Question

determine the molar solubility for baco₃ by constructing an ice table, writing the solubility constant expression, and solving for molar solubility. the value of ksp for baco₃ is 5.0 × 10⁻⁹.
assume the solution already contains 1.4 × 10⁻⁴ m co₃²⁻ ions. fill in the ice table with the appropriate value for each involved species to determine concentrations of all reactants and products.
complete parts 1 - 3 before submitting your answer.
1
2
3
baco₃(s)
⇌
ba²⁺(aq)
+
co₃²⁻(aq)

Explanation:

Step1: Write the solubility - product expression

The dissolution reaction of \(BaCO_3(s)\) is \(BaCO_3(s)
ightleftharpoons Ba^{2 + }(aq)+CO_3^{2 - }(aq)\). The solubility - product constant expression is \(K_{sp}=[Ba^{2 + }][CO_3^{2 - }]\).

Step2: Set up the ICE table

• For \(BaCO_3(s)\), it is a solid, and its concentration is not included in the \(K_{sp}\) expression.
• Initial (\(M\)):
• \([Ba^{2+}]=0\) (before dissolution of \(BaCO_3\) in this context, considering the common - ion effect from the pre - existing \(CO_3^{2 - }\)).
• \([CO_3^{2 - }]=1.4\times10^{-4}\ M\) (given).
• Change (\(M\)):
• Let the molar solubility of \(BaCO_3\) be \(x\). Then \([Ba^{2+}]\) increases by \(x\) (\([Ba^{2+}]\) change \(= + x\)), and \([CO_3^{2 - }]\) change due to \(BaCO_3\) dissolution is \(+x\) (but we already have a pre - existing \(CO_3^{2 - }\) concentration).
• Equilibrium (\(M\)):
• \([Ba^{2+}]=x\)
• \([CO_3^{2 - }]=(1.4\times 10^{-4}+x)\ M\)

Step3: Substitute into the \(K_{sp}\) expression

Given \(K_{sp}=5.0\times10^{-9}\). Substitute \([Ba^{2 + }]=x\) and \([CO_3^{2 - }]=(1.4\times 10^{-4}+x)\) into \(K_{sp}=[Ba^{2 + }][CO_3^{2 - }]\). Since \(K_{sp}\) is very small (\(5.0\times10^{-9}\)), \(x\) is very small compared to \(1.4\times10^{-4}\) (by the rule of \(K_{sp}\) for sparingly - soluble salts with a common ion: if \(K_{sp}\ll c_{common - ion}\), \(x\ll c_{common - ion}\)). So we can approximate \((1.4\times 10^{-4}+x)\approx1.4\times 10^{-4}\).
Then \(K_{sp}=x\times(1.4\times 10^{-4})\)

Step4: Solve for \(x\)

$$x=\frac{K_{sp}}{1.4\times 10^{-4}}=\frac{5.0\times 10^{-9}}{1.4\times 10^{-4}}$$

$$x = 3.6\times10^{-5}\ M$$

Answer:

The molar solubility of \(BaCO_3\) is \(3.6\times 10^{-5}\ M\)