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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 ? 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.

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Answer #1
Concepts and reason

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.

Fundamentals

Consider a general reaction:

M.X.) 2 nM (aq) +mX+ (aq)

The relation between solubility product and molar solubility is as follows:

K. = [mm*] [x]

Here,

The solubility product of salt is.

The molar solubility of M ion = M .

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:

The initial quantities is denoted as I. The change in each quantity is denoted as C. The equilibrium quantities is denoted as

All quantities are expressed in terms of concentrations or moles.

The dissociation of the given compound is written below.

BaF, Ba2+ + 2F

Solubility product of the given reaction is shown below.

BaF, = Ba²+ + 2F K = [Ba²+ ][F]

The total concentration of solution is as follows:

[LiF]=0.0750 M

LiF- Lit +F

[F]=0.0750 M

Now construct ICE table to find the equilibrium concentration of barium ion in the solution.

BaF (3) I(M) C(M) E(M) Ba?* (aq) + 2F (aq) 0 0.075 +2x 0.075 + 2x + IX

1.7*10*=(x) (2x +0.0750) 1.7x10º =(x)(4x² +0.005625+0.3x) 1.7*10= 4x +0.3x² +0.005625x 4x +0.3x² +0.005625x -1.7x10=0

On solving the cubic expression above (online),

x=0.000297 [Ba²+] = 0.000297M [Ba²+] = 2.97*10^M [Ba2+] = 3.0x10^M (two significant figures)

Therefore, the molar solubility of in 0.0750M of is 3.0x10*M

Ans:

The molar solubility of in 0.0750M of is 3.0x10*M

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