25.0 mL of 0.050 M Ba ( NO 3 ) 2 is mixed with 25.0 mL of 0.020 M NaF. K sp of BaF 2 is 0 . 5 × 10 - 6 at 298 K. The ratio of [ Ba 2 + ] [ F - ] 2 and K sp is _______. (Nearest integer) [2023]

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Solution

Q.
25.0 mL of 0.050 M $Ba {({NO}_{3})}_{2}$ 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 $[{Ba}^{2 +}] {[F^{-}]}^{2}$ and $K_{sp}$ is _______. (Nearest integer) [2023]

Ans.
(5)

On mixing equal volume concentration is Half.

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

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

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

$K_{s p} = 5 \times 10^{- 7} (given)$

$so,$ $\frac{[{Ba}^{2 +}] [F^{-}]}{K_{s p}} = 5$

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