Chemistry / Formation Constant
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Question

Consider an amphoteric hydroxide, M(OH)2(s), where M is a generic metal.

M(OH)2 (s) ⇌ M2+ (aq) + 2OH- (aq) Ksp = 2 x 10-16

M(OH)2 (s) + 2OH- (aq) ⇌ M(OH)42- (aq)  Kf = 0.060

Estimate the solubility of M(OH)2 in a solution buffered at pH = 7.0, 10.0, and 14.0.

Solution: Consider an amphoteric hydroxide, M(OH)2(s), where M is a ge...

$M(OH)_{2(s)} ⇌ M^{2+}_ {(aq)} + 2 OH^- _{(aq)} $

From this balanced equation we can see that equilibrium expression for Ksp:

$Ksp= [M^{2+}][OH^-]^2 = 2x10^{-16}.$

This is so because we have 1 mole of $M(OH)^2$ producing 1 mole of $M^{2+}$ and 2 moles of $OH^-$.

We are looking for Molar solubility, or x. In our equation, $[M^{2+}] = x$, so we need to solve for $[M^{2+}]$.

 **At pH= 7:**

$pH + pOH = 14$

$7 + pOH = 14$

$pOH= 14-7 =7 $

$ [OH^-] = 10^{-pOH} $

$ [OH^-] = 10^{-7}= 1x10^{-7}$

We take a look at our previous equation again and plug in what we found for [OH-].

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