The formate ion, (CHO2-), is related to the acetate ion and forms ionic salts with many metal ions. Assume that 9.7416 g of M(CHO2)2 (where M represents the atomic symbol for a particular metal) are dissolved in water. When a solution of 0.200 M sodium sulfate is added, a white precipitate forms. The sodium sulfate solution is added until no more precipitate forms, then a few excess milliliters are added. The precipitate is filtered, washed, and dried. It has a mass of 9.389 g. The filtrate is placed aside.
A potassium permanganate solution is standardized by dissolving 0.9234 g of sodium oxalate in dilute sulfuric acid, which is then titrated with the potassium permanganate solution. The principal products of the reaction are manganese(II) ion and carbon dioxide gas. It requires 18.55 mL of the potassium permanganate solution to reach the end point, which is characterized by the first permanent, but barely perceptible, pink (purple) color of the permanganate ion.
The filtrate from the original reaction is diluted by pouring all of it into a 250-mL volumetric flask, diluting to the mark with water, then mixing thoroughly. Then 10.00 mL of this diluted solution is pipetted into a 125-mL Erlenmeyer flask, approximately 25 mL of water is added, and the solution is made basic. What volume of the standard permanganate solution will be needed to titrate this solution to the end point? The principal products of the reaction are carbonate ion and manganese(IV) oxide. Find M.
Is anyone here a chemistry genuis? - DrBob222, Friday, April 9, 2010 at 11:33am
I haven't worked the problem; however, I have a few observations to make.
Paragraph 1 is all you need to identify M; but, I question the numbers. The formate ion has an ionic mass of 45 and two of them makes 90. Thus the molar mass of M(CHO2)2 is M + 90. When M forms the sulfate the new molar mass of the sulfate is M + 96. The way I see it; the sulfate must weigh more than the formate. According to the post the sulfate weighs less. Unless I've missed something the numbers just don't add up. That plus the fact that the mass of the formate is given to five places while the sulfate ppt is given to only four makes me suspicious that one digit may have been omitted from the sulfate ppt mass.
The purpose of paragraph 2 is to determine the molarity of KMnO4.
The purpose of paragraph 3 is to determine the mL KMnO4 needed to titrate the formate placed there by the actions in paragraph 1.
I would be interested in knowing what others think.
Is ayone here a chemistry genuis? - Nn, Wednesday, April 11, 2012 at 4:55pm
I have the same problem and the mass is larger after the precipitate is filtered washed and dried. I have 9.9389 on my paper. I think the person who posted the question mistyped the number