Question

Calculate the solubility in \(\text{mol dm}^{-3}\) of sparingly soluble salt BaBr if its solubility product is \(4.9 \times 10^{-13}\) at the same temperature.

• A. \(7 \times 10^{-7}\)
• B. \(7.5 \times 10^{-7}\)
• C. \(8 \times 10^{-7}\)
• D. \(4.9 \times 10^{-7}\)

Hint:

For \(1:1\) sparingly soluble salts: \[ K_{sp} = s^2 \] For \(1:2\) salts like \(CaF_2\): \[ K_{sp} = 4s^3 \] Always remember to adjust the expression according to ionic stoichiometry.

Answer 1

The Correct Option is A

Concept: The solubility product constant (\(K_{sp}\)) represents the equilibrium between a sparingly soluble salt and its ions in solution. For a simple \(1:1\) electrolyte: \[ AB(s) \rightleftharpoons A^+(aq) + B^-(aq) \] If the molar solubility is \(s\), then: \[ [A^+] = s \] \[ [B^-] = s \] Hence, \[ K_{sp} = s \times s = s^2 \]

Step 1: Writing the \(K_{sp}\) expression.
Given: \[ K_{sp} = 4.9 \times 10^{-13} \] For the sparingly soluble salt: \[ K_{sp} = s^2 \] Therefore, \[ s^2 = 4.9 \times 10^{-13} \]

Step 2: Simplifying the scientific notation.
Rewrite the number: \[ 4.9 \times 10^{-13} = 49 \times 10^{-14} \] Thus, \[ s = \sqrt{49 \times 10^{-14}} \]

Step 3: Taking square root.
\[ s = \sqrt{49} \times \sqrt{10^{-14}} \] \[ s = 7 \times 10^{-7} \] Therefore, the molar solubility is: \[ 7 \times 10^{-7}\ \text{mol dm}^{-3} \]

Step 4: Final conclusion.
Hence, the correct option is: \[ \boxed{(1)\ 7 \times 10^{-7}} \]