Question

A 25.00-mL sample containing Fe³⁺ and Cu²⁺ required 16.06 mL of 0.050 83 M EDTA for complete titration. A 50.00-mL sample of the unknown was treated with NH₄F to protect the Fe³⁺. Then Cu²⁺ was reduced and masked by thiourea. Addition of 25.00 mL of 0.050 83 M EDTA liberated Fe³⁺ from its fluoride complex to form an EDTA complex. The excess EDTA required 19.77 mL of 0.018 83 M Pb²⁺ to reach a xylenol orange end point. Find [Cu²⁺] in the unknown.

Short answer

The amount of Cu²⁺ in the unknown will be 0.015M.

Step-by-step solution

Given information

Amount of sample containing Fe³⁺ and Cu²⁺ = 25 mL

Amount of unknown sample treated with NH₄F to protect the Fe³⁺=50.00 mL

Amount required to form EDTA complex =25.00 mL of 0.05083 M EDTA liberated Fe³⁺ from its fluoride complex

Excess EDTA required to achieve xynol orange end point= 19.77 mL of 0.01883

EDTA

EDTA (Ethylenediaminetetraacetic acid) is a chelating compound used in different medicinal management like treatment of heavy metal toxicity, lead poisoning, neurotoxicity, coronary artery disease etc. This compound has several applications in industries, and laboratories also

Determine Cu2+

No of moles of Fe³⁺ and Cu²⁺ in 25.00 mL solution

$$\begin{gathered} =16.06mL\times 0.05083M \\ =0.82mmol \end{gathered}$$

For 2^nd titration

No of moles of EDTA used = $\left(25.00mL\right)\left(0.05083M\right)=1.271mmol$

No of moles of Pb²⁺ required =$\left(19.77mL\right)\left(0.01883M\right)=0.372mmol$

No of moles of Fe³⁺ present = $(1.271-0.372)=0.899mmol$

As 50 mL of unknown sample were used in 2^nd titration

No of moles of Fe³⁺ present in 25 mL will be = $\frac{1}{2}\left(0.899\right)mmol=0.449mmol$

Therefore, no of moles of Cu²⁺ present in 25 mL will be

$$\begin{gathered} =\left(0.82-0.449\right)mmol \\ =0.37mmol \end{gathered}$$

Concentration of Cu²⁺

$$\begin{gathered} =\frac{0.37mmol}{25mL} \\ =0.0148M \\ \approx 0.15M \end{gathered}$$

Therefore, the amount of Cu²⁺ in the unknown will be 0.015M.