JEE Main 2023 — Ionic Equilibrium Question with Solution

Initial concentrations before mixing:

$$[\mathrm{Ba}(\mathrm{N}{\mathrm{O}}_3{)}_2]=0.050\mathrm{M}$$

$$[\mathrm{NaF}]=0.020\mathrm{M}$$

Volumes of the solutions:

$$V_{\mathrm{Ba}(\mathrm{N}{\mathrm{O}}_3{)}_2}=V_{\mathrm{NaF}}=25.0\mathrm{mL}$$

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:

$$Q=[{\mathrm{Ba}}^{2+}][{\mathrm{F}}^{-}{]}^2$$

$$Q=(0.025\mathrm{M})(0.010\mathrm{M}{)}^2=2.5\times 10^{-6}$$

K_sp of BaF₂ is given as $$5\times 10^{-7}$$.

Now, we find the ratio of $$[{\mathrm{Ba}}^{2+}][{\mathrm{F}}^{-}{]}^2$$ to K_sp:

$$\text{Ratio}=\frac{(2.5\times 10^{-6})}{(5\times 10^{-7})}=5$$

So, the correct ratio of $$[{\mathrm{Ba}}^{2+}][{\mathrm{F}}^{-}{]}^2$$ to K_sp is 5.