Question

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 $BaCO_3 = 5.1 \times 10^{-9}$)

• A. $5.1 \times 10^{-5}\, M$
• B. $7.1 \times 10^{-8}\,M$
• C. $4.1 \times 10^{-5}\, M$
• D. $8.1 \times 10^{-7}\,M$

Answer 1

The Correct Option is A

To determine the concentration of Ba^{2+} at which the precipitate of BaCO_3 begins to form, we need to use the concept of the solubility product constant (K_{sp}). The K_{sp} provides the condition under which a salt starts to precipitate from the solution.

Given:

• K_{sp}\, \text{for}\, BaCO_3 = 5.1 \times 10^{-9}
• Concentration of CO_3^{2-}\, \text{in Na}_2\text{CO}_3\, \text{solution} = 1.0 \times 10^{-4}\, M

The reaction for the formation of BaCO_3 is:

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

The K_{sp} expression for barium carbonate is:

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

We need to find the concentration of Ba^{2+} at which the precipitate starts to form, so set:

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

Solving for [\text{Ba}^{2+}], we get:

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

This simplifies to:

[\text{Ba}^{2+}] = 5.1 \times 10^{-5}\, \text{M}

The precipitate of BaCO_3 begins to form when the concentration of Ba^{2+} reaches 5.1 \times 10^{-5}\, \text{M}.

Therefore, the correct answer is:

5.1 \times 10^{-5}\, M