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

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

Step-by-step explanation:

The signal radio is given by the expression:

`(Fe-wavelength)/(Mn-wavelength) =(0.185)/(0.128) =1.4453`

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

`(Mn/mL*V_(Mn) )/(V_(total) ) =(13.5*1)/(6) =2.25mgMn/mL`

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

`(1.05*2.25)/(2) =1.1813`

The concentration of Fe is:

`(2.5*1.4453)/(1.1813) =3.0587`

Now, you need to calculate the molarity:

`M=3.0587(mgFe)/(mL) *(1g)/(1x10^(6)mg ) *(1molFe)/(56g) *(1000mL)/(1L) =5.462x10^(-5) mol/L`

Answer 2

0.0693M Fe

Step-by-step 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:

`(3.87 mg Fe)/(mL) (1g)/(1000mg) (1mol)/(55.845g) (1000mL)/(1L)` = 0.0693M Fe