JEE MAIN - Chemistry (2023 - 13th April Morning Shift - No. 22)
$$25.0 \mathrm{~mL}$$ 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)
Answer
5
Explanation
Initial concentrations before mixing:
$$[\mathrm{Ba(NO_3)_2}] = 0.050\, \mathrm{M}$$
$$[\mathrm{NaF}] = 0.020\, \mathrm{M}$$
Volumes of the solutions:
$$V_{\mathrm{Ba(NO_3)_2}} = V_{\mathrm{NaF}} = 25.0\, \mathrm{mL}$$
After mixing, the total volume becomes:
$$V_{\mathrm{total}} = 25.0\, \mathrm{mL} + 25.0\, \mathrm{mL} = 50.0\, \mathrm{mL}$$
Now, we calculate the initial concentrations after mixing:
$$[\mathrm{Ba}^{2+}] = \frac{25.0\, \mathrm{mL} \times 0.050\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.025\, \mathrm{M}$$
$$[\mathrm{F}^{-}] = \frac{25.0\, \mathrm{mL} \times 0.020\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.010\, \mathrm{M}$$
Next, we calculate the reaction quotient (Q) for the precipitation of BaF₂:
$$Q = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$$
$$Q = (0.025\, \mathrm{M})(0.010\, \mathrm{M})^2 = 2.5 \times 10^{-6}$$
Ksp of BaF₂ is given as $$5 \times 10^{-7}$$.
Now, we find the ratio of $$[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$$ to Ksp:
$$\text{Ratio} = \frac{(2.5 \times 10^{-6})}{(5 \times 10^{-7})} = 5$$
So, the correct ratio of $$[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$$ to Ksp is 5.
$$[\mathrm{Ba(NO_3)_2}] = 0.050\, \mathrm{M}$$
$$[\mathrm{NaF}] = 0.020\, \mathrm{M}$$
Volumes of the solutions:
$$V_{\mathrm{Ba(NO_3)_2}} = V_{\mathrm{NaF}} = 25.0\, \mathrm{mL}$$
After mixing, the total volume becomes:
$$V_{\mathrm{total}} = 25.0\, \mathrm{mL} + 25.0\, \mathrm{mL} = 50.0\, \mathrm{mL}$$
Now, we calculate the initial concentrations after mixing:
$$[\mathrm{Ba}^{2+}] = \frac{25.0\, \mathrm{mL} \times 0.050\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.025\, \mathrm{M}$$
$$[\mathrm{F}^{-}] = \frac{25.0\, \mathrm{mL} \times 0.020\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.010\, \mathrm{M}$$
Next, we calculate the reaction quotient (Q) for the precipitation of BaF₂:
$$Q = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$$
$$Q = (0.025\, \mathrm{M})(0.010\, \mathrm{M})^2 = 2.5 \times 10^{-6}$$
Ksp of BaF₂ is given as $$5 \times 10^{-7}$$.
Now, we find the ratio of $$[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$$ to Ksp:
$$\text{Ratio} = \frac{(2.5 \times 10^{-6})}{(5 \times 10^{-7})} = 5$$
So, the correct ratio of $$[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$$ to Ksp is 5.
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