Copper-I reacts with neocuproine to form (neocuproine)2Cu+, with an absorption maximum at 454nm and a molar absorbtivity of 7.90×103 M-1cm-1 when measured in isoamyl alcohol.A rock containing copper is pulverized and digested with strong acid dissolving copper and other metals.The acid is neutralized with base and the resulting solution is diluted to a total volume of 250.0 mL (solution A).10.00 mL of solution A is transferred to a new flask and treated with 10.00 mL of a reducing agent to convert all of the copper to copper-I. Then 10.00 mL of buffer is added to adjust the pH (solution B).15.00 mL of solution B is withdrawn and placed in a new flask. 10.00 mL of neocuproine and 20.00 mL of isoamyl alcohol are added to the flask and it is shaken well.After settling, the isoamyl alcohol layer containing all of the copper-I complex is removed and the absorbance is measured at 454nm in a 1.0 cm cell. A blank carried through the same procedure gave an absorbance of 0.056. Calculate the milligrams of copper metal in an unknown that gives an absorbance of 0.874. I recommend that you start by calculating the concentration of copper in the isoamyl alcohol layer and then perform dilution calculations until you reach the concentration in solution A.

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Answer:

14,8mg of Copper

Explanation:

Lambert-Beer law says:

A = abC

Where A is absorbance (0,874-0,056 = 0,818), a is molar absorption coefficient (7,90x10³M⁻¹cm⁻¹), b is path length (1,0 cm) and C is molar concentration.

Solving for C:

C = 1,04x10⁻⁴M of (neocuproine)₂Cu+ = 1,04x10⁻⁴M Cu

This concentration is the concentration of Copper measured using the absorbance.

The concentration of Copper in solution B is:

1,04x10⁻⁴M Copper×[tex]\frac{20,00mL+10,00mL+15,00mL}{15,00mL}[/tex] = 3,11x10⁻⁴M Copper

Copper in solution A is:

3,11x10⁻⁴M Copper×[tex]\frac{10,00mL+10,00mL+10,00mL}{10,00mL}[/tex] = 9,32x10⁻⁴M Copper

As the total volume of solution A is 250,0mL, moles of Copper in A are:

9,32x10⁻⁴M Cu × 0,2500L = 2,33x10⁻⁴ moles of Copper

In grams:

2,33x10⁻⁴ moles of Cu×[tex]\frac{63,546g}{1mol}[/tex] = 0,0148 g of Copper ≡ 14,8 mg of Copper in an unknown

I hope it helps!