### Problem Statement:A chemist has a 50.0 mL sample of 0.010 M Na₂C₂O₄ at 25 °C (solution A, oxalate ion = C₂O₄²⁻). An aqueous solution of Ca(NO₃)₂ with a volume of 50.0 mL (solution B) is added to solution A, at the same temperature.Above what minimum concentration must Ca(NO₃)₂ in solution B be, *prior to mixing*, to cause CaC₂O₄ to precipitate given that for CaC₂O₄, Kₛₚ = 2.7 × 10⁻⁹ at 25 °C?### Instructions:Enter your answer with two **significant figures**, without units, and in millimolarity (mM) where:\[ \text{millimolarity} = (\text{molarity})(1000) \]### Possible Answers:- 0.001 (with margin: 0)- 0.0011 (with margin: 0)

We know if Ionic product is more than solubility product then precipitate forms.If ionic product is less than solubility product then precipitate does not form.In first step, we calculate total volume of the solution obtained by mixing both the solutions.next we make use of standard molarity equation of dilutions & calculate concentration of oxalate ion in final solution.We write solubility product expression of calcium oxalate and calculate concentration of calcium ion in final dilute solution to ensure precipitation begins.Next making use of standard molarity equation of dilution, we calculate concentration of stock calcium nitrate solution required.Thus, stock solution of calcium nitrate is found to be 0.0011mM