Question

The solubility product of \(\mathrm{AgBr}\) is \(4.9\times10^{-13}\) at a certain temperature. Calculate the solubility.

• A. \(4\times10^{-6}\ \text{mol dm}^{-3}\)
• B. \(4\times10^{-7}\ \text{mol dm}^{-3}\)
• C. \(7\times10^{-7}\ \text{mol dm}^{-3}\)
• D. \(3\times10^{-8}\ \text{mol dm}^{-3}\)

Hint:

For salts of type: \[ AB \rightleftharpoons A^+ + B^- \] the relation is: \[ K_{sp}=S^2 \] So, \[ S=\sqrt{K_{sp}} \]

Answer 1

The Correct Option is C

Concept: Silver bromide dissociates as: \[ \mathrm{AgBr(s)\rightleftharpoons Ag^+ + Br^-} \] If solubility is \(S\), then: \[ [\mathrm{Ag^+}]=S \] \[ [\mathrm{Br^-}]=S \] Thus, \[ K_{sp}=S^2 \]

Step 1: Writing the solubility product expression.
Given: \[ K_{sp}=4.9\times10^{-13} \] Since: \[ K_{sp}=S^2 \] therefore, \[ S=\sqrt{4.9\times10^{-13}} \]

Step 2: Evaluating the square root.
\[ S=\sqrt{4.9}\times\sqrt{10^{-13}} \] \[ S\approx2.21\times10^{-6.5} \] Using: \[ 10^{-6.5}=3.16\times10^{-7} \] Therefore, \[ S\approx2.21\times3.16\times10^{-7} \] \[ S\approx6.98\times10^{-7} \] \[ S\approx7\times10^{-7}\ \text{mol dm}^{-3} \] Hence, the correct answer is: \[ \boxed{(C)\ 7\times10^{-7}\ \text{mol dm}^{-3}} \]