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Solubility Product

The salt AlOH...

Question

The salt $A l (O H)_{3}$ is involved in the following two equilibria,
$A l (O H)_{3} (s) ⇌ A l^{3 +} (a q .) + 3 O H^{-} (a q .); K_{s p}$
$A l (O H)_{3} (s) + O H^{-} (a q) ⇌ A l (O H)_{4}^{-} (a q .); K_{c}$
Which of the following relationships is correct when the solubility is minimum?

[O H^{-}] = {(\frac{3 K_{s p}}{K_{c}})}^{\frac{1}{4}}

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[O H^{-}] = {(\frac{K_{c}}{K_{s p}})}^{\frac{1}{4}}

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[O H^{-}] = \sqrt{{(\frac{K_{s p}}{K_{c}})}^{\frac{1}{4}}}

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None of these

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Solution

The correct option is A $[O H^{-}] = {(\frac{3 K_{s p}}{K_{c}})}^{\frac{1}{4}}$
Dissolved $A l (O H)_{3}$ present in solution as $A l^{3 +}$ and $A l (O H)_{4}^{-}$ , lets say solubility of $A l (O H)_{3}$ is $S$ then
$S = [A l^{3 +}] + [A l (O H)_{4}^{-}]$
$= \frac{K_{s p}}{[O H^{-}]^{3}} + K_{c} [O H^{-}]$
for minimum solubility $\frac{d S}{d [O H^{-}]} = 0$ that is
$\frac{- 3 K_{s p}}{[O H^{-}]^{4}} + K_{c} = 0$ hence ,
$[O H^{-}] = {(\frac{3 K_{s p}}{K_{c}})}^{\frac{1}{4}}$

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Similar questions

A l (O H)_{3}

A l (O H)_{3} (s) ⇌ A l^{3 +} (a q .) + 3 O H^{-} (a q .); K_{s p}

A l (O H)_{3} (s) + O H^{-} (a q) ⇌ A l (O H)_{4}^{-} (a q .); K_{c}

A l (O H)_{3}

A l (O H)_{3} (s) ⇌ A l^{3 +} (a q) + 3 O H^{-} (a q); K_{s p}

A l (O H)_{3} (s) + O H (a q) ⇌ A l (O H)_{4}^{-} (a q); K_{c}

Z n (O H)_{2}

Z n (O H)_{2} (s) ⇌ Z n^{2 +} (a q) + 2 O H^{-} (a q); K_{s p} = 1.3 \times 10^{- 17}

Z n (O H)_{2} (s) + 2 O H^{-} (a q) ⇌ [Z n (O H)_{4}]^{2 -} (a q); K_{c} = 0.13

Z n {(O H)}_{2}

Z n {(O H)}_{2} (s) ⇌ {Z n}^{2 +} (a q) + 2 {O H}^{-} (a q)

K_{s p} = 1.2 \times 10^{- 17}

Z n {(O H)}_{2} (s) + 2 {O H}^{-} (a q) ⇌ {[Z n ({O H)}_{4}]}^{2 -} (a q)

K_{c} = 0.12

p H

Z n (O H)_{2}

A g C N

x

C N^{-}

s

A g C N

A g C N (s)_{s p} A g^{+} (a q) + C N^{-} (a q)

C N^{-} (a q) + H_{2} O (l)_{h} H C N (a q) + O H^{-} (a q)

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