Answer:
Explanation:
From the information given:
[tex]CaF_2 \to Ca^{2+} + 2F^-[/tex]
[tex]Ksp = 3.2 \times 10^{-11}[/tex]
no of moles of [tex]Ca^{2+}[/tex] = 0.01 L × 0.0010 mol/L
no of moles of [tex]Ca^{2+}[/tex] = [tex]1 \times 10^{-5} \ mol[/tex]
no of moles of [tex]F^-[/tex] = 0.01 L × 0.00010 mol/L
no of moles of [tex]F^-[/tex] = [tex]1 \times 10^{-6}\ mol[/tex]
Total volume = 0.02 L
[tex][Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L[/tex]
[tex][F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}[/tex]
[tex][F^{-}] = 5 \times 10^{-5} \ mol/L[/tex]
[tex]Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}[/tex]
Since Q<ksp, then there will no be any precipitation of CaF2
