Answered: EDTA and Mg

Science

Chemistry

EDTA and Mg

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter25: Instruments For Optical Spectrometry

Section: Chapter Questions

Problem 25.5QAP

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A 17.50 mL solution containing Zn2+ was treated wth 40.00 mL of 0.0600 M EDTA to complex all of the Zn2+ and leave excess EDTA in solution. Titation of the excess EDTA required 20.00 mL of 0.0500 M Ca2+ What was [Zn2+] molar concentration in the original solution? a) 1.600 b)0.0800 c)0.0400 d)0.0200
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4. The %Mn in steel can also be determined spectrophotometrically by oxidizing the manganese to the intensely colored permanganate, MnO4-. Standard solutions of permanganate gave the following absorbances (in a 1 cm cell): ABSORBANCE CONC. OF MnO4 (g/mL) 0.210 1.05 x 10-5 0.315 1.61 x 10-5 0.429 2.21 x 10-5 0.599 2.98 x 10-5 0.801 4.04 x 10-5 The following samples of steel were thoroughly reacted to convert the manganese into permanganate and then diluted to 500.00 mL. WT. OF STEEL (g) ABSORBANCE 0.5886 0.611 0.3498 0.359 0.4555 0.482 Again using Excel, plot the calibration data and calculate the %Mn in the steel.
Based on the UV Spectra, need help filling out the following table. Thank you :) λmax (nm) Absorbance e.g. 540 e.g. 0.05
10. A 50.0ml solution containing Ni2+ and Zn2+ was treated with 25.0 ml 0.0452M EDTA to bind all the metals. The excess unreacted EDTA required 12.4 ml 0.0123M Mg2+ for complete reaction. An excess reagent is added again to replace EDTA from Zn2+. Another 29.2 ml of 0.0123M Mg2+ were required for reaction with the liberated EDTA. Calculate the molarity of Ni2+ and Zn2+ in the original solution. O a) Ni2+ = 0.0110 M, Zn2+ = 7,1832 x 10^-3 M O b) Ni2+ =7.18 x10-^3 M, Zn2+ = 5.57x 10^-3 M O C) Ni2+ = 9.09 x10 ^-4 M, Zn2+ = 1.0625x 10^-3 M O d) none of the above results Option 5
Answer part i ii
A 25.0 mL aliquot of 0.0700M EDTA was added to a 32.0 mL solution containing an unknown concentration of V3+. All of the V3+ present in the solution formed a complex with EDTA, leaving an excess of EDTA in solution. This solution was back- titrated with a 0.0380 M Ga+ solution until all of the EDTA reacted, requiring 15.0 mL of the Ga+ solution. What was the original concentration of the V³+ solution? [V3+] = M
A 7.13 g sample of an ant-control preparation was decomposed by wet-ashing with H₂SO and HNO₂. The As in the residue was reduced to the trivalent state with hydrazine. After removal of the excess reducing agent, the As(III) required 31.46 mL titration with 0.03142 M I₂ in a faintly alkaline medium. What is the % As in the original sample? Select one: O 1.20% 2.08% O 1.04% O 00 0.519%