Step 1: Understanding the Concept:
Solubility is defined as the maximum amount of solute that can dissolve in a specific volume of solvent at a given temperature to form a saturated solution.
The solubility product constant (\( K_{sp} \)) is an equilibrium constant describing the extent to which a sparingly soluble ionic compound dissociates in water.
For a solid substance \( AB \), the equilibrium established in a saturated solution is:
\[ AB(s) \rightleftharpoons A^+(aq) + B^-(aq) \]
As more solute is added beyond the saturation point, the rate of dissolution equals the rate of precipitation.
Molar solubility (\( s \)) is usually expressed in units of \( \text{mol/L} \) or \( \text{mol/dm}^3 \).
Step 2: Key Formula or Approach:
For a binary salt like \( AgBr \), which dissociates into one cation and one anion (1:1 electrolyte):
The expression for the solubility product is:
\[ K_{sp} = [Ag^+][Br^-] \]
If the molar solubility is \( s \), then at equilibrium, the concentration of each ion is \( s \).
\[ K_{sp} = (s)(s) = s^2 \]
Step 3: Detailed Explanation:
Given the solubility product constant \( K_{sp} = 4.9 \times 10^{-13} \).
The formula relating \( s \) and \( K_{sp} \) for \( AgBr \) is:
\[ s = \sqrt{K_{sp}} \]
Substituting the numerical value provided:
\[ s = \sqrt{4.9 \times 10^{-13}} \]
To simplify the square root calculation, adjust the power of 10 to an even exponent by shifting the decimal point:
\[ 4.9 \times 10^{-13} = 49 \times 10^{-14} \]
Now, calculate the square root of the coefficient and the power separately:
\[ s = \sqrt{49} \times \sqrt{10^{-14}} \]
\[ \sqrt{49} = 7 \]
\[ \sqrt{10^{-14}} = 10^{-7} \]
Thus, the molar solubility is:
\[ s = 7 \times 10^{-7} \text{ mol dm}^{-3} \]
Step 4: Final Answer:
The calculation shows that the molar solubility of silver bromide is \( 7 \times 10^{-7} \text{ M} \).
This matches perfectly with the value given in option (C).