$$M(OH)x$$ has $$Ksp=4$$×$$10$$−$$12$$ and solubility $$1$$×$$10$$−$$4M$$. What is the value of X ?

$$M(OH)x$$⇌$$Mx++xOH$$−$$Ksp=xx(s)x+14$$×$$10$$−$$12=xx1$$×$$10$$−$$4x+1$$ If $$x=1$$, $$4$$×$$10$$−$$12=1$$×$$10$$−$$42=1$$×$$10$$−$$8$$ (Incorrect) If $$x=2$$, $$4$$×$$10$$−$$12=221$$×$$10$$−$$42+1=4$$×$$1012$$ (Correct)

Ionic equilibrium

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Question

$M (OH)_{x} has K_{sp} = 4 \times 10^{- 12} and solubility 1 \times 10^{- 4} M . What is the value of X ?$

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Solution

$\begin{array}{l} M (OH)_{x} ⇌ M^{x +} + x {OH}^{-} \\ K_{sp} = x^{x} (s)^{x + 1} \\ 4 \times 10^{- 12} = x^{x} {(1 \times 10^{- 4})}^{x + 1} \end{array}$
$\begin{array}{l} If x = 1, 4 \times 10^{- 12} = {(1 \times 10^{- 4})}^{2} = 1 \times 10^{- 8} (Incorrect) \\ If x = 2, 4 \times 10^{- 12} = 2^{2} {(1 \times 10^{- 4})}^{2 + 1} = 4 \times 10^{12} (Correct) \end{array}$

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Ionic equilibrium

Unlock the full solution & master the concept.

M(OH)x has Ksp=4×10−12 and solubility 1×10−4M. What is the value of X ?

M(OH)x⇌Mx++xOH−Ksp=xx(s)x+14×10−12=xx1×10−4x+1 If x=1, 4×10−12=1×10−42=1×10−8 (Incorrect) If x=2, 4×10−12=221×10−42+1=4×1012 (Correct)

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$M (OH)_{x} has K_{sp} = 4 \times 10^{- 12} and solubility 1 \times 10^{- 4} M . What is the value of X ?$