Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: sapphiregirl on September 04, 2009, 01:32:50 AM
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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.
I don't even know where to begin. Some guidance of some sort would be greatly appreciated. Thank you.
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You know masses of equimolar amounts of M(CHO2)2 and MSO4 - won't it be enough to find out molar mass of M?
Edit: hm, should be, but with these numbers (9.7416 & 9.389) molar mass seems to be negative...
Second part requires reaction equations and number of moles from the first part.
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i'm still confused. i don't get what you just said. what's the first step?
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Assume M is a molar mass of the metal and n is number of moles of the metal. Try to express masses of the salts in terms of these numbers. That will give you two equations in two unknowns.
Unfortunately, while this seems to be the correct approach, it leads nowhere - are you sure there is no typo in the data you have entered, especially in the masses of the salts?
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there were some typos i realized. here's the problem again:
The formate ion, (CHO2)1-, forms ionic compounds with many metal ions. Assume that 9.746 g M(CHO2)2 (where M represents the atomic symbol for a particular metal) is dissolved in water. When a solution of .200 M (molarity) sodium sulfate is added, a white precipitate forms. The sodium sulfate solution is added until no more precipitate forms; then a few milliliters are added in excess. The precipitate is filtered, dried, and weighed. It has a mass of 9.9389 g. The filtrate is saved for further use.
A potassium permanganate solution is standardized by dissolving 0.9234 g sodium oxalate in dilute sulfuric acid and then titrating with the potassium permanganate solution. The principal products of the reaction are manganese (II) ion and carbon dioxide gas. The titration requires 18.55 mL of the potassium permanganate solution to reach the endpoint, which is indicated by the first permanent, but barely perceptible, pink color of the permanganate ion.
The filtrate from the original reaction is diluted by pouring it into a 250-mL volumetric flask, diluting to the mark with water, and then mixing thoroughly. An aliquot consisting of 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 equivalence point? The principal products of the reaction are carbonate ion and manganese (IV) oxide. Identify M.
I'm still very confused.
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borek, i don't get what you said about that first step. i don't understand what i'm supposed to do. please help somebody. i'm very lost.
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If M is molar mass of the metal, and if chloride formula is MCl2, mass of the n moles of this chloride is n*(M+2*35.5).
Now try to write identical formulas for masses of both sulfate and formate. You know that in both equations M and n will be the same - just solve.
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ok where's the Cl2 coming from? The first part CHO2. And also what about all the titration and filter stuff?
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ok where's the Cl2 coming from? The first part CHO2. And also what about all the titration and filter stuff?
Cl is just an example, to make you understand how to deal with sulfate and formate.
Wrire equations for all reactions taking place. Start with preicipitation, as it will tell you what has been filtered and what was left in the solution.
http://www.titrations.info/
http://www.titrations.info/permanganate-titration
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Ok here are the equations, though I'm still confused.
The formate ion, (CHO2)1-, forms ionic compounds with many metal ions. Assume that 9.746 g M(CHO2)2 (where M represents the atomic symbol for a particular metal) is dissolved in water. When a solution of .200 M (molarity) sodium sulfate is added, a white precipitate forms. The sodium sulfate solution is added until no more precipitate forms; then a few milliliters are added in excess. The precipitate is filtered, dried, and weighed. It has a mass of 9.9389 g. The filtrate is saved for further use.
M(CHO2)2 + Na2SO4 -> 2NaCHO2 + MSO4 (which one is the precipitate???)
A potassium permanganate solution is standardized by dissolving 0.9234 g sodium oxalate in dilute sulfuric acid and then titrating with the potassium permanganate solution. The principal products of the reaction are manganese (II) ion and carbon dioxide gas. The titration requires 18.55 mL of the potassium permanganate solution to reach the endpoint, which is indicated by the first permanent, but barely perceptible, pink color of the permanganate ion.
