$$25.0$$ mL of $$0.050$$ M Ba(NO$$_3$$)$$_2$$ is mixed with $$25.0$$ mL of $$0.020$$ M NaF. K$$_{sp}$$ of BaF$$_2$$ is $$0.5 \times 10^{-6}$$ at 298K. The ratio of $$[\text{Ba}^{2+}][\text{F}^-]^2$$ and K$$_{sp}$$ is _____.

25.0 mL of 0.050 M Ba(NO$$_3$$)$$_2$$ is mixed with 25.0 mL of 0.020 M NaF. $$K_{sp}$$ of BaF$$_2 = 0.5 \times 10^{-6}$$.

Total volume = 25.0 + 25.0 = 50.0 mL.

$$[\text{Ba}^{2+}] = \frac{0.050 \times 25.0}{50.0} = 0.025 \text{ M}$$

$$[\text{F}^-] = \frac{0.020 \times 25.0}{50.0} = 0.010 \text{ M}$$

$$Q = [\text{Ba}^{2+}][\text{F}^-]^2 = 0.025 \times (0.010)^2$$

$$Q = 0.025 \times 10^{-4} = 2.5 \times 10^{-6}$$

$$\frac{Q}{K_{sp}} = \frac{2.5 \times 10^{-6}}{0.5 \times 10^{-6}} = 5$$

The ratio of $$[\text{Ba}^{2+}][\text{F}^-]^2$$ to $$K_{sp}$$ is $$\mathbf{5}$$.