25.0,mL of 0.050,M,Ba left( NO ₃right)₂ is mixed with 25.0,mL of 0.020,M NaF , K _{ sp } of BaF

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6-2.Equilibrium-II (Ionic Equilibrium)

hard

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

A

$2$

B

$3$

C

$5$

D

$4$

(JEE MAIN-2023)

Solution

${\left[ Ba ^{+2}\right]=\frac{25 \times 0.05}{50}=0.025 M }$

${\left[ F ^{-}\right]=\frac{25 \times 0.02}{50}=0.01 M }$

${\left[ Ba ^{+2}\right]\left[ F ^{-}\right]^2=25 \times 10^{-7}}$

$K _{ sp }=5 \times 10^{-7} \text { (given) }$

$\text { Ratio }=\frac{\left[ Ba ^{+2}\right]\left[ F ^{-}\right]^2}{ K _{ sp }}=5$

Standard 11

Chemistry

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