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The solubility product of SrF₂ in water is 8 × 10^–10, then the solubility in 0.1 M NaF aqueous solution is

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$\large 8\, \times {10^{ - 8}}m/l$

$8\, \times {10^{ - 6}}m/l$

$8\, \times {10^{ - 4}}m/l$

$8\, \times {10^{ - 2}}m/l$

answer is A.

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Detailed Solution

Solubility of SrF₂ is suppressed in presence of 0.1 M NaF solution due to common-ion effect and the following equilibrium gets established.

$\large Sr{F_2}\left( s \right) \rightleftharpoons \mathop {Sr{F_2}\left( {aq} \right)}\limits_{S'} \xrightarrow{{}}\mathop {S{r^{ + 2}}\left( {aq} \right)}\limits_{S'} + \mathop {2{F^ - }\left( {aq} \right)}\limits_{2S' + 0.1}$

$\large NaF\xrightarrow{{}}\mathop {N{a^ + }\left( {aq} \right)}\limits_{0.1} + \mathop {{F^ - }\left( {aq} \right)}\limits_{0.1 + 2S'}$

K_sp = S' x (2S' + 0.1)²

⇒ 8 x 10^-10 = S' x 10^-2

⇒ S' = 8 x 10^-8

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