Question:
The solubility of \(BaCO_3\) (molar mass \(197\,g\,mol^{-1}\)) is \(1.4\times10^{-3}\) g per 100 mL. The solubility product constant of \(BaCO_3\) is \(x\times10^{-9}\,mol^2L^{-2}\). The value of \(x\) (nearest integer) is:
The solubility of \(BaCO_3\) (molar mass \(197\,g\,mol^{-1}\)) is \(1.4\times10^{-3}\) g per 100 mL. The solubility product constant of \(BaCO_3\) is \(x\times10^{-9}\,mol^2L^{-2}\). The value of \(x\) (nearest integer) is:
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For salts of the type \(AB\),
\[
K_{sp}=s^2.
\]
For \(AB_2\) or \(A_2B_3\), always include the stoichiometric coefficients while writing \(K_{sp}\).
Updated On: Jun 18, 2026
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The Correct Option is B
Solution and Explanation
Concept:
For
\[
BaCO_3(s)\rightleftharpoons Ba^{2+}+CO_3^{2-}
\]
If molar solubility is \(s\),
\[
K_{sp}=s^2
\]
because both ions are produced in equal amounts.
Step 1: Convert solubility into g/L.
Given: \[ 1.4\times10^{-3}g \] per \[ 100mL \] Therefore per litre: \[ 1.4\times10^{-2}gL^{-1} \]
Step 2: Find molar solubility.
\[ s=\frac{1.4\times10^{-2}}{197} \] \[ s\approx7.1\times10^{-5} \] mol L\(^{-1}\)
Step 3: Calculate \(K_{sp}\).
\[ K_{sp}=s^2 \] \[ =(7.1\times10^{-5})^2 \] \[ \approx5.0\times10^{-9} \] Hence, \[ x\approx5 \] Therefore, \[ \boxed{5} \]
Step 1: Convert solubility into g/L.
Given: \[ 1.4\times10^{-3}g \] per \[ 100mL \] Therefore per litre: \[ 1.4\times10^{-2}gL^{-1} \]
Step 2: Find molar solubility.
\[ s=\frac{1.4\times10^{-2}}{197} \] \[ s\approx7.1\times10^{-5} \] mol L\(^{-1}\)
Step 3: Calculate \(K_{sp}\).
\[ K_{sp}=s^2 \] \[ =(7.1\times10^{-5})^2 \] \[ \approx5.0\times10^{-9} \] Hence, \[ x\approx5 \] Therefore, \[ \boxed{5} \]
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