A 1.000-g steel sample was analyzed for its Cr and Mn content by dissolving the
ID: 830835 • Letter: A
Question
A 1.000-g steel sample was analyzed for its Cr and Mn content by dissolving the sample in concentrated nitric acid, adding concentrated sulfuric and phosphoric acids, and oxidizing the Cr and Mn with potassium periodate. Final dilution to 250.0 mL was made with 0.5 M H2SO4. This solution had absorbances of 0.065 at 440 nm and 0.457 at 545 nm.
a. Using the information below, calculate the percentages of Cr and Mn in the steel samples.
b. What is the purpose of adding the concentrated sulfuric and phosphoric acids to the sample solution?
Ion ? @ 440 nm ? @ 545 nm MnO4- 99.3 2.32 x 103 Cr2O72- 412 7.41Explanation / Answer
Let's get famaliar with Beer lambert's law:
A=ebc
Where A is absorbance (no units, since A = log10 P0 / P )
e is the molar absorbtivity with units of L mol-1 cm-1
b is the path length of the sample - that is, the path length of the cuvette in which the sample is contained. We will express this measurement in centimetres.
c is the concentration of the compound in solution, expressed in mol L-1
It is clear that absorbance is directly proportional to concentration and molar absorbity.
Now let's take concentration of MnO4- =x and Cr2O7-2 = y
99.3x +412y = 0.065* k ( k some constant)
2320x + 7.41y = 0.457*k
Similary total moles of MnO4- = .250*x , Cr2O7-2 -= .250y
Because 1 moles of MnO4- contains 55 gm Mn.
.250x moles will have .250x*55 ,
Similary
Because 1 moles of Cr2O7-2 contains 52*2 = 104 gm
250x moles will have .250y*104
Total weight should be equal to 1 ,
so .250y*104+ .250x*55 =1
104y+55x = 4 --(1)
99.3x +412y / 2320x + 7.41y = 65/457 --- (2)
Solving these we get x = 0.035, y = 0.019
weight = (0.035*55)*0.25 = 0.48
y = (0.019*52)*0.25= 0.247
Concentrated Sulfuric acid and Phosphoric is added to prevent oxidation by air of steel or rusting.
Cheers
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