The solubility product of calcium fluoride (CaF
) is 3 x 10
. Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?
The Correct Answer and Explanation is:
To determine whether a fluoride concentration of 1.0 mg/L will remain soluble in water containing 200 mg/L of calcium, we must compare the ionic product with the solubility product constant (Ksp) of calcium fluoride (CaF₂).
Step 1: Write the dissociation equation
CaF2(s)↔Ca2+(aq)+2F−(aq)\text{CaF}_2 (s) \leftrightarrow \text{Ca}^{2+} (aq) + 2\text{F}^- (aq)CaF2(s)↔Ca2+(aq)+2F−(aq)
Step 2: Use the Ksp value
The solubility product constant (Ksp) of CaF₂ is:
Ksp=[Ca2+][F−]2=3×10−11K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = 3 \times 10^{-11}Ksp=[Ca2+][F−]2=3×10−11
Step 3: Convert concentrations to mol/L
- Calcium (Ca²⁺):
200 mg/L=20040.08=4.99 mmol/L=4.99×10−3 mol/L200 \, \text{mg/L} = \frac{200}{40.08} = 4.99 \, \text{mmol/L} = 4.99 \times 10^{-3} \, \text{mol/L}200mg/L=40.08200=4.99mmol/L=4.99×10−3mol/L - Fluoride (F⁻):
1.0 mg/L=1.018.998=5.26×10−5 mol/L1.0 \, \text{mg/L} = \frac{1.0}{18.998} = 5.26 \times 10^{-5} \, \text{mol/L}1.0mg/L=18.9981.0=5.26×10−5mol/L
Step 4: Calculate the ionic product
Ionic Product=[Ca2+][F−]2=(4.99×10−3)(5.26×10−5)2\text{Ionic Product} = [\text{Ca}^{2+}][\text{F}^-]^2 = (4.99 \times 10^{-3})(5.26 \times 10^{-5})^2Ionic Product=[Ca2+][F−]2=(4.99×10−3)(5.26×10−5)2
=(4.99×10−3)(2.77×10−9)= (4.99 \times 10^{-3})(2.77 \times 10^{-9})=(4.99×10−3)(2.77×10−9)
=1.38×10−11= 1.38 \times 10^{-11}=1.38×10−11
Step 5: Compare the ionic product with the Ksp
- Ionic Product = 1.38×10−111.38 \times 10^{-11}1.38×10−11
- Ksp = 3.0×10−113.0 \times 10^{-11}3.0×10−11
Since the ionic product is less than the Ksp, the solution is unsaturated with respect to calcium fluoride. This means no precipitation will occur, and all of the fluoride will remain dissolved.
Conclusion:
Yes, a fluoride concentration of 1.0 mg/L will be soluble in water that contains 200 mg/L of calcium. The current ion concentrations do not exceed the solubility limit of CaF₂, and therefore, the solution remains stable without forming a precipitate. This calculation ensures water quality compliance where both fluoride and calcium levels coexist safely.
