Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?

This is what i found for ksp
CaF2<--> Ca(2+) +2F(-) Ksp=(x)(4x^2) where x=solubility Therefore, Ksp=3.7 x 10^-11

and for Qsp I am not sure how to go on about calculating that.

I answered this below.

thank you sir.

To determine if a fluoride concentration of 1.0 mg/L will be soluble in water containing 200 mg/L of calcium, you should compare the Qsp (the reaction quotient) to the Ksp (the solubility product constant) of calcium fluoride (CaF2).

First, let's calculate the Qsp using the given concentrations. The equation for the dissolution of calcium fluoride is:

CaF2 ⟶ Ca2+ + 2F-

The Qsp expression is given as:

Qsp = [Ca2+][F-]²

Based on the provided concentrations, [Ca2+] = 200 mg/L and [F-] = 1.0 mg/L.

However, it's important to convert the concentrations from mg/L to mol/L (or M) before calculating Qsp and comparing it to Ksp.

To convert the concentrations, you need to know the molar masses of calcium (40.08 g/mol) and fluoride (19.00 g/mol).

[Ca2+] = 200 mg/L * (1 g/1000 mg) / (40.08 g/mol) = 0.004995 M (approximately 0.005 M)

[F-] = 1.0 mg/L * (1 g/1000 mg) / (19.00 g/mol) = 0.0526 M (approximately 0.053 M)

Now, we can calculate Qsp:

Qsp = (0.005 M)(0.053 M)² = 1.4 x 10^-6

Next, compare the calculated Qsp value to the Ksp value. The Ksp for calcium fluoride is given as 3.7 x 10^-11. If Qsp is greater than Ksp, it means the solution is supersaturated and any additional calcium fluoride added will not dissolve completely. On the other hand, if Qsp is less than Ksp, it means the solution is unsaturated, and more calcium fluoride can dissolve.

In this case, Qsp (1.4 x 10^-6) is significantly greater than Ksp (3.7 x 10^-11), indicating that the solution is saturated or possibly supersaturated. Therefore, a fluoride concentration of 1.0 mg/L will generally be soluble in water containing 200 mg/L of calcium.