a. Calculate the molar solubility of \(\mathrm{Sr} \mathrm{F}_{2}\) in water, ignoring the basic properties of \(\mathrm{F}^{-} .\left(\right.\) For \(\mathrm{SrF}_{2}, K_{\text {sp }}=7.9 \times 10^{-10}\).) b. Would the measured molar solubility of \(\mathrm{Sr} \mathrm{F}_{2}\) be greater than or less than the value calculated in part a? Explain. c. Calculate the molar solubility of \(\mathrm{SrF}_{2}\) in a solution buffered at \(\mathrm{pH}=2.00 .\left(K_{\mathrm{a}}\right.\) for \(\mathrm{HF}\) is \(\left.7.2 \times 10^{-4} .\right)\)

First, let's write the dissociation equilibrium for SrF2: \[ \text{SrF}_2 \rightleftharpoons \text{Sr}^{2+} + 2 \text{F}^{-} \] Now, let the value of solubility of SrF2 in water be represented as 's' mol/L. At equilibrium, we may write: \[ [\text{Sr}^{2+}] = s \ \text{mol/L} \] \[ [\text{F}^{-}] = 2s \ \text{mol/L} \] Also, we know that Ksp = 7.9 x 10^{-10} is equal to the product of their concentration at equilibrium: \( K_{sp} = [\mathrm{Sr}^{2+}] [\mathrm{F}^{-}]^2 \) Substitute the expressions for the molar concentrations and solve for 's': \( 7.9 \times 10^{-10} = s(2s)^2 \)

In answering this part, we should consider the basic properties of F-. Since F- can act as a base, it can form HF with H+ ions in water: \[ \text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^- \] The presence of additional ions (OH-) can lead to a common ion effect if the solubility product is exceeded, precipitating more SrF2 and decreasing the actual solubility. So we can conclude that the measured molar solubility of SrF2 will be less than the calculated value.

At pH = 2, [H+] in the buffered solution is 10^{-2} mol/L. The formation of HF will dominate the reaction due to the abundance of H+: \[ \text{F}^- + \text{H}^+ \rightleftharpoons \text{HF} \] The values of [HF] and [H+] are buffered, so they remain constant. We can assume that all of the F- ions react with the H+ ions to form HF. Therefore, the concentration of F- remains almost constant in this case. Also, we know that K_a for HF is 7.2 × 10^{-4}: \( K_{a} = \dfrac{[\mathrm{HF}]}{[\mathrm{F}^{-}][\mathrm{H}^{+}]} \) Using this relationship, we can find the new equilibrium concentration of F- in the buffered solution: \( 7.2 \times 10^{-4} = \dfrac{[\mathrm{HF}]}{[\mathrm{F}^{-}](10^{-2}) \) Now, we can use the new concentration of F- to find the solubility of SrF2 in the buffered solution using the expression for K_sp: \( K_{sp} = [\mathrm{Sr}^{2+}] [\mathrm{F}^{-}]^2 \) From this equation, combined with the relationship given by K_a, we can find the molar solubility of SrF2 in the pH=2.00 buffered solution.