Question

Solubility product of AgBr is $4.9 \times 10^{-13}$. What is its solubility?

• A. $2.4 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3}$
• B. $3.2 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3}$
• C. $4.9 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3}$
• D. $7.0 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3}$

Hint:

When computing a square root involving scientific notation, always shift the decimal point to make the exponent an even number first. Taking half of an even exponent ($10^{-14} \rightarrow 10^{-7}$) can be performed effortlessly in your head!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks us to compute the molar solubility of silver bromide (AgBr) in water given its equilibrium solubility product constant ($K_{sp}$).

Step 2: Key Formula or Approach:
Silver bromide is a 1:1 binary electrolyte ($AB$ type) that dissociates completely in an aqueous solution according to the equilibrium equation: $$\text{AgBr}_{(s)} \rightleftharpoons \text{Ag}^+_{(aq)} + \text{Br}^-_{(aq)}$$ If $S$ represents the molar solubility of AgBr in $\text{mol}\ \text{dm}^{-3}$, the equilibrium concentrations are $[\text{Ag}^+] = S$ and $[\text{Br}^-] = S$. The solubility product expression is: $$K_{sp} = [\text{Ag}^+][\text{Br}^-] = (S)(S) = S^2$$ Rearranging to solve for solubility gives: $$S = \sqrt{K_{sp}}$$

Step 3: Detailed Explanation:
Given value: $$K_{sp} = 4.9 \times 10^{-13}$$ To easily evaluate the square root, rewrite the scientific notation so that the exponent on 10 is an even integer: $$K_{sp} = 49 \times 10^{-14}$$ Now, substitute this into our solubility formula: $$S = \sqrt{49 \times 10^{-14}}$$ $$S = \sqrt{49} \times \sqrt{10^{-14}}$$ $$S = 7.0 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3}$$

Step 4: Final Answer:
The solubility of AgBr is $7.0 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3}$, which maps to option (D).