6-2.Equilibrium-II (Ionic Equilibrium)
hard

$0.050\,M\,Ba \left( NO _3\right)_2$ ના $25.0\,mL$ ને $0.020\,M,NaF$,ના $25.0\,mL$ સાથે મિશ્ર કરવામાં આવે છે.$298\,K$ પર $BaF _2$ ની $Ksp\, 0.5 \times 10^{-6}$ છે.$\left[ Ba ^{2+}\right]\left[ F ^{-}\right]^2$ અને $Ksp$ નો ગુણોત્તર $........$ છે.

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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