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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: narutoverse13 on September 16, 2014, 04:29:35 PM

Title: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 16, 2014, 04:29:35 PM

Hello! I am literally stumped as to what I should for the following problem:

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.

I've figured out the first part. M is Lead. But I am not sure what to do for the second and third part :(.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: Borek on September 16, 2014, 05:18:54 PM

Start writing reaction equations. In both cases it is just a simple stoichiometry.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 16, 2014, 05:23:07 PM

Borek, I definitely need your help

1st equation (NIE)= Mn²⁺ + SO4^- > MSO4. (After doing my equation for M, I found M to be Pb).
2nd Equation (NIE)= C2O4^- + MnO4^- > Mn²⁺ + CO2
3rd Equation (lost completely): CHO2^- + MnO4^- > CO3^2- + MnO2

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 16, 2014, 05:50:40 PM

Okay, I used the 9.7416 g and the 9.389 grams to figure out what M is. I am honestly just not sure what I have to use the .9234 grams & 18.55 mL in equation 2 & all the numbers in equation 3. What else should be done?

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: Borek on September 17, 2014, 02:46:57 AM

Quote from: narutoverse13 on September 16, 2014, 05:23:07 PM

2nd Equation (NIE)= C2O4^- + MnO4^- > Mn²⁺ + CO2

Close. C₂O₄^2-, otherwise looks good. Now balance this equation and use the information given to find concentration of the permanganate. (Beware: you were told how much sodium oxalate you have; and your NIE uses oxalate ions).

Quote

3rd Equation (lost completely): CHO2^- + MnO4^- > CO3^2- + MnO2

Correct - just balance it. Once balanced - use the information about the amount of formate (if you know the original metal was lead, you can easily calculate how many moles of formate was in the lead formate sample) to calculate volume of permanganate solution (you know permanganate concentration from the previous step).

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 17, 2014, 12:50:45 PM

Correction* The First Equation should've been (NIE)= M²⁺ + SO4^2- > MSO4. (After doing my equation for M, I found M to be Pb).

So would it be a good idea to replace the C2O4^2- in the second equation with NaC2O4? If I did that, then I am not sure how I would balance that since there is Na only on one side. Also, how do we know the concentration (mols/L) of permanganate? We were just given the mL of potassium permanganate needed for the solution to reach endpoint.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: Borek on September 17, 2014, 02:04:24 PM

Quote from: narutoverse13 on September 17, 2014, 12:50:45 PM

So would it be a good idea to replace the C2O4^2- in the second equation with NaC2O4? If I did that, then I am not sure how I would balance that since there is Na only on one side.

Then add Na⁺ between products. It is a spectator anyway.

But you don't have to - you can use oxalate anion as you did, just calculate number of moles from known mass od the molar mass of sodium oxalate (you have the same number of moles of oxalate anions as of sodium oxalate, don't you?)

Quote

Also, how do we know the concentration (mols/L) of permanganate? We were just given the mL of potassium permanganate needed for the solution to reach endpoint.

You can calculate number of moles from stoichiometry, you are given volume. What is the definition of the concentration? Just plug and chug.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 19, 2014, 12:20:58 PM

Okay, so I balanced the equations from  the second and third ones.
2nd Equation: 5C₂O₄^2- + 2MnO₄^- > Mn²⁺ + 2CO₂
3rd Equation: 3CHO₂^- + 2MnO₄^- > CO₃^2- + MnO₂^2-.
I found the following: .0069 moles of C₂O₄^- ; .0655 moles of CHO₂^- ; .0328 moles SO₄^2- ; .40 M Na⁺.  Based on this information, how exactly can I find the volume of permanganate (MnO₄^-) needed to titrated the solution to the equivalence point of 18.55 mL? I know that concentration is moles of solute/L of solution. The only thing I know is the volume of potassium permanganate needed to the equivalence point. I have NO idea how many grams of permanganate solution there is. How then could I possibly find concentration of the potassium permanganate solution, Borek?
What equations should I set up next? My chemistry professor is so terrible at teaching, and I have to learn all of this by myself while using a chemistry textbook that is literally 15 years old. I am literally trying to solve this problem by trying to decipher this textbook.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: Borek on September 19, 2014, 05:04:25 PM

Quote from: narutoverse13 on September 19, 2014, 12:20:58 PM

Okay, so I balanced the equations from  the second and third ones.
2nd Equation: 5C₂O₄^2- + 2MnO₄^- > Mn²⁺ + 2CO₂

This is not balanced - oxygen is not balanced, charge is not balanced. Surprisingly the ratio of coefficients that you need to use is right.

