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Analytical chemistry

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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.

  • Analytical chemistry - ,

    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
    0.6334 g

    = 85.78%

  • Analytical chemistry - ,

    bump

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