To find the molar solubility, [tex]\( S \)[/tex], of barium carbonate ([tex]\(BaCO_3\)[/tex]), given its solubility product constant ([tex]\(K_{sp}\)[/tex]) of [tex]\(2.58 \times 10^{-9}\)[/tex], here are the steps we need to follow:
1. Write the dissolution equation for [tex]\(BaCO_3\)[/tex]:
[tex]\[
\text{BaCO}_3 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{CO}_3^{2-} (aq)
\][/tex]
2. Define the expressions for the concentrations of the ions in terms of the molar solubility [tex]\( S \)[/tex]:
[tex]\[
[\text{Ba}^{2+}] = S
\][/tex]
[tex]\[
[\text{CO}_3^{2-}] = S
\][/tex]
3. Write the expression for the solubility product constant ([tex]\(K_{sp}\)[/tex]):
[tex]\[
K_{sp} = [\text{Ba}^{2+}] \cdot [\text{CO}_3^{2-}]
\][/tex]
4. Substitute the concentrations in the [tex]\(K_{sp}\)[/tex] expression:
[tex]\[
K_{sp} = S \cdot S
\][/tex]
[tex]\[
K_{sp} = S^2
\][/tex]
5. Solve for [tex]\( S \)[/tex]:
[tex]\[
S = \sqrt{K_{sp}}
\][/tex]
[tex]\[
S = \sqrt{2.58 \times 10^{-9}}
\][/tex]
6. Calculate the value of [tex]\( S \)[/tex]:
[tex]\[
S \approx 5.079370039680118 \times 10^{-5} \ \text{M}
\][/tex]
Therefore, the molar solubility [tex]\( S \)[/tex] of [tex]\( BaCO_3 \)[/tex] is:
[tex]\[
S \approx 5.079 \times 10^{-5} \ \text{M}
\][/tex]