For the titration calculations in general see http://www.titrations.info/titration-calculation - while it doesn't address specifically your problem, all titration calculations are based on the same principles.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 19, 2014, 11:43:02 PM

I think the overall answer is 11.2 mL of permanganate. Is this correct Borek? As I said Borek, my teacher hasn't even gone over titration; he expects us to do the problem without balancing the equation properly. I figured out the 5 on the oxalate and the 2 on the permanganate using a trick with oxidation numbers. He said not to worry about balancing the oxygens and just use the oxidation number trick. This obviously goes against what i normally do, but I just simply followed what he instructed me to do. The trick is that since Mn on the left side has a +7 charge and the Mn on the right has a +2 charge, the oxalate will have a 5 in front of it. Since C₂ has an overall charge of 6 (3 x 2) on the left side and the Carbon on the right has a charge of +4, the Mn has a 2 in front of it. I am not sure why this worked, but he showed me this method. I figured out everything from there, so I just want to confirm if 11.2 mL is the volume of permanganate needed

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: Borek on September 20, 2014, 03:26:50 AM

Quote from: narutoverse13 on September 19, 2014, 11:43:02 PM

my teacher hasn't even gone over titration; he expects us to do the problem without balancing the equation properly. I figured out the 5 on the oxalate and the 2 on the permanganate using a trick with oxidation numbers.

OK, that's a correct approach of finding the ratio at which main molecules in redox system react - just don't call such reactions balanced (you may write "from oxidation numbers we know they will react this way" or something like that).

I wanted to check your result, but it is impossible. Please check if the data you have entered is correct:

Quote from: narutoverse13 on September 16, 2014, 04:29:35 PM

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.

Quote

I've figured out the first part. M is Lead.

Lead doesn't fit the data given.

Actually there is no element that fits the data given.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 20, 2014, 12:38:07 PM

Originally, I also wondered what "M" was. I got "M" to equal 206.5 grams. I asked my teacher what to do about this, and  he said to use the element closes to the number I got. Since Lead is 207.2, I used lead. My teacher also said that Lead was correct for the identity of M.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: Borek on September 20, 2014, 02:21:51 PM

Quote from: narutoverse13 on September 16, 2014, 04:29:35 PM

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.

When 9.7416 g of Pb(CH₂O)₂ reacts with sulfate, it produces 9.939 g of PbSO₄, not 9.389 g.

To get 9.389 g of sulfate (which is less than the mass of formate, despite molar mass of sulfate being larger than the mass of formate) you need a metal M with a negative molar mass.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 20, 2014, 02:39:11 PM

Oh wow! I didn't realize that I typed that part in incorrectly. You're right, it should have a mass of 9.9389 grams (I guess I forgot to type the 9 before the 3). But I still got the element to be Pb.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: Borek on September 20, 2014, 02:51:16 PM

For 9.9389 g Pb is a reasonable answer.

Show how you got 11.2 mL, that's not what I found.

First of all - what have you got for the concentration of permanganate solution?

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 20, 2014, 02:54:56 PM

I actually got 11.6 mL of permanganate.
I got the concentration of permanganate to be .15 M. I did this calculations by finding moles of permanganate first and then dividing that by 18.55 mL.

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: Borek on September 20, 2014, 03:10:14 PM

Don't round down intermediate values (or rather: report them rounded, but use full precision or at least several guard digits in further calculations).

11.76 mL is what I got (note I am using EBAS, so my molar masses can be slightly different).

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: narutoverse13 on September 20, 2014, 03:12:31 PM

Oh wow! I can't believe that I got so close to what you got!!!!! I am so excited!!!!!!!!! Thank you so much for your help Borek! You are right when you said that I may have rounded excessively. I couldn't have done this without your assistance!

Title: Re: Second Marathon Problem (Ridiculously Hard)
Post by: KungKemi on March 11, 2017, 06:44:00 PM

Sorry to resurrect an old forum post, however, just for future reference, I did this question a few days ago and I found that for the second and third reactions you needed to balance these using the half-reaction method (which doesn't make a whole lot of sense because the Zumdahl textbooks only cover this for Electrochemistry later on). Anyway, in doing so you would get the equations:

2MnO₄^-_(aq) + 16H⁺_(aq) + 5C₂O₄^2-_(aq) ::equil:: 2Mn²⁺_(aq) + 8H₂O_(l) + 10CO_{2 (g)}

2MnO₄^-_(aq) + OH^-_(aq) + 3CHO₂^-_(aq) ::equil:: 3CO₃^2-_(aq) + 2H₂O_(l) + 2MnO_{2 (s)}

The first reaction occurs in acidic solution (because of the sulfuric acid), and the second reaction occurs in basic solution (because of the hydrolysis of water by the formate ion).

KungKemi