The option that identifies the molar solubility of CdF₂ in pure water is;
A: The molar solubility of CdF₂ in pure water is 0.0585M, and adding NaF decreases this solubility because the equilibrium shifts to favor the precipitation of some CdF₂.
The equation of this reaction is;
CdF₂ (s) ⇄ Cd²⁺ (aq) + 2F⁻ (aq)
We are told that;
[Cd²⁺] (eq) = 0.0585m and [F⁻] eq = 0.117m.
Also, we see that 0.90m NaF is added to the saturated solution.
Now, molar solubility is defined as the number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturated.
Thus, we can conclude that The molar solubility of CdF₂ in pure water is 0.0585M, and adding NaF decreases this solubility because the equilibrium shifts to favor the precipitation of some CdF₂.
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