At \( 25^{\circ} \mathrm{C} \) calolum fluoride has a solubility product constant \( K_{\text {sp }}=3.0 \times 10^{-1 t} \). The solubility of \( \mathrm{CaF}_{2} \) at this temperature is _ moli

To find the solubility of calcium fluoride (\( \text{CaF}_2 \)) in terms of molarity, we start with the dissociation equation for \( \text{CaF}_2 \): \[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \] Let \( s \) be the solubility of \( \text{CaF}_2 \) in mol/L. When \( \text{CaF}_2 \) dissolves, it produces: - \( s \) moles of \( \text{Ca}^{2+} \) - \( 2s \) moles of \( \text{F}^- \) The solubility product constant \( K_{sp} \) for \( \text{CaF}_2 \) is given by: \[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \] Substituting the concentrations in terms of \( s \): \[ K_{sp} = (s)(2s)^2 = s(4s^2) = 4s^3 \] Given that \( K_{sp} = 3.0 \times 10^{-1t} \), we can set up the equation: \[ 4s^3 = 3.0 \times 10^{-1t} \] To find \( s \), we rearrange the equation: \[ s^3 = \frac{3.0 \times 10^{-1t}}{4} \] \[ s^3 = 0.75 \times 10^{-1t} \] Now, we take the cube root of both sides: \[ s = \left(0.75 \times 10^{-1t}\right)^{1/3} \] Calculating this gives: \[ s = 0.75^{1/3} \times 10^{-\frac{1t}{3}} \] Now, we need to evaluate \( 0.75^{1/3} \): \[ 0.75^{1/3} \approx 0.9086 \] Thus, we have: \[ s \approx 0.9086 \times 10^{-\frac{1t}{3}} \] To express the solubility in terms of \( t \), we need the value of \( t \). However, since \( t \) is not specified, we can only express the solubility in terms of \( t \): \[ s \approx 0.9086 \times 10^{-\frac{1t}{3}} \text{ mol/L} \] If you have a specific value for \( t \), you can substitute it in to find the numerical solubility.