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Titration with NaOH and oxalic acid

  1. May 18, 2015 #1
    So Im doing a titration with NaOH and oxalic acid, prepared by adding 14g of the NaOH to 5dm^3 of H2O, the acid is (COOH)2.2H2O (2.5g in 0.25dm^3) . I seek to calculate the percentage purity of the NaOH.

    Now i'm confused, usually the percentage purity is based on products?? As in you have to know how many grams were formed. Any help on how to do this or if any information is missing? Any help apreciated,
     
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  3. May 18, 2015 #2

    Borek

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    You are trying to determine how many grams of NaOH is present in 14 g of the substance labeled "NaOH" (and express it as percentage) - does this way of wording the problem help?
     
  4. May 18, 2015 #3

    James Pelezo

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    Why are you using Oxalic Acid - Dihydrate*? It's not my titration, but I'd titrate the NaOH with ~0.50M KHPh (Potassium Hydrogen Phthalate). It is much safer and easier to use for experiments like this and is more accurate. It is a widely accepted reagent for calibrating Strong Base solution preps. Oxalic Acid is diprotic. Have you estimated the volume of O.70M NaOH you will use to titrate/neutralize 25-ml of a 0.80M (= 14-g/500-ml) Oxalic Acid solution through 2-equivalence points? (Hint: Vol > 50-ml) Here's a couple of notes that you should review (these are from Fisher Scientific MSDS on each chemical):

    Oxalic Acid hazards: EMERGENCY OVERVIEW Appearance: white powder. Danger! Causes burns by all exposure routes. Harmful if swallowed, inhaled, or absorbed through the skin. Possible risk of harm to the unborn child. May cause kidney damage. Target Organs: Kidneys, heart, eyes, skin, brain, nerves, mucous membranes.

    KHPh: Non-Hazardous Reagent. MSDS: EMERGENCY OVERVIEW Appearance: white crystals. Caution! May cause eye, skin, and respiratory tract irritation. Target Organs: None known.

    You can obtain KHPh from chem suppliers in small application quantities from most chem suppliers. I'd suggest googling chemical suppliers, Fisher, Sigma-Aldrich, Sargent-Welch, etc., find a phone number and call them. They will be glad to give you information on KHPh. If you plan to proceed with the oxalic acid, please wear eye protection, rubber gloves and if available a corrosion resistant apron. Be safe ... Have a good day. :-)
     
    Last edited: May 18, 2015
  5. May 18, 2015 #4

    Borek

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    This is fear mongering. Most MSDSs from Fisher Scientific make every reagent look like a chemical warfare agent, they are doing doing that most likely to not get sued if something goes wrong. Yes, oxalic acid can be unpleasant, but it is not much more dangerous than most compounds typically handled in the lab.

    That being said I agree KHP is a better standard substance in this case.
     
  6. May 18, 2015 #5

    James Pelezo

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    Thank you, I agree that the MSDS is 'over kill', but where else (and this is a ligit question from me) are we to learn (before use) of the 'potential' hazards and handling of a reagent? I've always suggest error on the side of the safer option. You are a wise & learned mentor. Keep up the good work.
     
  7. May 20, 2015 #6
    No it doesn't help, because I don't know what to compare it to. purity theoretical/what you have * 100. I can get work out the grams but I don't know how to get the theoretical. Any ideas?
     
  8. May 20, 2015 #7

    James Pelezo

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    Are you still planning to titrate the NaOH solution preparation into the Oxalic Acid solution? This reagent will have two equivalence points because it is diprotic. I'm not saying that you won't be able to determine the NaOH percent, but this approach (if you have limited experience at titrating multiprotic acids) will greatly complicate your your calculations. I still say you need to go online and review titrations with KHPh (Potassium Hydrogen Phthalate), develop a model problem and solve it before attempting a titration with H2C2O4. I'll be glad to offer suggestions as you go through the process.
     
  9. May 20, 2015 #8
    I already have, this is for a school practical. The only percentage purity questions we've ever done are the type of the likes "you have a 100g lump of NaOH, 30g of it are contaminants" sort of thing.
    So im very confused about how to do this, even if you don't do this specific example could you guide me with a different example?
     
    Last edited: May 20, 2015
  10. May 20, 2015 #9

    Borek

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    If you determine 9 g of something in 10 g sample, purity is 90%. No idea where your problem is - it is not much different from the problem you have just listed.

