If I read the problem right the third part titrates the formate ion from part 1. Write the equation for formate + KMnO4 in basic solution.
2. You know how many mols formate from part 1. Convert that to mols MnO4^- in part 3.
3. Convert mols in part 3 for the dilution.
4. Then M KMnO4 = mols KMnO4/L KMnO4. You know mols and M, solve for L and convert to mL.
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 9.9389 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.
I already figured out the first part that M= Pb and I think I got the second part that KMnO4 molarity= .1793693, but I'm completely lost on part three. Any help is appreciated.
2 answers
CHO2^- + KMnO4 > CO3^2- + MnO2
How are you supposed to balance the equations since there is no K in the products?
How are you supposed to balance the equations since there is no K in the products?
2 answers