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Standard X

Chemistry

Solubility

The solubilit...

Question

The solubility product of $B a C r O_{4} i s 2.4 \times 10^{- 10} M_{2}$ . The maximum concentration of $B a (N O_{3})$ ; possible without, precipitation in a $6 \times 10^{- 4} M K_{2} C r O_{4}$ solution is :

4 \times 10^{- 7} M

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1.2 \times 10^{10} M

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6 \times 10^{- 4} M

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3 \times 10^{- 4} M

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Solution

The correct option is A $4 \times 10^{- 7} M$
$K s p = [B a^{2 +}] [C r O 4^{2 -}] = 2.4 \times 10^{- 10} M^{2}$
In $6 \times 10^{- 4} M K_{2} C r O_{4}$ the $[C r O_{4}^{2 -}] = 6 \times 10^{-} 4 M$
So $[B a^{2 +}] \times 6 \times 10^{- 4} = 2.40 \times 10^{- 10}$
Hence $[B a^{2 +}] = 4 \times 10^{- 7} M$

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Similar questions

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The solubility product of BaCrO4is2.4×10−10M2. The maximum concentration of Ba(NO3) ; possible without, precipitation in a 6×10−4MK2CrO4 solution is :

The correct option is A 4×10−7MKsp=[Ba2+][CrO42−]=2.4×10−10M2In 6×10−4MK2CrO4 the [CrO2−4]=6×10−4MSo [Ba2+]×6×10−4=2.40×10−10 Hence [Ba2+]=4×10−7M

The solubility product of $B a C r O_{4} i s 2.4 \times 10^{- 10} M_{2}$ . The maximum concentration of $B a (N O_{3})$ ; possible without, precipitation in a $6 \times 10^{- 4} M K_{2} C r O_{4}$ solution is :