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Standard XII

Chemistry

Solubility Product

Given the sol...

Question

Given the solubility product of $C r (O H)_{2}$ as $2.0 \times 10^{- 16}$ at $25^{\circ} C$ . Calculate the minimum $p H$ when $C r^{2 +}$ ions start precipitating in the form of $C r (O H)_{2}$ from a solution having $2 \times 10^{- 4} M C r^{2 +}$ concentration ?

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Solution

The correct option is C 8
The equilibrium is established between undissolved $C r (O H)_{2}$ and its ions. The equilibrium established will be,
$C r (O H)_{2} (s) ⇌ C r^{2 +} (a q) + 2 O H^{-} (a q)$
$K_{s p} = 2.0 \times 10^{- 16} K_{s p} = [C r^{2 +}] [O H^{-}]^{2}$
$2.0 \times 10^{- 16} = [2 \times 10^{- 4}] [O H^{-}]^{2}$
$[O H^{-}]^{2} = 1 \times 10^{- 12} [O H^{-}] = 1 \times 10^{- 6}$
$p O H = - log [O H^{-}]$
$p O H = 6$
We know that,
$p H + p O H = 14$ at $25^{o} C$
$p H = 14 - 6 = 8$

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Solubility Product

Standard XII Chemistry

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Given the solubility product of $C r (O H)_{2}$ as $2.0 \times 10^{- 16}$ at $25^{\circ} C$ . Calculate the minimum $p H$ when $C r^{2 +}$ ions start precipitating in the form of $C r (O H)_{2}$ from a solution having $2 \times 10^{- 4} M C r^{2 +}$ concentration ?