At T(K) in a saturated solution of MgCO_3 and | vedclass

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(D) For the saturated solution of $MgCO_3$, the solubility product expression is $K_{sp}(MgCO_3) = [Mg^{2+}][CO_3^{2-}]$.Given $[Mg^{2+}] = 3.2 	imes

Vedclass · Accepted Answer

$\sqrt{1.6} 	imes 10^{-5} \ M$

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(D) For the saturated solution of $MgCO_3$, the solubility product expression is $K_{sp}(MgCO_3) = [Mg^{2+}][CO_3^{2-}]$.Given $[Mg^{2+}] = 3.2 	imes 10^{-5} \ M$ and $K_{sp}(MgCO_3) = 1.6 	imes 10^{-6}$.Substituting these values: $1.6 	imes 10^{-6} = (3.2 	imes 10^{-5}) 	imes [CO_3^{2-}]$.Therefore, $[CO_3^{2-}] = \frac{1.6 	imes 10^{-6}}{3.2 	imes 10^{-5}} = 0.5 	imes 10^{-1} = 0.05 \ M$.Now, for the saturated solution of $Ag_2CO_3$, the solubility product expression is $K_{sp}(Ag_2CO_3) = [Ag^{+}]^2[CO_3^{2-}]$.Given $K_{sp}(Ag_2CO_3) = 8.0 	imes 10^{-12}$ and $[CO_3^{2-}] = 0.05 \ M$.Substituting these values: $8.0 	imes 10^{-12} = [Ag^{+}]^2 	imes (0.05)$.$[Ag^{+}]^2 = \frac{8.0 	imes 10^{-12}}{0.05} = 160 	imes 10^{-12} = 1.6 	imes 10^{-10}$.$[Ag^{+}] = \sqrt{1.6 	imes 10^{-10}} = \sqrt{1.6} 	imes 10^{-5} \ M$.

At $T(K)$ in a saturated solution of $MgCO_3$ and $Ag_2CO_3$, if the concentration of $Mg^{2+}$ ion is $3.2 \times 10^{-5} \ M$, then the concentration of $Ag^{+}$ ion in the solution will be [Given: $K_{sp}(MgCO_3) = 1.6 \times 10^{-6}$ and $K_{sp}(Ag_2CO_3) = 8.0 \times 10^{-12}$ at $T(K)$]

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