Question

Calculate the volume of 0.0500 M EDTA needed to titrate 23.37 mL of 0.0741 M Mg(NO3)2

Answer 1

Answer on Question #81815 – Chemistry – Other

Task:

Calculate the volume of 0.0500 M EDTA needed to titrate 23.37 mL of 0.0741 M Mg(NO₃)₂.

Solution:

Reagents for EDTA titrations: EDTA, H₄Y and Na₂H₂Y · 2H₂O.

Complexes of EDTA and metal ions (1:1).

$$Mg^{2+} + H_2Y^{2-} = MgH_2Y$$ $$n(Mg^{2+}) = n(H_2Y^{2-});$$ $$C = \frac{n}{V}; \Rightarrow n = C*V$$ $$C(Mg^{2+}) * V(Mg^{2+}) = C(H_2Y^{2-}) * V(H_2Y^{2-});$$ $$0.0741M * 23.37 \, \text{mL} = 0.0500M * V(H_2Y^{2-});$$ $$V(H_2Y^{2-}) = \frac{0.0741M * 23.37 \, \text{mL}}{0.0500M} = 34.63 \, \text{mL};$$ $$V(H_2Y^{2-}) = 34.63 \, \text{mL}.$$

Answer: 34.63 mL of EDTA.

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