Question

21-B. Mn was used as an internal standard for measuring Fe by atomic absorption. A standard mixture containing 2.00 mg Mn/mL and 2.50 mg Fe/mL gave a quotient (Fe signal/Mn signal) 5 1.05/1.00. A mixture with a volume of 6.00 mL was prepared by mixing 5.00 mL of unknown Fe solution with 1.00 mL containing 13.5 mg Mn/mL. The absorbance of this mixture at the Mn wave- length was 0.128, and the absorbance at the Fe wavelength was 0.185. Find the molarity of the unknown Fe solution.

Answer 1

Answer:

0.0693M Fe

Explanation:

It is possible to quantify Fe in a sample using Mn as internal standard using response factor formula:

F = A(analyte)×C(std) / A(std)×C(analyte) (1)

Where A is area of analyte and std, and C is concentration.

Replacing with first values:

F = 1.05×2.00mg/mL / 1.00×2.50mg/mL

F = 0.84

In the unknown solution, concentration of Mn is:

13.5mg/mL × (1.00mL/6.00mL) = 2.25 mg Mn/mL

Replacing in (1) with absorbances values and F value:

0.84 = 0.185×2.25mg/mL / 0.128×C(analyte)

C(analyte) = 3.87 mg Fe / mL

As molarity is moles of solute (Fe) per liter of solution:

$\frac{3.87 mg Fe}{mL} \frac{1g}{1000mg} \frac{1mol}{55.845g} \frac{1000mL}{1L}$ = 0.0693M Fe

Answer 2

Answer:

the molarity of the unknown Fe solution is 5.462x10⁻⁵mol/L

Explanation:

The signal radio is given by the expression:

$\frac{Fe-wavelength}{Mn-wavelength} =\frac{0.185}{0.128} =1.4453$

You need to calculate the concentration of Mn, applying the following expression:

$\frac{Mn/mL*V_{Mn} }{V_{total} } =\frac{13.5*1}{6} =2.25mgMn/mL$

Now, the influence of this concentration on initial signal is equal:

$\frac{1.05*2.25}{2} =1.1813$

The concentration of Fe is:

$\frac{2.5*1.4453}{1.1813} =3.0587$

Now, you need to calculate the molarity:

$M=3.0587\frac{mgFe}{mL} *\frac{1g}{1x10^{6}mg } *\frac{1molFe}{56g} *\frac{1000mL}{1L} =5.462x10^{-5} mol/L$