$25.0 \mathrm{~mL}$ of $0.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ is mixed with $25.0 \mathrm{~mL}$ of $0.020 ~\mathrm{M} ~\mathrm{NaF} . \mathrm{K}_{\mathrm{Sp}}$ of $\mathrm{BaF}_{2}$ is $0.5 \times 10^{-6}$ at $298 \mathrm{~K}$. The ratio of $\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}$ and $\mathrm{K}_{\mathrm{sp}}$ is ___________.

(Nearest integer)

Initial concentrations before mixing:<br/><br/> $[\mathrm{Ba(NO_3)_2}] = 0.050\, \mathrm{M}$<br/><br/> $[\mathrm{NaF}] = 0.020\, \mathrm{M}$ <br/><br/> Volumes of the solutions:<br/><br/> $V_{\mathrm{Ba(NO_3)_2}} = V_{\mathrm{NaF}} = 25.0\, \mathrm{mL}$ <br/><br/> After mixing, the total volume becomes:<br/><br/> $$V_{\mathrm{total}} = 25.0\, \mathrm{mL} + 25.0\, \mathrm{mL} = 50.0\, \mathrm{mL}$$ <br/><br/> Now, we calculate the initial concentrations after mixing: <br/><br/> $$[\mathrm{Ba}^{2+}] = \frac{25.0\, \mathrm{mL} \times 0.050\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.025\, \mathrm{M}$$ <br/><br/> $$[\mathrm{F}^{-}] = \frac{25.0\, \mathrm{mL} \times 0.020\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.010\, \mathrm{M}$$ <br/><br/> Next, we calculate the reaction quotient (Q) for the precipitation of BaF₂: <br/><br/> $Q = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$ <br/><br/> $Q = (0.025\, \mathrm{M})(0.010\, \mathrm{M})^2 = 2.5 \times 10^{-6}$ <br/><br/> K_sp of BaF₂ is given as $5 \times 10^{-7}$. <br/><br/> Now, we find the ratio of $[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$ to K_sp: <br/><br/> $\text{Ratio} = \frac{(2.5 \times 10^{-6})}{(5 \times 10^{-7})} = 5$ <br/><br/> So, the correct ratio of $[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$ to K_sp is 5.