150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . How could the concentration of \(\mathrm{Sr}^{2+}\) ions in solution be decreased? (A) Adding some \(\operatorname{NaF}(s)\) to the beaker (B) Adding some \(\operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}(s)\) to the beaker (C) By heating the solution in the beaker (D) By adding a small amount of water to the beaker, but not dissolving all the solid

First, it is crucial to understand that by adding a reactive component to this system or changing its conditions, the system will adapt to maintain the solubility equilibrium. This is aligned with Le Chatelier's principle.

Now, evaluate each option individually to see how it would impact the system. (A) Adding some \(NaF(s)\) would increase the concentration of \(F^{-}\) ions in the solution. This in turn would push the equation to the left, reducing the concentration of \(Sr^{2+}\). (B) Adding \(Sr(NO_{3})_{2}(s)\) to the beaker would increase the concentration of \(Sr^{2+}\), shifting the equilibrium to the left but increasing the \(Sr^{2+}\) ion concentration. (C) Heating the solution would typically favor the endothermic direction. Without information regarding the enthalpy change (\(ΔH\)), we cannot tell which direction would be favored by heat. (D) Adding water would increase the volume of solution, causing dilution. The concentration of ions would initially decrease, but the system would reestablish equilibrium by dissolving more \(SrF_{2}\), potentially leading to no real change in \(Sr^{2+}\) concentration.

Based on the analysis, option (A) adding \(NaF(s)\) would effectively decrease the concentration of \(Sr^{2+}\) ions in solution without reestablishing equilibrium.