The compound M ( OH ) x has a K sp of 4 × 1 0 − 6 and has a solubility of 1 0 − 2 mol L − 1 . The value of x should be

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ChemistryIonic EquilibriumSolubility ProductMedium2 minQB

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The compound $M (OH)_{x}$ has a $K_{sp}$ of $4 \times 1 0^{- 6}$ and has a solubility of $1 0^{- 2} mol L^{- 1}$ . The value of $x$ should be

Answer:

Solution:

M(OH)ₓ ⇌ Mˣ⁺(aq) + xOH⁻(aq)

Solubility s = 10⁻² mol L⁻¹, so [Mˣ⁺] = 10⁻² M and [OH⁻] = 10⁻²x M.

Ksp = 10⁻² × (10⁻²x)ˣ = 4 × 10⁻⁶ ⟹ x = 2.

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

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