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JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Equilibrium

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 1 0^{- 6}

at 298 K. The ratio of

[B a^{2 +}] [F^{-}]^{2}

and

K_{sp}

is ______ (Nearest integer)

Answer: 5

Solution

After mixing, the concentrations are:

[Ba^{2 +}] = \frac{25 \times 0.050}{50} = 0.025

M and

[F^{-}] = \frac{25 \times 0.020}{50} = 0.010

M. The ionic product is

[Ba^{2 +}] [F^{-}]^{2} = 0.025 \times (0.010)^{2} = 25 \times 1 0^{- 7}

. Given

K_{s p} = 0.5 \times 1 0^{- 6} = 5 \times 1 0^{- 7}

, the ratio is

\frac{25 \times 1 0 ^{- 7}}{5 \times 1 0 ^{- 7}} = 5

.

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer

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