Lets start by writing a balanced chemical equation for the
titration/ neutralization process:
                   Â
H2C2O4.2H2O +
Ba(OH)2 = BaC2O4 +
4H2O
1 mole H2C2O4.2H2O =
1 mole Ba(OH)2 = 126g
H2C2O4.2H2O
Equality of mass and volume:
0.896g H2C2O4.2H2O = 0.0382 L Ba(OH)2
Our goal = mol Ba(OH)2/L Ba(OH)2 =
Molarity of Ba(OH)2
Now lets set up the given values and calculate the unknown:
     1mol
Ba(OH)2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.896
g H2C2O4.2H2O
-------------------------------
x  ---------------------------- = 0.186 mol/L
Ba(OH)2
126 g
H2C2O4.2H2OÂ Â Â Â Â Â Â Â Â Â Â Â Â Â
0.0382 L Ba(OH)2
The molarity of barium hydroxide is 0.186 M.
