Question

The solubility product of 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 a simple binary salt (AB type, producing 2 ions), the relationship is always $K_{sp} = s^2$. For an AB2 or A2B salt (3 ions), it's $K_{sp} = 4s^3$. Knowing these shortcuts saves setup time!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a 1:1 salt like Silver Bromide ($AgBr$), the solubility ($S$) is the square root of the solubility product ($K_{sp}$).

Step 2: Formula Application:
$AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$ $K_{sp} = [Ag^+][Br^-] = (S)(S) = S^2$ $S = \sqrt{K_{sp}}$

Step 3: Explanation:
Given $K_{sp} = 4.9 \times 10^{-13}$. $S = \sqrt{4.9 \times 10^{-13}} = \sqrt{49 \times 10^{-14}}$ $S = 7 \times 10^{-7}$ mol dm$^{-3}$.

Step 4: Final Answer:
The solubility is $7 \times 10^{-7}$ mol dm$^{-3}$.