At 25⁰C, in the saturated solution of Ag₂CrO₄. If concentration of Ag⁺ ion is 1 × 10^–5mol/l, then the solubility product of Ag₂CrO₄ at 25⁰C is

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$1\, \times \,{10^{ - 14}}{m^3}/{l^3}$

$\large 5\, \times \,{10^{ - 16}}{m^3}/{l^3}$

$5\, \times \,{10^{ - 9}}{m^3}/{l^3}$

$5\, \times \,{10^{ - 10}}{m^3}/{l^3}$

answer is A.

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Detailed Solution

Following equilibrium is established in the saturated solution.

$\large A{g_2}Cr{O_4}\left( s \right) \rightleftharpoons \mathop {A{g_2}Cr{O_4}}\limits_S \left( {aq} \right) \to \mathop {2A{g^ + }}\limits_{2S} \left( {aq} \right) + \mathop {CrO_4^{ - 2}}\limits_S \left( {aq} \right)$

$\large {K_{sp}} = {\left( {2S} \right)^2} \times S = 4{S^3} = {\left( {5 \times {{10}^{ - 6}}} \right)^3} \times 4$

$\large {K_{sp}} =4 \times 5 \times 5 \times 5 \times 10^{-18}$

$\large {K_{sp}} = 5 \times 10^{-16}$

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