A sample of nickel is to be determined using absorption spectrum analysis.

Question

. A sample of nickel is to be determined using absorption spectrum analysis. A 0.286 g sample of nickel ore is crushed and digested in 1M HNO3. The stock solution is then diluted to 250.0 mL in a volumetric flask (assuming the solution has a density = 1.00 g/mL).

The analysis then proceeds as follows:

Sample Solution. Std Ni solution (11.2ppm)

15.0ml. 0.00ml

15.0ml. 5.00ml

Final volume. Transmittance

100.0ml. 54.1%

100.0ml. 19.9%

Answer 1

Step 1: Calculate the concentration of Ni in the diluted standard solution

First, determine the concentration of Ni in the standard solution after dilution.

The standard solution of Ni is 11.2 ppm. When 5.00 mL of this standard solution is diluted to a final volume of 100.0 mL, the concentration of Ni in the diluted standard can be calculated using the dilution formula: C1V1=C2V2C1​V1​=C2​V2​

Where:

1. C1C1​ is the initial concentration (11.2 ppm)
2. V1V1​ is the initial volume (5.00 mL)
3. C2C2​ is the final concentration
4. V2V2​ is the final volume (100.0 mL)

C2=C1×V1V2C2​=V2​C1​×V1​​ C2=11.2 ppm×5.00 mL100.0 mLC2​=100.0mL11.2ppm×5.00mL​ C2=0.56 ppmC2​=0.56ppm

Step 2: Calculate the absorbance from transmittance

Transmittance (T) and absorbance (A) are related by the equation: A=−log⁡10(T)A=−log10​(T)

Convert the transmittance percentages to decimals:

1. For 54.1% transmittance: T=0.541T=0.541 A=−log⁡10(0.541)≈0.267A=−log10​(0.541)≈0.267
2. For 19.9% transmittance: T=0.199T=0.199 A=−log⁡10(0.199)≈0.701A=−log10​(0.199)≈0.701

Step 3: Create a calibration curve

We have two data points for our calibration curve:

1. 0.0 ppm Ni with absorbance corresponding to 54.1% transmittance (0.267 absorbance)
2. 0.56 ppm Ni with absorbance corresponding to 19.9% transmittance (0.701 absorbance)

Step 4: Determine the concentration of Ni in the sample solution

To find the concentration of Ni in the sample solution, we can assume a linear relationship between absorbance and concentration (Beer-Lambert law): A=ϵ⋅c⋅lA=ϵ⋅c⋅l

Given the calibration points, we can set up a linear equation of the form: A=m⋅c+bA=m⋅c+b

Using the points (0, 0.267) and (0.56, 0.701), we calculate the slope mm: m=0.701−0.2670.56−0=0.4340.56≈0.775m=0.56−00.701−0.267​=0.560.434​≈0.775

Now, since we are using a linear fit and our intercept is the absorbance at 0 ppm (0.267), our equation becomes:A=0.775⋅c+0.267A=0.775⋅c+0.267

Given the absorbance of the sample solution (0.267), we solve for cc: 0.267=0.775⋅c+0.2670.267=0.775⋅c+0.2670=0.775⋅c0=0.775⋅c c=0c=0

Since our result of c=0c=0 does not make sense (indicating an error in our earlier assumption), we need to reconsider our initial values or how the solution was treated.

Let's re-examine the setup to ensure accurate dilution and preparation details are correct. Assume correct preparation:

1. The original absorbance value for the pure sample should be different from the one used for calibration.