Question

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?

Answer 1

Answer:

$m_{CaCO_3}=1.28x10^{-3}gCaCO_3$

Explanation:

Hello,

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

$M_{MnSO_4}=\frac{0.435g*\frac{1mol}{151g} }{0.470L}=6.13x10^{-3}M$

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

$n_{MnSO_4}=6.13x10^{-3}\frac{mol}{L} *0.0465L=2.85x10^{-4}mol$

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

$n_{MnSO_4}=n_{EDTA}$

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

$M_{EDTA}=\frac{2.85x10^{-4}mol}{0.0419L}=6.80x10^{-3}M$

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

$n_{EDTA}^{2^{nd} titration}=6.80x10^{-3}\frac{mol}{L} *0.00188L=1.28x10^{-5}mol$

Finally, the grams of calcium carbonate turn out:

$n_{CaCO_3}=n_{EDTA}^{2^{nd}titration}=1.28x10^{-5}mol\\m_{CaCO_3}=1.28x10^{-5}mol*\frac{100.09g}{1mol} \\m_{CaCO_3}=1.28x10^{-3}gCaCO_3$

Best regards.