Question: In The Analysis Of Sodium Using FAAS, Explain Why The Absorbance Of Sodium Increases In The Presence Of A High Concentration Of Potassium. (C) [6

In The Analysis Of Sodium Using FAAS, Explain Why The Absorbance Of Sodium Increases In The Presence Of A High Concentration Of Potassium. (C) [6 Marks] The Sulphate Concentration In An Unknown Sample Can Be Determined Using An Indirect Method Of EDTA Complexation Titration As Follows: The Sulphate In A 247.1 Mg Sample Was Precipitated As BaSO4

(b) [3 marks] In the analysis of sodium using FAAS, explain why

 

(b) [3 marks] In the analysis of sodium using FAAS, explain why the absorbance of sodium increases in the presence of a high concentration of potassium. (c) [6 marks] The sulphate concentration in an unknown sample can be determined using an indirect method of EDTA complexation titration as follows: The sulphate in a 247.1 mg sample was precipitated as BaSO4 by addition of 25.00 mL of 0.03992 M BaCl2. The precipitate was removed by filtration and the remaining BaCl2 consumed 36.09 mL of 0.02017 M EDTA for titration to the Calmagite end point. Calculate the % sulphate (SO4) in the sample. (d) [8 marks] [Atomic masses: S = 32.07 [0] = 16.00] The recommendation procedure for preparing a very dilute solution is not to weigh out a very small mass of a compound or measure a very small volume of stock solution. Instead, it is done by a series of dilutions. A sample of 0.8214 g of KMNO4 was dissolved in water and made up to a 500-mL volumetric flask. A 2.000-mL sample of this solution was transferred to a 1000-mL volumetric flask and diluted to the mark with water. Next, 10.00 mL of the diluted solution were transferred to a 250-mL volumetric flask and diluted to the mark with distilled water. (i) Calculate the concentration (in molarity) of the final solution. (ii) Calculate the mass of KMnO4 needed to directly prepare a 250-mL of the final solution and comment on the accurate of this method as compared to the procedures stated above. [Molecular mass of KMnO4 = 158.04 g mol] 2

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