posted by Anonymous .
A 0.6334 gram of sample of impure mercury (II) oxide was dissolved in an unmeasured excess of potassium iodide.
Reaction: HgO + 4I- + H2O ------→HgI42- + 2OH-
Calculate the % HgO in the sample if titration of the liberated hydroxide required 42.59 ml of 0.1178 M HCl.
This is what I worked out, just wondering if someone can tell me if I have gone wrong.
%HgO = liter HgO x M HgO x mole ratio(HgO/OH) x FW HgO x 100
weight of sample
= 0.04259 L x 0.1178 mol/L x ½ x 216.589 g/mol x100