(22 pts) Palladium II and gold (II) can be determined simul ltaneously by comple
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Question
(22 pts) Palladium II and gold (II) can be determined simul ltaneously by complexing the two ions with methiomepraziop (C19H24N2Sz. F.W. 344.54). The absorption maximum for the palladium complex occurs at 480 nm, and that for the gold complex is at635 mm. Molar absorptivity data at these wavelengths are: Molar Absorptivity 480 mm 635 nm Pd Complex 3,550 Au complex 2,960 14,500 A25.0 mL sample was treated with an excess of methiomeprainc and subsequently diluted to 50.0 ml. a) Calculate the molar concentrations of Pd(ll and Au(II) if the diluted solution had an absorbance of 0.533 at 480 mm and 0.590 at 635 mm when measured in a 1.000 cm cell. [Au(III)] b) Report these concentrations in ppm Au (FW. 1969665) and Pd FW. 106.40. c) What would the absorbance at 480 nm be for a 25.0 mLsample, treated and diluted to 500.0 mLas above, if the sample contained 4.24 x 105 M PdCI) and 5.66 x 10 M Au(II)?Explanation / Answer
7. Absorbance of mixture = absorbance due to eaxch component
Absorbance = molar absorptivity x path length x concentration
with path length for all the sample values = 1cm
(a) at 480 nm
absorbance = absorbance due to Pd(II) + absorbance due to Au(III)
0.533 = 3550[Pd(II)] + 2960[Au(III)]
at 635 nm
0.590 = 564[Pd(II)] + 14500[Au(III)]
Solving the two equations,
[Pd(II)] concentration in dilute solution = 0.00012 M
[Pd(II)] concentration in 50 ml solution = 2 x 0.00012 = 2.4 x 10^-4 M
[Au(III)] concentration in dikute solution = 0.000036 M
[Au(III)] concentration in 50 ml solution = 2 x 0.000036 = 7.2 x 10^-5 M
b. concentrations in ppm units,
[Pd(II)] = 2.4 x 10^-5 mol/L x 1000 x 106.4 g/mol = 25.5 ppm
[Au(III)] = 7.2 x 10^-5 mol/L x 1000 x 196.9665 g/mol = 14.2 ppm
c. with,
[Pd(II)] = 4.24 x 10^-5 M
[Au(III)] = 5.66 x 10^-4 M
we get at 480 nm for diluted solution,
absorbance = (3550 x 4.24 x 10^-5 + 2960 x 5.66 x 10^-4)/2
= (0.15052 + 1.6754)/2
= 0.913
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