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Q.

The solubility of Pb(OH)₂ in water is 6 × 10^–6M and the solubility of Pb(OH)₂ in a buffer soluton of pH = 8 is

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a

$8.64\, \times {10^{ - 4}}$

b

$8.64\, \times {10^{ - 16}}$

c

$\large 8.64\, \times {10^{ - 6}}$

d

$8.64\, \times {10^{ - 2}}$

answer is B.

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Detailed Solution

The solubility of Pb(OH)₂ will be suppressed in a solution of P^H = 8 due to common ion effect and the following equilibrium gets established.

$\large Pb{\left( {OH} \right)_2}\left( s \right) \rightleftharpoons \mathop {Pb{{\left( {OH} \right)}_2}\left( {aq} \right)}\limits_{S'} \xrightarrow{{}}\mathop {P{b^{ + 2}}\left( {aq} \right)}\limits_{S'} + \mathop {2O{H^ - }\left( {aq} \right)}\limits_{2S' + {{10}^{ - 6}}}$

K_sp = S' x (2S' +10^-6)²

⇒ 4 x (6 x 10^-6)³ = S' x 10^-12

⇒ S' = 8.64 x 10^-4

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The solubility of Pb(OH)2 in water is 6 × 10–6M and the solubility of Pb(OH)2 in a buffer soluton of pH = 8 is