sebanyak 0.74 g basa kuat L(OH)2 tepat dinetralkan dengan 100 mL larutan HCL 0.2M. massa atom relatif L adalah

Reaksi:

L(OH)2(aq) + 2 HCl(aq) --> LCl2(aq) + 2 H2O(l)

mol HCl = [HCl] × V HCl
mol HCl = 0.2 × 100 × 10^-3
mol HCl = 2 × 10^-2 mol

mol L(OH)2 = 1/2 mol HCl = 10^-2 mol.
Mr L(OH)2 = massa L(OH)2 / mol L(OH)2
Mr L(OH)2 = 0.74 / 10^-2 = 74 g/mol.
Ar O = 16, Ar H = 1.

Maka,
Ar L + 2 × 1 + 2 × 16 = 74
Ar L + 34 = 74
Ar L = 40 g/mol