spectrophotometer analysis spectrophotometer analysis spectrophotometer analysis
ID: 543622 • Letter: S
Question
spectrophotometer analysis spectrophotometer analysis spectrophotometer analysis Spectrophotometrie Analysis Prelaboratory Assignment Name and Drawer Number 1. Convert the following transmittance and percent transmittance values to a Equation (17-4). bsorbance by using a. T 0.209 0.630 b.96T = 17.3 0,1 6R c.96T= 54.8 0.26 d. T = 0.615 2+ from a standard solution of copper (II) nitrate. In one experiment a standard solution was the addition of lumetric flask. After 5.00 mL of 6 M NH3(aq), the solution was diluted to the 50.00 mL mark The ammonia added to the flask is in excess of that required to prepare thec complex. a. Use the molarity and volume of the added copper (ID) nitrate solution to calculate the number of moles of Cu added to the volumetric flask, This is accomplished, as you recall, by multiplying the molarity of the solution (mol/L) by the volume of the solution (L) with deio iuon was diluted to the 50.00 modatr copper-ammonia ha b. Use the number of moles of copper (II) in the volumetric flask and the final volume of the flask to calculate the concentration (M) of copper (IT) in the filled volumetric flask. (In the presence of excess ammonia all of the copper (II) will be in the form of the copper-ammonia complex ion, Cu(NHs)', so the concentration calculated in part b is the concentration of Cu(NHs)) 3. What will you eaxis and on the x-axis when graphing your data from the spectrophotometric lab? y-axis X-axisExplanation / Answer
Ans. #1. Absorbance = 2 – log % T
Or, Abs = 2 – log (100T)
a. Absorbance = 2 – log (100 x 0.209) = 2 – log 20.9 = 2 – 1.3201 = 0.6798 = 0.680
b. Absorbance = 2 – log (17.3) = 2 – 1.2380 = 0.7619 = 0.761
c. Absorbance = 2 – log (54.8) = 2 – 1.7387 = 0.2612
d. Absorbance = 2 – log (100 x 0.615) = 2 - 1.7889 = 0.2111
#2. The standard solution is prepared as follow-
I. A 50.0 mL standard flask is taken.
II. To this 8.0 mL of 0.09864 M Cu(NO3)2 is added.
III. 5.00 mL of 6M NH3 is also added to the same flask.
IV. The final volume is made upto the mark with distilled water.
Note that this is one solution and the final volume is 50.0 mL.
#2a. Number of moles of Cu(NO3)2 = Molarity x Volume of solution in liters
= 0.09864 M x 0.008 L
= 0.00078912 mol
Dissociation of Cu(NO3)2: Cu(NO3)2(aq) ---------> Cu2+(aq) + 2NO3-(aq)
Since 1 mol Cu(NO3)2 yields 1 mol Cu2+ ions, the moles of Cu2+ in solution must be equal to that of Cu(NO3)2.
So,
Moles of Cu2+ = 0.00078912 mol = Moles of Cu(NO3)2
#2b. 1 mol Cu(NO3)2 dissociates into 1 mol Cu2+ and 2 moles NO3-.
So, moles of Cu2+ taken = 0.00078912 mol
Now,
[Cu2+] = Moles of Cu2+ / Volume of solution made upto in liters
= 0.00078912 mol / 0.050 L
= 0.0157824 mol/ L
= 0.0157824 M
Therefore, [Cu2+] in 50.0 mL solution = 0.0157824 M
#3. Y-axis Absorbance , x-axis concertation
The graph is prepared by plotting absorbance on y-axis and concertation on X-axis.
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