A 46.50 mL aliquot from a 0.470 L solution that contains 0.435 g of MnSO 4 ( MW = 151.00 g/mol) required 41.9 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO 3 ( MW = 100.09 g/mol) will react with 1.88 mL of the EDTA solution?

Explanation:

Hello,

In this case, for the first titration, the molarity of the MnSO₄ is:

Now, the moles that are neutralized by the EDTA for the 45.50-mL aliquot are:

In such a way, those moles equals the EDTA moles based on the titration main equation:

Next, the concentration of the EDTA 41.9-mL solution is:

Afterwards, for the second titration, the moles of EDTA that equal the neutralized grams of calcium carbonate are:

Finally, the grams of calcium carbonate turn out:

Best regards.

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