The solubility product of a sparingly soluble salt AB at room temperature is 1.21 x 10^-6 . Its molar solubility is

Explanation

To determine the molar solubility of a sparingly soluble salt AB, we need to consider its solubility product constant (Ksp). The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble salt in water. It is expressed as the product of the concentrations of the dissolved ions raised to their stoichiometric coefficients in the balanced chemical equation. For the salt AB, let's assume its balanced chemical equation for dissolution is: AB(s) ⇌ A+(aq) + B-(aq) The solubility product expression for this equilibrium is: Ksp = [A+][B-] Given that the solubility product constant (Ksp) for AB is 1.21 x 10-6 , we can use this information to find the molar solubility of AB. Let's assume the molar solubility of AB is "s" moles per liter. Since AB dissociates into one A+ ion and one B- ion, the concentrations of the dissolved ions are also "s" moles per liter. Therefore, we can write: Ksp = [A+][B-] = s * s = s2 Substituting the value of Ksp into the equation: 1.21 x 10-6= s2 Taking the square root of both sides: s = √(1.21 x 10-6) Calculating the value: s ≈ 1.10 x 10-3mol/L Therefore, the molar solubility of the salt AB at room temperature is approximately 1.10 x 10-3mol/L.