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.
See your other post below.
To solve the third part of the problem, we need to determine the volume of the standardized permanganate solution required to titrate the diluted filtrate solution to the end point.
First, let's calculate the concentration of the diluted filtrate solution. We know that 10.00 mL of the diluted solution was pipetted into a 125-mL Erlenmeyer flask and approximately 25 mL of water was added. Therefore, the total volume of the diluted solution is 10.00 mL + 25 mL = 35.00 mL = 0.035 L.
Now, let's calculate the moles of the substance being titrated in the diluted filtrate solution, which in this case is the carbonate ion.
Using the balanced chemical equation:
2 MnO4- + 5 C2O4^2- + 16 H+ -> 2 MnO2 + 10 CO2 + 8 H2O
We can see that the stoichiometric ratio between the permanganate ion (MnO4-) and the carbonate ion (C2O4^2-) is 2:5.
Since it requires 18.55 mL of the standardized potassium permanganate solution (0.1793693 M) to reach the end point, we can calculate the moles of permanganate ion used:
mol MnO4- = Molarity * Volume(L) = 0.1793693 mol/L * (18.55/1000) L = 0.003327725 mol
Using the stoichiometry between MnO4- and C2O4^2-, we can calculate the moles of carbonate ion in the diluted filtrate solution:
mol C2O4^2- = (5/2) * mol MnO4- = (5/2) * 0.003327725 mol = 0.0083193125 mol
Now, let's calculate the concentration of carbonate ion in the diluted filtrate solution:
Concentration of C2O4^2- (M) = moles/Liter = 0.0083193125 mol / 0.035 L = 0.2371225 M
Finally, to determine the volume of permanganate solution needed to titrate the diluted filtrate solution, we can use the equation:
Concentration of C2O4^2- * Volume of C2O4^2- solution (L) = Concentration of MnO4- * Volume of MnO4- solution (L)
Rearranging the equation to solve for the volume of MnO4- solution:
Volume of MnO4- solution (L) = (Concentration of C2O4^2- * Volume of C2O4^2- solution (L)) / Concentration of MnO4-
Substituting the known values:
Volume of MnO4- solution (L) = (0.2371225 M * 0.010 L) / 0.1793693 M ≈ 0.01322678 L
Converting the volume to milliliters:
Volume of MnO4- solution (mL) = 0.01322678 L * 1000 mL/L ≈ 13.23 mL
Therefore, approximately 13.23 mL of the standardized potassium permanganate solution will be needed to titrate the diluted filtrate solution to the end point.