Step 1: Understanding the Concept:
For a sparingly soluble salt, a saturated solution exists in dynamic equilibrium between the undissolved solid and its dissolved ions.
The solubility product constant (\( K_{sp} \)) represents the product of the concentrations of the ions, each raised to the power of its stoichiometric coefficient in the balanced equation.
Solubility (\( s \)) is the concentration of the dissolved salt in a saturated solution.
Step 2: Key Formula or Approach:
Based on the provided options and common problem structure for "BaBr" (treated here as a 1:1 electrolyte \( AB \)):
\[ AB(s) \rightleftharpoons A^+(aq) + B^-(aq) \]
If the molar solubility is \( s \), then:
\[ [A^+] = s \text{ and } [B^-] = s \]
\[ K_{sp} = [A^+][B^-] = s^2 \]
Thus, \( s = \sqrt{K_{sp}} \).
Step 3: Detailed Explanation:
Given \( K_{sp} = 4.9 \times 10^{-13} \).
We need to find \( s \):
\[ s^2 = 4.9 \times 10^{-13} \]
To calculate the square root easily, we shift the decimal point to make the exponent an even number:
\[ 4.9 \times 10^{-13} = 49 \times 10^{-14} \]
Now, taking the square root:
\[ s = \sqrt{49 \times 10^{-14}} \]
\[ s = \sqrt{49} \times \sqrt{10^{-14}} \]
\[ s = 7 \times 10^{-7} \text{ mol dm}^{-3} \]
The unit \( \text{mol dm}^{-3} \) is equivalent to Molarity (M).
Step 4: Final Answer:
The molar solubility of the salt is \( 7 \times 10^{-7} \text{ mol dm}^{-3} \).