The solubility constant is[tex]$6.4 \times 10^{-9}$[/tex].
The equilibrium for the solubility of the salt, [tex]$\mathrm{MgF}_2$[/tex] is,
[tex]$\mathrm{MgF}_2(s) \stackrel{\mathrm{H} \cdot \mathrm{O}}{\rightleftharpoons} \mathrm{Mg}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)$[/tex]
The solubility product constant for [tex]$\mathrm{MgF}_2$[/tex] is,
[tex]$K_{\text {吶 }}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{F}^{-}\right]^2$[/tex]
The molar solubility of [tex]$\mathrm{MgF}_2$[/tex] is taken as x, the x moles of [tex]$\mathrm{MgF}_2$[/tex]is dissolved in one litre water to produce x moles of [tex]$\mathrm{Mg}^{2+}$[/tex]and 2x moles of [tex]$\mathrm{F}^{-}$[/tex].
[tex]$\begin{array}{lccc} & \mathrm{MgF}_2(s) & \stackrel{\mathrm{H}_2 \mathrm{O}}{\rightleftharpoons} \mathrm{Mg}^{2+}(a q)+2 \mathrm{~F}^{-}(a q) \\ \text { Initial } & x & 0 & 0 \\ \text { Change } & & +x & +2 x \\ \text { Equilibrium } & x & 2 x\end{array}$[/tex]
The solubility product constant for[tex]$\mathrm{MgF}_2$[/tex] is,
[tex]$\begin{aligned}K_{u p} &=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{F}^{-}\right]^2 \\&=(x)(2 x)^2 \\&=4 x^3\end{aligned}$[/tex]
Substitute the solubility of [tex]$\mathrm{MgF}_2$[/tex] as [tex]$1.17 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$[/tex]in the above equation as shown below:
[tex]$\begin{aligned}K_{i p} &=4 x^3 \\&=4\left(1.17 \times 10^{-3}\right)^3 \\&=6.41 \times 10^{-9}\end{aligned}$[/tex]
The solubility constant is [tex]$6.4 \times 10^{-9}$[/tex].(using 2 significant figures).
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