me the foil method, if you have to add a number, show me. My brain is apparently not working right now! :/ Please show all steps.
Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.
Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).
Common ion effect is defined as the solubility of a partially soluble salt which will decrease solubility with the addition of a soluble salt that has an ion in common with them.
Consider a general reaction:
The relation between solubility product and molar solubility is as follows:
Here,
The solubility product of salt is.
The molar solubility of M ion = .
The molar solubility of X ion = .
An ICE table is used to find the concentrations or moles of individual reactants and products at equilibrium.
The construction of an ICE table is as follows:
All quantities are expressed in terms of concentrations or moles.
The dissociation of the given compound is written below.
Solubility product of the given reaction is shown below.
The total concentration of solution is as follows:
Now construct ICE table to find the equilibrium concentration of barium ion in the solution.
On solving the cubic expression above (online),
Therefore, the molar solubility of in of is
Ans:The molar solubility of in 0.0750M of is
Determine the molar solubility of BaF2 in a solution containing 0.0750 M LiF. Ksp (BaF2) = 1.7 ? 10^-6. Please do n...
Determine the molar solubility of BaF2 in a solution containing 0.0750 M LiF. Ksp (BaF2) = 1.7
Determine the molar solubility of BaF2 in a solution containing 0.0750 M LiF. Ksp, BaF2 = 1.7 X 10-6. Which one of these is the correct answer? BaF2 molar solubility = 2.3 x 10-5 M BaF2 molar solubility = 0.0750 M BaF2 molar solubility = 8.5 x 10-7 M BaF2 molar solubility = 1.2 x 10-2 M BaF2 molar solubility = 3.0 x 10-4 M
Determine the molar solubility of BaF2 in a solution containing 0.350 M NaF. Ksp (BaF2) = 9.8 × 10-6. a) 4.5 × 10-5 M b) 8.0× 10-5 M c) 3.6 × 10-4 M d) 2.3 × 10-5 M e) 8.2 × 10-9 M
1.) Determine the molar solubility of barium fluoride (BaF2, Ksp = 1.7 X 10-6) in a 0.500 M sodium fluoride solution.
Determine the molar solubility of CuCl in a solution containing 0.030 M LiCl. Ksp (CuCl) = 1.7×10-7. Determine the molar solubility of CuCl in a solution containing 0.030 M LiCl. Ksp (CuCl) = 1.7×10-7. 4.1×10-4 M 2.9×10-14 M 5.7×10−6 M 5.1×10−9 M 4.7×10−5 M
6. The Ksp for Barium Fluoride, BaF2, is 1.0 x 10-6. a. Write the Ksp expression for this salt. b1. Calculate the concentration of the anion. Do not do the algebra. Leave your answer in terms of x. b2. Calculate the molar solubility. Do not do the algebra. Leave your answer in terms of x. c. Calculate the concentration of the barium ion if 0.20 moles of KF is added to 1.0liter of the saturated barium fluoride solution. You do...
Calculate the molar solubility of CaF2 in a solution containing 0.325 M of Ca(NO3)2. The Ksp value for CaF2 is 1.46×10−10. can you also explain me how to solve for X toward the end of the problem
Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 × 10-13.
Determine the molar solubility of CaSO4 in a solution containing 9.0×10−2 M K2SO4. Ksp (CaSO4) = 7.1×10-5. 8.4×10-3 M 5.0×10-9 M 7.9×10−4 M 2.2×10−3 M 6.4×10−6 M
Determine the molar solubility of AgBr in a solution containing 0.200 M NaBr. Ksp (AgBr) = 7.7 × 10-13. a. 3.8 × 10-12 M b. 5.8 × 10-5 M c. 0.200 M d. 8.8 × 10-7 M e.1.54 × 10-13 M