    Or is your problem with the titration and calculating the result of the titration?

    http://www.titrations.info/titration-calculation
     
  11. May 20, 2015 #10

    James Pelezo

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    If I'm understanding your last post, you're having problems calculating percentages. However, and because you stated this problem to be a part of a 'school' practical, before I give counsel, I would like to stress - delicately and with the utmost respect - that since this is not under the 'Homework' forum, I assume (for administrative purposes) that I can give you more guidance than if it were posted in the 'Homework' forum. However, please understand that I am doing this to help you understand the concept and answer your question as a query under the Chemistry Form heading. If this is a homework assignment, then I would ask that, in the future, you please post these type questions under the homework forum. This is a tightly regulated forum, and I'm living proof that posting comments, questions or replies in the wrong forum, or in conflict with posting guidelines is a definite NO! NO! Please understand, I am not trying to be rude or impolite, but please be cautious as you use this site. It will help you, but only in the spirit in which it is intended. I assume you know that this segment of Physics Forums is not here to do your homework for you, but since it is under this heading I will present - to the best of my limited ability the determination of weight-% of component in a mix. I hope it will help.

    We'll use your data ... Let's take the 30 grams NaOH in a 100 gram mix (=> the lump). Percent(%) is defined as [(Part of interest / Total mix containing the part of interest] x100%. Applying this as Weight-% = (wt of part of interest / total wt of mix-lump) x 100%. Actually, you can obtain '2' answers;i.e.,(1) Wt-% Contaminant and/or (2) Wt-%NaOH. I'm sure you are interested in the NaOH, but for instructional purposes let's do both... Wt-%Contaminant = (30g/100g)100% = 30% by weight Contaminant. The remaining amount is the NaOH. Wt of NaOH(g) = (100g-lump) - (30g-Contaminant) = 70g NaOH. Then, Wt-% NaOH = (70g/100g)100% = 70% by Wt NaOH. The sum of percentages always must = 100%. You could have also calculated the %NaOH by subtracting the %-Contaminant from 100%-lump to get 70% NaOH. Then 70% of 100g = 0.70(100%) = 70% by weight NaOH.

    Note: This calculation was based on weight ratios => Wt-% results. However, 'Volume-%' (Vol-%) is also frequently used. Given 30-ml of Alcohol in 90-ml of Alcohol/water Solution, what is the Vol-% of Alcohol in the solution? Vol-% Alcohol = (30ml Alcohol/90ml soln) x 100% = 33.33 Vol-.% Alcohol and 66.67 Vol-% Water ( = 100% - 33.33% = 66.67% water by volume). I think this is what you are asking about. If not, restate your problem and I'll try again. All the best and good Luck. :-)
     
    Last edited: May 20, 2015
  12. May 20, 2015 #11
    No, I was saying those are the easy ones, iIget your last post, My point was that you have no mass of contaminant nor anything to compare the NaOH to. they are just saying "there is 14g of NaOH in this water. how pure is it?" essentially. I feel as though im missing data.
     
  13. May 20, 2015 #12

    James Pelezo

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    OK, looking back at your original post, the problem lists 14g NaOH in 5dm3 of H2O. Let's say for the sake of accuracy, the 5dm3 is the volume of solution and the density of the solution ~1.0g/ml. I think that would be reasonable relative to just 14g NaOH dissolved into 500,000ml of solution. Then, 14g NaOH in 5dm3 solution = 14g NaOH/500000ml Solution = 14g NaOH/500000g Solution => 0.000028 g-fraction or 0.00280% by wt NaOH. So purity of the NaOH/H2O solution = 0.00280% active in NaOH.
     
    Last edited: May 20, 2015
  14. May 21, 2015 #13

    Borek

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    Are you sure it is '14 g of NaOH' and not '14 g of substance labeled as NaOH'? I mean - they are taking 14 g of substance but 14 g is the total mass of the substance and impurities. You use titration to measure exact amount of NaOH, assuming impurities are inert and don't interfere. After that you can calculate mass of the 'real NaOH' in the 14 g, and if it is lower than 14, the difference are the impurities.
     
  15. May 21, 2015 #14
    That's what they mean
     
  16. May 21, 2015 #15
    How does one do that without the products?
     
  17. May 21, 2015 #16

    Borek

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    You seem to be thoroughly confused - actually you are so confused, I am confused as to what your confusion is o_O

    What is the reaction between oxalic acid and NaOH? What are products of this reaction? What are - in general - products of the reaction between an acid and a base?

    Or do you mean something else by "products"?

    Do you know what titration is, how it works and why it is used?
     
  18. May 21, 2015 #17

    James Pelezo

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    I'll second that, Borek!
     
    Last edited by a moderator: Apr 17, 2017
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