Chemistry - Ionic Equilibrium Question with Solution | TestHub

Question Explanation: Determine which ion concentrations cause ${\mathrm{CaF}}_2$ precipitation.

Concept: This question is based on Precipitation Condition (IP vs ${\mathrm{K}}_{\mathrm{sp}}$ )

Solution:   When equal volumes of two solutions are mixed, the concentration of each ion is immediately halved. Precipitation occurs when IP=[Ca²⁺ ][F^-]² > K_sp = 1.7 × 10^-10 .

(1): $\mathrm{IP}=\left(\frac{10^{-4}}{2}\right){\left(\frac{10^{-4}}{2}\right)}^2=1.25\times 10^{-13}$. $\mathrm{IP}<{\mathrm{K}}_{\mathrm{sp}}\Rightarrow$ No. ppt

(2): $\mathrm{IP}=\left(\frac{10^{-2}}{2}\right){\left(\frac{10^{-3}}{2}\right)}^2=1.25\times 10^{-9}.\mathrm{IP}>{\mathrm{K}}_{\mathrm{sp}}\Rightarrow$ Yes. ppt

(3): $\mathrm{IP}=\left(\frac{10^{-5}}{2}\right){\left(\frac{10^{-3}}{2}\right)}^2=1.25\times 10^{-12}.\mathrm{IP}<{\mathrm{K}}_{\mathrm{sp}}\Rightarrow$ No. ppt

(4): $\mathrm{IP}=\left(\frac{10^{-3}}{2}\right){\left(\frac{10^{-5}}{2}\right)}^2=1.25\times 10^{-14}$. $\mathrm{IP}<{\mathrm{K}}_{\mathrm{sp}}\Rightarrow$ No. ppt