What is the solubility of MgCO₃ in a solution that contains 0.055 M Mg²⁺ ions? (Ksp of MgCO₃ is 3.5 × 10⁻⁸)

The solubility of MgCO₃ in a solution that contains 0.055 M Mg²⁺ ions is 6.36 * 10⁻⁷.What is the solubility of a solution?The solubility of a solution is the amount of solute present in a given volume of solvent.The solubility of MgCO₃ in a solution that contains 0.055 M Mg²⁺ ions is calculated from the dissociation equation for the reaction given below:MgCO₃ (s) → Mg²⁺ (aq) + CO₃²⁻ (aq)The solubility product constant, Ksp =  [Mg²⁺] * [CO₃²⁻]At equilibrium, [Mg²⁺] = 0.055 + x[CO₃²⁻] = xAssuming x is <<< 0.055, hence 0.055 + x = 0.055Ksp = 0.055 * xx = Ksp / 0.055x = 3.5 × 10⁻⁸ / 0.055x = 6.36 * 10⁻⁷Hence, the solubility of MgCO₃ = 6.36 * 10⁻⁷Learn more about solubility at: https://brainly.com/question/24124376#SPJ1