A few questions from my pre lab, not sure what they\'re asking for? 1) A student

Question

A few questions from my pre lab, not sure what they're asking for?

1) A student prepares 4 working standard solutions from the stock solution provided by the instructor in order to plot a calibration curve of absorbance vs. concentration for nickel (II) nitrate, Ni(NO3)2.

a) what are the concentrations of Ni(NO3)3 in the working standards?

Sample 1: 2.00mL of 0.4000M Ni(NO3)2 diluted to 10.00mL   _____________

Sample 2:    5.00mL of 0.4000M Ni(NO3)2 diluted to 10.00mL   _____________

Sample 3:    7.00mL of 0.4000M Ni(NO3)2 diluted to 10.00mL   _____________

Sample 4:    10.00mL of 0.4000M Ni(NO3)2 not diluted              _____________

Show one sample calculation for the dilutions.

2) The student [re[ared a calibration curve from his working standards, and one obtained the equation for the line: y = 1.502x + 0.018.

The student then obtained 3mL of an unknown Ni(NO3)2 solution from his instructor, and diluted the sample to 10mL. The diluted sample had an absorbance regarding of 0.232

b) What is the concentration of the diluted unknown sample of Ni(NO3)2?

c) Wha it the concentration of the original unknown sample of Ni(NO3)3?

Explanation / Answer

(a) M1V1 = M2V2

Sample 1: 0.4000 x 2.00 = M2 x 10.00

Concentration M2 = 0.08000 M


Sample 2: 0.4000 x 5.00 = M2 x 10.00

Concentration M2 = 0.2000 M


Sample 3: 0.4000 x 7.00 = M2 x 10.00

Concentration M2 = 0.2800 M


Sample 4: 0.4000 x 10.00 = M2 x 10.00

Concentration M2 = 0.4000 M


(b) y = 1.502x + 0.018

where y is absorbance and x is concentration


When y = 0.232:

0.232 = 1.502x + 0.018


Diluted concentration x = (0.232 - 0.018)/1.502

= 0.1425 M (approximately 0.142 M)


(c) M1V1 = M2V2

M1 x 3.00 = 0.1425 x 10.00


Original concentration M1 = 0.4749 M (approximately 0.475 M)

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