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

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(A) The dissolution of $Na_2CO_3$ in water is given by: $Na_2CO_3 ightarrow 2Na^{+} + CO_3^{2-}$.Given the concentration of $Na_2CO_3$ is $1.0 	ime

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$5.1 	imes 10^{-5} \ M$

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(A) The dissolution of $Na_2CO_3$ in water is given by: $Na_2CO_3 ightarrow 2Na^{+} + CO_3^{2-}$.Given the concentration of $Na_2CO_3$ is $1.0 	imes 10^{-4} \ M$,the concentration of carbonate ions is $[CO_3^{2-}] = 1.0 	imes 10^{-4} \ M$.For the precipitation of $BaCO_3$ to begin,the ionic product must exceed the solubility product constant $(K_{sp})$.The condition for the start of precipitation is: $[Ba^{2+}][CO_3^{2-}] = K_{sp}(BaCO_3)$.Substituting the given values: $[Ba^{2+}] 	imes (1.0 	imes 10^{-4}) = 5.1 	imes 10^{-9}$.Solving for $[Ba^{2+}]$: $[Ba^{2+}] = \frac{5.1 	imes 10^{-9}}{1.0 	imes 10^{-4}} = 5.1 	imes 10^{-5} \ M$.

Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-4} \ M$ $Na_2CO_3$ 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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