A 50.0-mL sample of 0.100 M Ca(NO₃)₂ is mixed with 50.00 mL of 0.200 M NaF.When the system has come to equilibrium,which of the following sets of conditions will hold? The K_sp for CaF₂ is 4.0 $\times$ 10^-¹¹. Moles Solid CaF₂

Question

Quizplus · Accepted Answer

We can use the equilibrium constant expression for the reaction:

Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0 + 2 NaFS1U1B13S1U1B0 ⇌ CaFS1U1B12S1U1B0 + 2 NaNO3S1U1B10

K = [CaFS1U1B12S1U1B0][NaNO3S1U1B10]^2 / [Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0][NaFS1U1B13S1U1B0]^2

Since the initial concentrations of Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0 and NaFS1U1B13S1U1B0 are equal and the reaction stoichiometry is 1:2, we can assume that x moles of CaFS1U1B12S1U1B0 will be formed and 2x moles of NaNO3S1U1B10 will be formed at equilibrium. The initial moles of Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0 and NaFS1U1B13S1U1B0 in the mixture are:

moles Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0 = 0.100 mol/L x 0.050 L = 0.0050 mol
moles NaFS1U1B13S1U1B0 = 0.200 mol/L x 0.050 L = 0.0100 mol

After the reaction reaches equilibrium, the concentrations of CaFS1U1B12S1U1B0 and NaNO3S1U1B10 are:

[CaFS1U1B12S1U1B0] = x / (0.050 L + 0.050 L) = x / 0.100 L
[NaNO3S1U1B10] = 2x / (0.050 L + 0.050 L) = 2x / 0.100 L

[Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0] and [NaFS1U1B13S1U1B0] will be decreased by x and 2x moles, respectively, so their concentrations will be:

[Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0] = (0.0050 mol - x) / 0.100 L
[NaFS1U1B13S1U1B0] = (0.0100 mol - 2x) / 0.100 L

Substituting these expressions into the equilibrium constant expression and solving for x, we get:

4.0 F1F1F1S1F1F1F10 10S1U1P1-S1S1P0S1U1P111S1S1P0 = (x / 0.100)^1 (2x / 0.100)^2 / [(0.0050 - x) / 0.100]^1 [(0.0100 - 2x) / 0.100]^2

Simplifying and solving for x, we get:

x = 0.0557 mol

Therefore, the moles of CaFS1U1B12S1U1B0 formed at equilibrium is 0.0557 mol, which corresponds to choice C.

A 500-ML Sample of 0 ×\times× 10-11

A 500-ML Sample of 0 $\times$ 10^-¹¹