Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
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(C...Asked by bill
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.
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.
Answers
Answered by
DrBob222
I'll help you do the first part and leave the second part for you.
The first part is the formation of MSO4.
mass MSO4 = 9.9389
mass M(CHO2) = 9.7416.
You can identify M from this.
9.9389 x (molar mass of the Mformate)/(molar mass MSO4) = 9.7416
Substitute for molar mass of the Mformate (M + 90) which is the molar mass M + molar mass of 2*formate.
Substitute for MSO4 (M + 96) which is molar mass M + molar mass of SO4^2-.
Solve for M. I think it is Pb.
Now you do the second part. The permanganate will oxidize the formate. Post your work if you get stuck.
The first part is the formation of MSO4.
mass MSO4 = 9.9389
mass M(CHO2) = 9.7416.
You can identify M from this.
9.9389 x (molar mass of the Mformate)/(molar mass MSO4) = 9.7416
Substitute for molar mass of the Mformate (M + 90) which is the molar mass M + molar mass of 2*formate.
Substitute for MSO4 (M + 96) which is molar mass M + molar mass of SO4^2-.
Solve for M. I think it is Pb.
Now you do the second part. The permanganate will oxidize the formate. Post your work if you get stuck.
Answered by
confusedstudent
wait I found M to be Pb, but I don't know what to do next.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.