J.R. S. answered 04/07/22
Ph.D. University Professor with 10+ years Tutoring Experience
a) Concentration of standard HCl: Na2CO3 + 2HCl ==> 2NaCl + CO2 + H2O
In table below, I've converted mass Na2CO3 to moles using 99.8% and molar mass provided. Also calculated molarity of HCl using stoichiometry of balanced equation and the given volumes use.
Mass Na2CO3 V titrant used mols Na2CO3 M titrant
0.54g ...................21.15 mL.........0.002746.......0.2597 M
0.50g ..................19.56 mL.........0.002354........0.2407 M
0.64g ..................25.06 mL........0.003857........0.3078 M
Average molarity of HCl = 0.2694 M
b) Concentration of OH- in filtered solution: X(OH)2(s) ==> X2+ + 2OH-
2OH- + 2H+ ==> 2H2O
17.92 mls HCl x 1 L / 1000 ml x 0.2694 mol / L x 1 mol OH- / mol H+ = 0.004828 mols OH- in 20 ml
[OH-] = 0.004828 mol / 0.020 L = 0.2413 M OH-
c) Ksp = [X2+][OH-]2 = (0.1207)(0.2413)2
Ksp = 7.03x10-3
d) molar solubility
7.03x10-3 = (x)(2x)2 = 4x3
x = 0.1207 M = molar solubility
e) Closest to Ksp would be Ba(OH)2