Question

The solubility product of AgBr is \(4.9 \times 10^{-13}\) at a certain temperature. Calculate the molar solubility of AgBr.

• A. \(4 \times 10^{-6}\ \text{mol dm}^{-3}\)
• B. \(4 \times 10^{-7}\ \text{mol dm}^{-3}\)
• C. \(7 \times 10^{-7}\ \text{mol dm}^{-3}\)
• D. \(3 \times 10^{-8}\ \text{mol dm}^{-3}\)

Hint:

For salts of type \(AB\), \[ K_{sp} = s^2 \]

Answer 1

The Correct Option is C

Concept: For a 1:1 electrolyte: \[ K_{sp} = s^2 \] where \(s\) is molar solubility.

Step 1: Writing the dissociation equation.
\[ \text{AgBr}(s) \rightleftharpoons \text{Ag}^+ + \text{Br}^- \]

Step 2: Writing the \(K_{sp}\) expression.
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \] \[ K_{sp} = s^2 \]

Step 3: Calculating the solubility.
\[ s^2 = 4.9 \times 10^{-13} \] \[ s = \sqrt{4.9 \times 10^{-13}} \] \[ s = 7 \times 10^{-7}\ \text{mol dm}^{-3} \]