Question

A 50.0-mL aliquot of solution containing 0.450 g of MgSO₄ (FM 120.37) in 0.500 L required 37.6 mL of EDTA solution for titration. How many milligrams of CaCO₃ (FM 100.09) will react with 1.00 mL of this EDTA solution?

Short answer

0.995 milligrams of CaCO₃ (FM 100.09) will react with 1.00 mL of this EDTA solution

Step-by-step solution

Given Information

Amount of aliquot of solution taken= 50.0 mL

Amount of MgSO₄ in 0.5L solution = 0.45g

Amount of EDTA required for titration =37.6 mL

Determine the concentration of EDTA

Formula mass for MgSO₄

$$\begin{gathered} =\left(24.305+32.065+4\times 15.999\right) \\ =120.366 \\ \approx 120.37 \end{gathered}$$

The amount of Mg²⁺ in 50 mL aliquot

$=\left(\frac{50.0\mathrm{mL}}{500\mathrm{mL}}\right)\left(\frac{0.45\mathrm{g}}{120.37\mathrm{g}/\mathrm{mol}}\right)$

=0.3738mmol

According to the question 37.6 mL of EDTA reacts with 0.3738 mmol of Mg²⁺

Therefore, the concentration of EDTA

$$\begin{gathered} =\frac{0.3738\mathrm{mmol}}{37.6\mathrm{mL}} \\ =0.009943\mathrm{M} \end{gathered}$$

Determine the amount of CaCO3 reacted

Formula mass of CaCO₃

$$\begin{gathered} =\left(40.078+12.011+3\times 15.999\right) \\ =100.09 \end{gathered}$$

Amount of CaCO₃ reacted with 1.00 mL of this EDTA solution

$$\begin{gathered} =0.009943\mathrm{mmol}\times 100.09\frac{\mathrm{mg}}{\mathrm{mmol}} \\ =0.995\mathrm{mg} \end{gathered}$$