Question

A sample of hard water contains about 2.0 x 10^-3 M Ca^2+ . A soluble Fluoride containing salt such as NaF is added to fluoridate the water. what is the maximum concentratio0n of F- that can be presented without precipitation oc CaF2? Ksp for CaF2=5.3 x 10^ -11

Answer 1

Task:

A sample of hard water contains about $2.0 \times 10^{\wedge} - 3$ M Ca$^2+$. A soluble Fluoride containing salt such as NaF is added to fluoridate the water. what is the maximum concentration of F- that can be presented without precipitation or CaF2? Ksp for CaF2=5.3 x 10^4 - 11

Solution:

The chemical equation for this reaction is

$$\mathrm{Ca}^{2+}(\mathrm{aq}) + 2\mathrm{F}^{-}(\mathrm{aq}) = \mathrm{CaF}_{2}(\mathrm{s})$$

The equation for solubility constant is

$$\mathrm{Ksp} = [\mathrm{Ca}^{2+}] [\mathrm{F}^{-}]^{2}$$

The precipitation forms when the concentration of fluoride ion is

$$[\mathrm{F}^{-}] = (\mathrm{Ksp} / [\mathrm{Ca}^{2+}])^{1/2}$$

That's why the maximum concentration of F- must be less than

$$[\mathrm{F}^{-}] = (5.3 \cdot 10^{-11} / 2.0 \cdot 10^{-3})^{1/2} = 1.63 \cdot 10^{-4} \mathrm{M}$$

Answer: $[\mathrm{F}^{-}] < 1.63 \cdot 10^{-4} \mathrm{M}$