Solid Ba(NO_3)_2 is gradually dissolved in a | vedclass

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(B) The precipitation of $BaCO_3$ begins when the ionic product exceeds the solubility product constant $(K_{sp})$.The equilibrium expression is: $K_{

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$5.1 	imes 10^{-6} \ M$

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(B) The precipitation of $BaCO_3$ begins when the ionic product exceeds the solubility product constant $(K_{sp})$.The equilibrium expression is: $K_{sp} = [Ba^{2+}][CO_3^{2-}]$.Given: $K_{sp} = 5.1 	imes 10^{-9}$ and $[CO_3^{2-}] = 1.0 	imes 10^{-3} \ M$.Substituting the values: $5.1 	imes 10^{-9} = [Ba^{2+}] 	imes (1.0 	imes 10^{-3})$.Therefore,$[Ba^{2+}] = \frac{5.1 	imes 10^{-9}}{1.0 	imes 10^{-3}} = 5.1 	imes 10^{-6} \ M$.

Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-3} \ M \ Na_2CO_3$ solution. At what concentration of $Ba^{2+}$ will a precipitate begin to form? ($K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9}$)

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