Palladium (II) and gold (III) can be determined simultaneously by complexing the
ID: 1063293 • Letter: P
Question
Palladium (II) and gold (III) can be determined simultaneously by complexing the two ions with methiomeprazine. The absorption maximum for the palladium complex occurs at 480 nm, and that for the gold complex is at 635 nm. Molar absorptivity data at these wavelengths are given in the table below: A 25.0-mL sample was treated with an excess of methiomeprazine and subsequently diluted to 50.0-mL. Calculate (with the help of Excel, or otherwise) the molar concentrations of Pd (II) and Au (III) if the diluted solution had an absorbance of 0.533 at 480 nm and 0.590 at 635 nm when measured in a 1.00-cm cell. Show all your work, including your output if a computerized system was used.Explanation / Answer
Let the diluted solution contain c1 M Pd (II) and c2 M Au (III).
Set up Beer’s law as below:
0.533 = (3.55*103 M-1cm-1).c1.(1.00 cm) + (2.96*103M-1cm-1).c2.(1.00 cm) …..(1)
0.590 = (5.64*102 M-1cm-1).c1.(1.00 cm) + (1.45*104 M-1cm-1).c2.(1.00 cm) ….(2)
Multiply (1) by 5.64*102 and (2) by 3.55*103 and subtract
(0.533).(5.64*102) – (0.590)*(3.55*103) = (3.55*103M-1)(5.64*102).c1 + (2.96*103M-1).(5.64*102).c2 – (5.64*102M-1).(3.55*103).c1 – (1.45*104M-1).(3.55*103).c2
Cancel out the common terms to obtain
300.612 – 2094.5 = 1669440.c2 – 51475000.c2
====> -1793.888 = -49805506.c2
====> c2 = 1793.888/49805506 = 3.60*10-5
The molar concentration of Au(III) complex is 3.60*10-5 M.
Put the value of c2 in (1) and obtain
0.533 = (3.55*103M-1cm-1).c1.(1.00 cm) + (2.96*103M-1cm-1).(3.60*10-5 M).(1.00 cm)
====> 0.533 = (3.55*103 M-1).c1 + 0.10656
====> 0.42644 = (3.55*103 M-1).c1
====> c1 = 1.20*10-4 M
The molar concentration of Pd (II) is 1.20*10-4 M.
The concentrations which we obtained are for dilute 50.0 mL solutions; the concentrations in the original 25.0 mL solution are
Pd (II) = (1.20*10-4 M)*(50.0 mL/25.0 mL) = 2.4*10-4 M (ans).
Au (III) = (3.60*10-5 M)*(50.0 Ml/25.0 mL) = 7.2*10-5 M (ans).
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