Na2C2O4 + HSO4- + KMnO4 -> Mn2- + CO2 + ? (what happens to the Na, SO4, and K?)
The filtrate from the original reaction is diluted by pouring it into a 250-mL volumetric flask, diluting to the mark with water, and then mixing thoroughly. An aliquot consisting of 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 equivalence point? The principal products of the reaction are carbonate ion and manganese (IV) oxide. Identify M.
Filtrate + KMnO4 -> MnO2 + CO3-
help.
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M(CHO2)2 + Na2SO4 -> 2NaCHO2 + MSO4 (which one is the precipitate???)
Sulfate.
Na2C2O4 + HSO4- + KMnO4 -> Mn2- + CO2 + ? (what happens to the Na, SO4, and K?)
Nothing - they are just spectators.
By some magic trick you have changed NaCHO2 to Na2C2O4.
Mn2+. And you can ignore HSO4-, sulfuric acid is just a source of H+.
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it say sodium oxalate. isn't sodium oxalate Na2C2O4?
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M(CHO2)2 + Na2SO4 -> 2NaCHO2 + MSO4
is this the balanced equation for what they're talking about in the first part?
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is this final reaction when it says you take the filtrate from the original reaction, dilute it with water, then titrate it with potassium permanganate solution, and the products are manganese (IV) oxide and carbonate ion?
MSO4 (where M is the metal) + KMnO4 -> MnO2 + CO3-?
if so, where is the carbon coming from in the products?
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it say sodium oxalate. isn't sodium oxalate Na2C2O4?
Ah OK, there will be two reactions - one with oxalate, one with formate.
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is this final reaction when it says you take the filtrate from the original reaction, dilute it with water, then titrate it with potassium permanganate solution, and the products are manganese (IV) oxide and carbonate ion?
MSO4 (where M is the metal) + KMnO4 -> MnO2 + CO3-?
if so, where is the carbon coming from in the products?
What is filtrate? What was precipitated?
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these are the reactions that i have:
M(CHO2)2 + Na2SO4 -> 2NaCHO2 + MSO4 (this is from the first paragraph)
Na2C2O4 + H2SO4 + KMnO4 -> Mn2+ + 2CO2 (this is from the second paragraph)
MSO4 + MnO41- -> MnO2 + CO31- (this is from the third paragraph)
i don't know where to go from here.
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M(CHO2)2 + Na2SO4 -> 2NaCHO2 + MSO4 (this is from the first paragraph)
OK
Na2C2O4 + H2SO4 + KMnO4 -> Mn2+ + 2CO2 (this is from the second paragraph)
OK
MSO4 + MnO41- -> MnO2 + CO31- (this is from the third paragraph)
This is wrong. I have already signalled why, but instead of trying to understand the hints you prefer to write
i don't know where to go from here.
You won't get far this way. When you are thinking you are doing fine. Yes, you can.
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i thought the precipitate and the filtrate was the same which is why i was confused.
anyway, the filtrate is NaCHO2 so the third formula is NaCHO2 + MnO41- -> MnO2 + CO31-, right?
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anyway, the filtrate is NaCHO2 so the third formula is NaCHO2 + MnO41- -> MnO2 + CO31-, right?
Something like that - probably ignoring Na+ (spectator) and writing product as CO2 it will be easier to balance reaction equation.
Observe your chanrges - no such thing as CO3-
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The principal products of the reaction are carbonate ion and manganese (IV) oxide.
straight from the problem. isn't the carbonate ion CO3 (charge of 1-)?
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Carbonate is CO32-.
For the stoichiometry of permanganate/formate reaction it doesn't matter if the product is CO2 or CO32- - in both cases molar ratio of permanganate and formate is identical, and that's what is important for the final answer. IMHO reaction will be easier to balance with CO2 listed as a product.
No idea who will check your work and how - nitpicker may argue that you have not followed the problem verbatim.
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thank you for all your help. i just handed in my homework today to the teacher so we'll see what happens. thanks again.