The Ksp of AgBr is 4.9 × 10⁻¹³ mol² dm⁻⁶. What is its molar solubility in pure water?
Correct answer B. 7.0 × 10⁻⁷ mol dm⁻³
Explanation
AgBr gives one Ag⁺ and one Br⁻ ion, so if the solubility is s, Ksp = s². Hence s = √(4.9 × 10⁻¹³) = 7.0 × 10⁻⁷ mol dm⁻³.
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