25.0 mL 0.050 M Ba(NO3) mixed |TheBlockElementsJEEMAIN2023

JEE MAIN 2023

13-4-23_S1

Question

25.0 mL of 0.050 M Ba(NO3)2 is mixed with 25.0 mL of 0.020 M NaF. Ksp of BaF2 is 0.5 × 10–6 at 298 K. The ratio of [Ba2+][F–]2 and Ksp is ____. (Nearest Integer)

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Solution

Sol. $$ \begin{aligned} & {\left[\mathrm{Ba}^{+2}\right]=\frac{25 \times 0.05}{50}=0.025 \mathrm{M}} \\ & {\left[\mathrm{~F}^{-}\right]=\frac{25 \times 0.02}{50}=0.01 \mathrm{M}} \\ & {\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{F}^{-}\right]^2=25 \times 10^{-7}} \\ & \mathrm{~K}_{\mathrm{sp}}=5 \times 10^{-7}(\text { given }) \\ & \text { Ratio }=\frac{\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{F}^{-}\right]^2}{\mathrm{~K}_{\mathrm{sp}}}=5 \end{aligned} $$

Question Tags

JEE Main

Chemistry

Medium

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