Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?
I got Qsp 1.38 * 10^-8 and Ksp = 3.45*10^-11 does that seem right? and does this mean fluride concentration will be soluble in water containing calcium?
I'll bet good money that you didn't square the F^-
1 mg/L F = 0.001/19 = 5.26E-5 mols/L
200 mg/L Ca = 0.200/40.078 = 4.99E-3 M
Qsp = (Ca^2+)(F^-)^2 =
(4.99E-3)(5.26E-5)^2 = 1.38E-11. This is smaller than 3.45E-11; therefore, no ppt; i.e., Ksp is not exceeded. Ppts occur when Qsp exceeds Ksp.
By the way, 1.38E-8 is larger than Ksp by a factor of about 1000; using your figures Ksp is exceeded and you get a ppt.
oh alright thank you i found out my mistake. thank you sir.
To determine if the fluoride concentration of 1.0 mg/L will be soluble in water containing 200 mg/L of calcium, we need to compare the solubility product constant (Ksp) of calcium fluoride (CaF2) with the ionic product (Qsp) under the given conditions.
The solubility product constant (Ksp) for calcium fluoride (CaF2) is the equilibrium constant for its dissolution in water. It can be expressed as [Ca2+][F-]^2, where [Ca2+] is the concentration of calcium ions, and [F-] is the concentration of fluoride ions in the saturated solution. The Ksp value for calcium fluoride is 3.45 x 10^-11.
To determine if the fluoride concentration is soluble, we need to calculate the ionic product (Qsp). The ionic product is calculated similarly to the Ksp, but it uses the actual concentrations of the ions involved.
In this case, the concentration of calcium ions is 200 mg/L or 0.2 g/L. Since calcium fluoride has a 1:2 stoichiometric ratio between calcium ions and fluoride ions, the concentration of fluoride ions can be calculated as follows:
Fluoride concentration (F- concentration) = (0.2 g/L) / 2 = 0.1 g/L = 100 mg/L
Now that we have the actual concentrations of calcium and fluoride ions, we can calculate the ionic product (Qsp):
Qsp = [Ca2+][F-]^2 = (0.2 g/L) × (0.1 g/L)^2 = 0.002 g^3/L^3 = 2 x 10^-3 g^3/L^3
Comparing Qsp with the Ksp value, which is 3.45 x 10^-11, we can see that Qsp is larger. Therefore, Qsp > Ksp, indicating that the fluoride concentration of 1.0 mg/L will be soluble in water containing 200 mg/L of calcium.
In summary, your calculated Qsp value of 1.38 x 10^-8 is incorrect. However, based on the correct Qsp value calculated above (2 x 10^-3 g^3/L^3) being greater than the Ksp value (3.45 x 10^-11), it indicates that the fluoride concentration of 1.0 mg/L will be soluble in water containing 200 mg/L of calcium.