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 Question:
We are provided the solubility product constant ($K_{sp}$) of a sparingly soluble salt, Silver bromide (AgBr). We must calculate its molar solubility ($s$) in water.

Step 2: Detailed Explanation:
First, write the balanced dissociation equation for the salt in aqueous solution:
$\text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)$
Let the molar solubility of $\text{AgBr}$ be $s$ mol/L.
When $s$ moles of solid dissolve, they produce:
$[\text{Ag}^+] = s$
$[\text{Br}^-] = s$
The expression for the solubility product constant ($K_{sp}$) is the product of the ion concentrations:
$K_{sp} = [\text{Ag}^+] \times [\text{Br}^-]$
$K_{sp} = s \times s$
$K_{sp} = s^2$
We are given $K_{sp} = 4.9 \times 10^{-13}$. Substitute this into the equation:
$s^2 = 4.9 \times 10^{-13}$
To easily take the square root by hand, manipulate the scientific notation to secure an even exponent:
$s^2 = 49 \times 10^{-14}$
Take the square root of both sides to solve for $s$:
$s = \sqrt{49 \times 10^{-14}}$
$s = \sqrt{49} \times \sqrt{10^{-14}}$
$s = 7 \times 10^{-7} \text{ mol/L}$
The unit mol/L is identical to $\text{mol dm}^{-3}$.

Step 3: Final Answer:
The solubility is $7\times10^{-7}\text{ mol dm}^{-3}$, matching option (c).