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Coordination and Oxidation Number

The Ksp of Ba...

Question

The Ksp of $B a C r O_{4}$ is $2.4 \times 10^{- 10} M^{2}$ . The maximum concentration of $B a (N O_{3})_{2}$ possible without precipitation in a $6 \times 10^{- 4} M K_{2} C r O_{4}$ solution is ?

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Solution

$B a (N O_{3})_{2} + K r C r O_{4} ⇌ B a C r O_{4} + 2 K N O_{3}$
$K_{s p}$ of $B a C r O_{4} = 2.4 \times 10^{- 10} M^{2}$
$= [B a^{2 +}] [C r O_{4}^{2 -}]$
Concentration of $[C r O_{4}^{2 -}]$ ions
$= 6 \times 10^{- 4} M$
We get $C r O_{4}^{2 -}$ ions from $K r C r O_{4}$
$2.4 \times 10^{- 10} M^{2} = [B a^{2 +}] [C r O_{4}^{2 -}] 2.4 \times 10^{- 10} M = [B a^{2 +}] [6 \times 10^{- 4}] [B a^{2 +}] = \frac{2.47 \times 10^{- 10}}{6 \times 10^{- 4}} = 4 \times 10^{- 5} M$
Concentration of $B a^{2 +}$ ions $= 4 \times 10^{- 5} M$ .

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Similar questions

B a C r O_{4}

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B a (N O_{3})_{2}

6 \times 10^{- 4} M K_{2} C r O_{4}

B a C r O_{4}

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B a {(N O_{3})}_{2}

6 \times 10^{- 4} M K_{2} C r O_{4}

B a C r O_{4} i s 2.4 \times 10^{- 10} M_{2}

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