Given that the K sp for CaF2 at 25°C is 3.9 × 10–11, will a precipitate form when 0.10 L of 2.3 × 10–4 M CaCl2 is added to 0.50 L of 5.0 × 10–3 M NaF? Assume volumes are additive.
Please help me I have no clue where to start.
Thanks!
Calculate concentration of CaCl2 and NaF. Total volume is 0.1L + 0.5L = 0.6 L. Each salt is diluted from the initial volume to the final volume.
(CaCl2) = 2.3E-4 M x (0.1L/0.6L) = ?
(NaF) = 5E-3 x (0.5/0.6) = ?
Calculate Qsp = (Ca^2+)(F^-)^2 and plug in the values from above.
Compare Qsp with Ksp. If Qsp>Ksp a ppt forms. If Qsp<Ksp a ppt will not form.
Post your work if you get stuck.
To determine whether a precipitate will form when CaCl2 is added to NaF, we need to compare the Q (the reaction quotient) to the Ksp (the solubility product constant) for CaF2.
The equation for the dissolution of CaF2 is given as:
CaF2(s) ⇌ Ca2+(aq) + 2F^-(aq)
According to the balanced equation, the molar ratio of CaF2 to Ca2+ is 1:1. This means that the concentration of Ca2+ will be equal to the concentration of CaF2 when CaF2 is completely dissolved.
Step 1: Calculate the initial concentrations of Ca2+ and F^- ions:
For Ca2+:
We are adding 0.10 L of a 2.3 × 10–4 M solution of CaCl2, so the initial concentration of Ca2+ is 2.3 × 10–4 M.
For F^-:
We have 0.50 L of a 5.0 × 10–3 M solution of NaF, so the initial concentration of F^- is 5.0 × 10–3 M.
Step 2: Calculate the reaction quotient, Q, by multiplying the concentrations of the ions:
Q = [Ca2+][F^-]^2
Q = (2.3 × 10–4)(5.0 × 10–3)^2
Step 3: Compare the Q value to the Ksp value:
If Q < Ksp, then no precipitate will form because the system is not saturated.
If Q > Ksp, then a precipitate will form because the system is already saturated.
In this case, we compare the Q value to the given Ksp value of CaF2, which is 3.9 × 10–11.
To determine if a precipitate will form when 0.10 L of 2.3 × 10–4 M CaCl2 is added to 0.50 L of 5.0 × 10–3 M NaF, we need to compare the ion concentrations required for precipitation with the ion concentrations provided.
First, let's write the balanced equation for the dissociation of CaF2:
CaF2(s) ⇌ Ca2+(aq) + 2F–(aq)
From the equation, we can see that one mole of CaF2 produces one mole of Ca2+ and two moles of F–.
Next, let's consider the ions present in the solution:
For CaCl2:
CaCl2(s) ⇌ Ca2+(aq) + 2Cl–(aq)
For NaF:
NaF(s) ⇌ Na+(aq) + F–(aq)
Now, calculate the ion concentrations in the mixed solution by using the equation: concentration (M) = moles/volume (L)
For Ca2+:
Concentration = (0.10 L) × (2.3 × 10–4 M) = 2.3 × 10–5 M
For F–:
Concentration = (0.50 L) × (5.0 × 10–3 M) = 2.5 × 10–3 M
From the stoichiometry of the balanced equation, we know that for every Ca2+ ion, two F– ions are required for the formation of CaF2.
To determine if precipitation will occur, we need to compare the ion product (Q) to the solubility product constant (Ksp):
Q = [Ca2+][F–]^2
Now, substitute the ion concentrations calculated earlier into the ion product equation:
Q = (2.3 × 10–5 M)(2.5 × 10–3 M)^2 = 1.4 × 10–9
Since Q (1.4 × 10–9) is less than Ksp (3.9 × 10–11), a precipitate will not form.