What is the molar solubility of silver(I)bromide at 25°C? The solubility product constant for silver(I)bromide is 5.0 $\times$ 10^-13 at 25°C. A) 7.1 $\times$ 10^-7 M B) 2.5 $\times$ 10^-13 M C) 4.2 $\times$ 10^-4 M D) 5.0 $\times$ 10^-5 M E) 5.0 $\times$ 10^-13 M

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The molar solubility of silver(I) bromide (AgBr) can be calculated using its solubility product constant (Ksp). For AgBr, Ksp = [Ag⁺][Br⁻] = 5.0 × 10⁻¹³. Since AgBr dissociates into one Ag⁺ and one Br⁻ ion, let the molar solubility be 's'. Then, Ksp = s². Solving for 's', s = √(5.0 × 10⁻¹³) = 7.1 × 10⁻⁷ M.

What Is the Molar Solubility of Silver(I)bromide at 25°C? the Solubility