Question

The United States Public Health Service (USPHS) recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is $\mathbf{1}\mathit{m}\mathit{g}{\mathit{F}}^{\mathbf{-}}\mathbf{/}\mathit{L}\mathbf{.}$. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level?$\left(K_{sp}forCa{F}_2=4.0\times {10}^{-11}\right)$

Short answer

The$Ca{F}_2$ will precipitate when${\left[{\text{Ca}}^{\text{2 +}}\right]}_{\text{0}}$ is greater than $\text{0.02mol/L}$.

Step-by-step solution

Definition for solubility product

Solubility product $\left(K_{sp}\right)$is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation.

• The value of$K_{sp}$ indicates the solubility of an ionic compound-the smaller the value, the less soluble the compound in water.
• For concentrations of ions that do not correspond to equilibrium conditions, we use the ion product $\left(Q\right)$, to predict whether a precipitate will form.
• Note that$Q$ has the same form as except $K_{sp}$that the concentrations of ions are not equilibrium concentrations.
• If ${\text{Q >K}}_{\text{sp}}$, the solution is supersaturated and salt will precipitate out until the product of the ionic concentration is equal to$K_{sp}$

Calculation of [F-]0

The recommended concentration of fluoride ions in water is $1{\text{mgF}}_2/\text{L}$. The concentration of fluoride ions can be calculated from a given concentration.

$$\begin{gathered} n\left(F_2\right)=\frac{mass\left(F_2\right)}{molarmass\left(F_2\right)}=\frac{1\times {10}^{-3}g}{37.996g/mol} \\ n\left(F_2\right)=2.6\times {10}^{-5}mol \\ n\left(F^{-}\right)=2\times n\left(F_2\right) \\ =5.2\times {10}^{-5}mol \end{gathered}$$

$$\begin{gathered} {\left[{\text{F}}^{\text{-}}\right]}_{\text{0}}=\frac{\text{n}\left({\text{F}}^{\text{-}}\right)}{\text{V}}=\frac{5.2\times {10}^{-5}}{1L} \\ {\left[{\text{F}}^{\text{-}}\right]}_{\text{0}}=52\times {10}^{-5}mol/L \end{gathered}$$

Reaction of ion product

For the reaction shown below the ion product is given by:

$\text{Q =}{\left[{\text{Ca}}^{\text{2 +}}\right]}_{\text{0}}{\left[{\text{F}}^{\text{-}}\right]}_{\text{0}}^{\text{2}}$

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

Calculation of [Ca2+]0

As noted above, salt will precipitate if $Q>K_{sp}$.

${\text{K}}_{\text{sp}}\text{=}{\left[{\text{Ca}}^{\text{2 +}}\right]}_{\text{0}}{\left[{\text{F}}^{\text{-}}\right]}_{\text{0}}^{\text{2}}$

role="math" localid="1663828756661" ${\left[{\text{Ca}}^{\text{2 +}}\right]}_{\text{0}}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\left[{\text{F}}^{\text{-}}\right]}_{\text{0}}^{\text{2}}}\text{=}\frac{4.0\times {10}^{-11}}{{\left(5.2\times {10}^{-5}\right)}^2}$

${\left[{\text{Ca}}^{\text{2 +}}\right]}_{\text{0}}=\text{0.015mol/L}$

${\text{K}}_{\text{sp}}\text{=}{\left[{\text{Ca}}^{\text{2 +}}\right]}_{\text{0}}{\left[{\text{F}}^{\text{-}}\right]}_{\text{0}}^{\text{2}}$

${\left[{\text{Ca}}^{\text{2 +}}\right]}_{\text{0}}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\left[{\text{F}}^{\text{-}}\right]}_{\text{0}}^{\text{2}}}\text{=}\frac{4.0\times {10}^{-11}}{{\displaystyle {\left(5.2\times {10}^{-5}\right)}^2}}$

${\left[{\text{Ca}}^{\text{2 +}}\right]}_{\text{0}}=\text{0.015mol/L}$

Hence,$Ca{F}_2$ will precipitate when${\left[{\text{Ca}}^{\text{2 +}}\right]}_{\text{0}}$ is greater than $0.02\text{mol/L}$.