Question

The Zn in 0.7556-g sample of foot powder was titrated with 21.27 mL of 0.01645 M EDTA. Calculate the percentage of Zn in this sample.

Answer 1

Answer on Question #81816 – Chemistry – Other

Task:

The Zn in 0.7556-g sample of foot powder was titrated with 21.27 mL of 0.01645 M EDTA. Calculate the percentage of Zn in this sample.

Solution:

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

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

$$Zn^{2+} + H_2Y^{2-} = ZnH_2Y$$ $$n(Zn^{2+}) = n(H_2Y^{2-});$$ $$C = \frac{n}{V}; \Rightarrow n = C*V$$ $$V(H_2Y^{2-}) = 21.27 \, \text{mL} = 0.02127 \, \text{L};$$ $$n = \frac{m}{M}; \quad M(Zn) = 65.38 \, \text{g/mol};$$ $$\frac{m(Zn)}{M(Zn)} = C(H_2Y^{2-}) * V(H_2Y^{2-});$$ $$\frac{m(Zn)}{65.38 \, \text{g/mol}} = 0.01645 \, M * 0.02127 \, \text{L};$$ $$m(Zn) = 65.38 \, \text{g/mol} * 0.01645 \, M * 0.02127 \, \text{L} = 0.022876 \, g$$ $$w(Zn) = \frac{m(Zn)}{m(sample)} * 100\% = \frac{0.022876 \, g}{0.7556 \, g} * 100\% = 3.0275\%$$ $$w(Zn) = 3.0275\%$$

Answer: 3.0275% of Zn.

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