The Ksp for calcium fluoride (CaF2) is 3.9 x 10^–11 at 25°C. How many moles of solid NaF must be added to 1.00 L of 0.100 M CaCl2 in order to precipitate CaF2?
Please help me set up this ICE table and understand how to solve this!! thank you!!
(Ca^2+) = 0.1M
Ksp = (Ca^2+)(F^-)2
Substitute 0.1M for Ca^2+ and solve for (F^-) in mols/L.
You have 1.00 L so that is the number of mols F^- (= mols NaF) that must be added.
To set up the ICE table for this problem, we need to consider the balanced chemical equation for the precipitation of CaF2:
CaCl2(aq) + 2NaF(aq) -> CaF2(s) + 2NaCl(aq)
The molar ratio between CaCl2 and CaF2 is 1:1, meaning for every 1 mole of CaCl2, we will form 1 mole of CaF2. Therefore, the initial concentration of CaF2 is 0 M.
Now, let's construct the ICE table:
Initial:
CaCl2: 0.100 M
NaF: x M (to be determined)
CaF2: 0 M
Change:
CaCl2: -x M
NaF: -2x M (since the molar ratio between CaCl2 and NaF is 1:2)
CaF2: +x M
Equilibrium:
CaCl2: 0.100 - x M
NaF: x M
CaF2: x M
To find the value of x, we need to set up a Ksp expression for CaF2:
Ksp = [Ca2+][F-]^2 = (0.100 - x)(2x)^2 = 3.9 x 10^–11
Simplify the expression:
4x^3 - 0.2x^2 + 3.9 x 10^–11 = 0
This equation has no simple analytical solution, so we need to solve it numerically. Using a graphing calculator or software, we find that x ≈ 9.8 x 10^–5.
Now, we can calculate the number of moles of solid NaF added to the solution:
moles of NaF = concentration × volume = x × 1.00 L = (9.8 x 10^–5) × 1.00 = 9.8 x 10^–5 moles
Therefore, approximately 9.8 x 10^–5 moles of solid NaF must be added to precipitate CaF2 in 1.00 L of 0.100 M CaCl2 solution.
To solve this problem, we can use an ICE table (Initial, Change, Equilibrium) to track the changes in concentration of the reactants and products.
Step 1: Write the balanced chemical equation for the precipitation reaction:
CaCl2 + 2NaF -> CaF2 + 2NaCl
Step 2: Set up the ICE table:
In this case, the initial concentration of CaCl2 is given as 0.100 M, and we want to find the amount of NaF needed to precipitate CaF2. Let's assume x moles of NaF are added.
Initial concentrations:
CaCl2: 0.100 M
NaF: x M (unknown)
CaF2: 0 M (since it's a solid)
NaCl: 0 M
Change in concentration:
CaCl2: -x (reacts with x moles of NaF to form CaF2)
NaF: -2x (reacts with CaCl2 to form CaF2)
CaF2: +x (product formed)
NaCl: +2x (product formed)
Equilibrium concentrations:
CaCl2: 0.100 - x
NaF: x - 2x = -x
CaF2: x
NaCl: 2x
Step 3: Set up the expression for the equilibrium constant Ksp:
The solubility product constant (Ksp) is the product of the equilibrium concentrations of the products, raised to the power of their stoichiometric coefficients, divided by the product of the equilibrium concentrations of the reactants, raised to the power of their stoichiometric coefficients.
For CaF2: Ksp = [CaF2]
For CaCl2: [CaCl2] = 0.100 - x
For NaF: [NaF] = -x
Ksp = [CaF2][NaCl]² / [CaCl2]²[NaF]²
3.9 x 10^–11 = x * (2x)² / (0.100 - x)²
Step 4: Solve the equation to find the value of x:
Now we can solve the equation to find the value of x, which corresponds to the number of moles of NaF needed.
Multiply both sides of the equation by (0.100 - x)²:
(0.100 - x)² * 3.9 x 10^–11 = 4x²
Expand and simplify:
0.039 x 10^-11 - 0.2x + x² * (3.9 x 10^-11) = 4x²
Combine like terms:
x² * (3.9 x 10^-11) - 4x² + 0.2x - 0.039 x 10^-11 = 0
Rearrange the equation:
-0.039 x 10^-11 - 4x² + x² * (3.9 x 10^-11) + 0.2x = 0
Simplify and rewrite:
x² * (3.9 x 10^-11 - 4) + 0.2x - 0.039 x 10^-11 = 0
Step 5: Solve the quadratic equation:
To solve the quadratic equation, we can use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
In this case, a = (3.9 x 10^-11 - 4), b = 0.2, and c = -0.039 x 10^-11.
Substitute these values into the quadratic formula and calculate x.
Step 6: Calculate the moles of NaF:
Once you have the value of x, you can calculate the moles of NaF added, which is equal to the value of x.
Finally, convert the moles of NaF to the volume of NaF needed by using the molarity:
Volume (L) = moles / molarity
This will give you the volume of NaF needed to precipitate CaF2.