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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