A student prepared a set of standard nickel(II) solutions, varying in concentrat

Question

A student prepared a set of standard nickel(II) solutions, varying in concentration, and processed them spectroscopically. The Beer’s Law curve that was obtained is shown below.

y=0.1347x+0.0178

R^2=0.9982

y axis is absorbance (AU)

x axis is Ni(II) concentration (ppm)

a. An unknown sample of nickel(II) was prepared in the exact same way as the standard solutions and analyzed spectroscopically. The sample produced an absorbance of 0.948 AU. What is the concentration of nickel(II) in the sample?

b. Suppose the measured unknown sample was actually a dilution that had been prepared from a more concentrated sample to yield a solution that would measure within the range of the standard calibration curve. The dilution was prepared by adding 4.00 mL of the concentrated nickel(II) solution to a 10-mL volumetric flask and diluting to the mark. What is the concentration of the original nickel(II) solution?

c. Suppose the original concentrated nickel(II) solution had been prepared by dissolving a solid unknown sample, with a mass of 0.0785 g, into 500.00 mL. What is the % (wt/wt) of nickel(II) in the solid sample?

Explanation / Answer

a)
y=0.1347x+0.0178
here y =0.948 AU
0.948 = 0.1347x+0.0178
x= 6.91 ppm = 6.91 mg/L
Answer: 6.91 ppm or 6.91 mg/L

b)
use:
Pi*Vi = Pf*Vf
Pi*4 = 6.91*10
Pi = 17.3 ppm = 17.3 mg/L
Answer: 17.3 ppm = 17.3 mg/L

c)
In original sample,
17.3mg is there in 1 L
in 500 mL, Ni = 17.3/2 = 8.6 mg
total weight of sample = 0.0785 g = 78.5mg
%wt = 8.6*100/78.5 = 11%
Answer: 11%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.