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The solubility product of a rate earth metal hydroxide $M {(OH)}_{3}$ at room temperature is $4.32 \times 10^{- 14}$ . Its solubility is

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$2 \times 10^{- 3} M$

$2.0 \times 10^{- 4} M$

$2 \times 10^{- 5} M$

$2.0 \times 10^{- 6} M$

answer is A.

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Detailed Solution

M(OH)₃⇌M³⁺+3OH⁻

K_sp for AB₃ type molecules is K_sp=27s⁴

4.32×10⁻¹⁴=27s⁴

s⁴= $\frac{4.32 \times 10^{- 14}}{27}$ =16×10⁻¹²

s= 2×10⁻³M

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