How many grams of MgF2 will dissolve in 150 ml of 0.100 M NaF SOLUTION ? Ksp for MgF2 IS 6.4 X 10-9
i need help PLEASE
To determine the amount of MgF2 that will dissolve in the given NaF solution, we need to use the concept of the solubility product constant (Ksp).
The balanced equation for the dissociation of MgF2 is:
MgF2 (s) ↔ Mg2+ (aq) + 2F- (aq)
From the balanced equation, we can see that each mole of MgF2 produces one mole of Mg2+ ions and two moles of F- ions upon dissociation.
Given:
Volume of NaF solution = 150 ml = 0.150 L
Concentration of NaF solution = 0.100 M
Ksp for MgF2 = 6.4 x 10^-9
Now, let's assume "x" grams of MgF2 dissolves in the solution. This will result in the formation of "x" moles of Mg2+ ions and "2x" moles of F- ions.
Considering the equation MgF2 (s) ↔ Mg2+ (aq) + 2F- (aq), the solubility product expression for MgF2 can be written as:
Ksp = [Mg2+] [F-]^2
Substituting the concentrations:
Ksp = x * (2x)^2 = 4x^3
Now, we can set up an equation to solve for "x":
Ksp = 4x^3
6.4 x 10^-9 = 4x^3
Solving for "x":
x^3 = (6.4 x 10^-9) / 4
x^3 = 1.6 x 10^-9
Taking the cube root of both sides:
x ≈ 1.26 x 10^-3
Therefore, approximately 1.26 x 10^-3 grams of MgF2 will dissolve in 150 ml of 0.100 M NaF solution.
To determine the number of grams of MgF2 that will dissolve in 150 mL of 0.100 M NaF solution using the given Ksp value, we need to set up an ICE table and solve for the concentration of Mg2+ and F- ions.
1. Write the balanced equation for the dissociation of MgF2:
MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
2. Set up the ICE table:
| MgF2(s) | ↔ | Mg2+(aq) | + | 2F-(aq) |
| Initial moles | | | | |
| Change (mol) | -x | +x | | +2x |
| Equilibrium moles | | | | |
3. Express the equilibrium concentration of each ion in terms of x. The concentration of Mg2+ is equal to x (since the stoichiometry is 1:1), and the concentration of F- is equal to 2x.
4. Write the expression for the solubility product constant (Ksp) of MgF2:
Ksp = [Mg2+][F-]^2
Substitute the equilibrium concentrations into the Ksp expression:
Ksp = (x)(2x)^2 = 6.4 x 10^-9
5. Simplify and solve for x:
(4x^3) = 6.4 x 10^-9
x^3 = (6.4 x 10^-9) / 4
x^3 = 1.6 x 10^-9
x ≈ ∛(1.6 x 10^-9)
6. Calculate the value of x using a scientific calculator or computer software. The approximate value of x is determined to be 1.26 x 10^-3 M.
7. Calculate the number of moles of MgF2 dissolved:
Moles of MgF2 = (150 mL) x (0.100 mol/L) = 15.0 mmol = 0.015 moles
8. Calculate the mass of MgF2 dissolved:
Mass of MgF2 = (0.015 moles) x (62.3 g/mol) = 0.9345 grams
Therefore, approximately 0.9345 grams of MgF2 will dissolve in 150 mL of 0.100 M NaF solution.
MgF2 is slightly soluble and has a Ksp so---
......................MgF2 ==> Mg^2+ + 2F^-
I......................solid.............0...........0
C.....................solid..............x..........x
E......................solid..............x...........x
NaF is very soluble and dissociates completely so---
.........................NaF==> Na^+ + F^-
I.........................0.1M.......0..........0
C......................-0.1M........0.1M....0.1M
E.........................0............0.1M.....0.1M
Therefore, in solution you have xM of Mg^2+ and 2x M of F^- from the MgF2. In addition you have 0.1 M F^- from the NaF. Ksp looks like this.
Ksp = (Mg^2+)(F^-)^2
6.4E-9 = (x)(2x+0.1)
Solve for x = molar solubility (moles/L) of MgF2
Convert to grams/L. mols x molar mass = grams MgF2.
Then convert g/L to g/150 mL.
Post your work if you run into trouble.