The 3 pre-lab questions below -please write questions and answers in your notebo
ID: 1043844 • Letter: T
Question
The 3 pre-lab questions below -please write questions and answers in your notebook . 1. Using a Spectronic 10 spectrophotometer to measure absorbance at 418 nm, the following data was obtained for these known concentrations a) Find the concentration of an unknown sample with an absorbance of 16.2 abs units? Concentrat 0.010 M 0.020 M 0.030 M 0.040 M 0.050 M Absorbance 5.0 abs units 10.0 abs units 15.0 abs units 20.0 abs units 25.0 abs units The unknown sample had a volume of 2.00 mL. Using the molecular weight of Ni to be 58.693 g/mol, what mass of nickel is present in the sample? The unknown sample was prepared fron a 0.200 grams nickel complex compound. What is the nickel in this sample(% mass Nimass compound X 100)? b) c) 2. A 103.0 mL of a different unknown nickel complex sample was titrated with EDTA. If 13.8 mL of 0.007891 M EDTA was needed to obtain a color change: a) Find the moles and molar concentration of the unknown nickel complex? b) Using the initial volume of nickel complex and the molecular weight of Ni to be 58.693 gmol, what mass (in grams) of nickel is present in the sample? c) The unknown sample was prepared from a 0.300 gram nickel complex compound. What is the % nickel in this sample (% mass Nimass compound X 100)? 3. Answer the following questions. a) How will you prepare your burette for the titration? b) What will you do with the burette once you are finished with the titration?Explanation / Answer
1
a)
Plot a graph of absorbance vs concentration of the standard solution.
Use the regression equation. Note that since 1.0*10-14 is extremely small compared to 500x, we can ignore the latter term and write
16.2 = 500x
=> x = 16.2/500
= 0.0324
The concentration of the unknown sample is 0.0324 M (ans).
b)
Volume of the unknown sample = 2.00 mL.
Moles of the unknown sample present = (volume of unknown sample in L)*(concentration of unknown sample in mol/L) = (2.00 mL)*(1 L/1000 mL)*(0.0324 mol/L) (1 M = 1 mol/L)
= 6.48*10-5 mole.
Assume that1 mole of the unknown sample contains 1 mole of Ni. Molar mass of Ni = 58.693 g/mol.
Therefore, mass of Ni in the unknown sample = (moles of Ni)*(molar mass of Ni) = (6.48*10-5 mole)*(58.693 g/mol) = 0.00380 g
c)
Mass percent of Ni in the unknown sample = (mass of Ni)/(mass of unknown) = (0.00380 g)/(0.200 g)*100
= 1.90%
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