Solid $$Ba(NO_3)_2$$ is gradually dissolved in a $$1.0 \times 10^{-4}$$ M $$Na_2CO_3$$ solution. At which concentration of $$Ba^{2+}$$, precipitate of $$BaCO_3$$ begins to form? ($$K_{sp}$$ for $$BaCO_3 = 5.1 \times 10^{-9}$$)

To determine the concentration of $$ \text{Ba}^{2+} $$ at which precipitation of $$ \text{BaCO}_3 $$ begins, we start by understanding the solubility equilibrium. The dissolution of $$ \text{BaCO}_3 $$ is given by:

$$ \text{BaCO}_3(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{CO}_3^{2-}(aq) $$

The solubility product constant ($$ K_{sp} $$) for this reaction is $$ 5.1 \times 10^{-9} $$. The $$ K_{sp} $$ expression is:

$$ K_{sp} = [\text{Ba}^{2+}][\text{CO}_3^{2-}] = 5.1 \times 10^{-9} $$

The solution initially contains $$ \text{Na}_2\text{CO}_3 $$ at a concentration of $$ 1.0 \times 10^{-4} $$ M. Since $$ \text{Na}_2\text{CO}_3 $$ dissociates completely in water:

$$ \text{Na}_2\text{CO}_3 \rightarrow 2\text{Na}^+ + \text{CO}_3^{2-} $$

the initial concentration of $$ \text{CO}_3^{2-} $$ ions is $$ 1.0 \times 10^{-4} $$ M.

Solid $$ \text{Ba}(\text{NO}_3)_2 $$ is gradually dissolved. It dissociates completely as:

$$ \text{Ba}(\text{NO}_3)_2 \rightarrow \text{Ba}^{2+} + 2\text{NO}_3^- $$

so the concentration of $$ \text{Ba}^{2+} $$ comes from the dissolved $$ \text{Ba}(\text{NO}_3)_2 $$.

Precipitation of $$ \text{BaCO}_3 $$ begins when the ion product equals the $$ K_{sp} $$. At this point, the solution is just saturated, and any further addition of $$ \text{Ba}^{2+} $$ will cause precipitation.

Since we are adding $$ \text{Ba}(\text{NO}_3)_2 $$ gradually and the precipitation just begins, the amount of $$ \text{Ba}^{2+} $$ added is very small. Therefore, the concentration of $$ \text{CO}_3^{2-} $$ remains approximately equal to its initial concentration because very little $$ \text{CO}_3^{2-} $$ is consumed to form the precipitate. Thus, we can approximate:

$$ [\text{CO}_3^{2-}] \approx 1.0 \times 10^{-4} \, \text{M} $$

Substituting into the $$ K_{sp} $$ expression:

$$ K_{sp} = [\text{Ba}^{2+}] \times [\text{CO}_3^{2-}] $$

$$ 5.1 \times 10^{-9} = [\text{Ba}^{2+}] \times (1.0 \times 10^{-4}) $$

Solving for $$ [\text{Ba}^{2+}] $$:

$$ [\text{Ba}^{2+}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} $$

Performing the division:

$$ [\text{Ba}^{2+}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} = 5.1 \times \frac{10^{-9}}{10^{-4}} = 5.1 \times 10^{-9 + 4} = 5.1 \times 10^{-5} $$

Thus, the concentration of $$ \text{Ba}^{2+} $$ at which precipitation begins is $$ 5.1 \times 10^{-5} $$ M.

Comparing with the options:

A. $$ 5.1 \times 10^{-5} $$ M

B. $$ 7.1 \times 10^{-8} $$ M

C. $$ 4.1 \times 10^{-5} $$ M

D. $$ 8.1 \times 10^{-7} $$ M

Option A matches the calculated concentration.

Hence, the correct answer is Option A.