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Homework Help: Titration -- What went wrong?

  1. May 17, 2018 #1
    1. The problem statement, all variables and given/known data

    We had a titration lab where we took 1.0 g of NaOH and added it to 50 mL of water. We then took that solution and added 10 mL of it to 500 mL of water, producing a 510 mL solution in total (this became the titrant). We then took 0.4 g of potassium hydrogen phthalate and added it to 50 mL of water (this was the analyte).

    Our group filled the burette 4 times and didn't get any colour change in the phenolphthalein. Where did we go wrong?

    2. Relevant equations

    C1V1=C2V2 and unit conversions

    3. The attempt at a solution

    I calculated that you need 199.7 mL of NaOH to titrate the KHP; can someone please tell me whether I'm right or not?

    This is how I calculated it
    (0.4 g KHP) * (1 mol KHP / 204.22 g KHP) = 0.019587 mol KHP and since it's monoprotic also 0.0019587 mol NaOH

    next the NaOH

    1 g NaOH * (1 mol NaOH / 39.997 g NaOH) = 0.025002 mol NaOH

    Now using C1V1=C2V2,

    (0025002 mol NaOH)/(50 mL*1L/1000mL) <--- this is the initial concentration, C1
    10 mL * 1L / 1000 mL <----- this is the initial volume, V1
    the final volume is 510 mL NaOH or 0.510 L
    rearranging and solving, I got 0.009805 M for the concentration

    C = n/V
    V = n/C
    V = (0.0019587 mol NaOH)/(0.009805 mol/L NaOH) = 0.1997 L or 199.7 L

    Any help would be appreciated. Thanks!
     
    Last edited: May 17, 2018
  2. jcsd
  3. May 18, 2018 #2
    From first hand experience, I found that a student took the sample to be titrated from a reagent of concentrated sample, rather than the dilute sample meant for titration. When the teacher found out what bottle they took the samples from , the teacher said he could have been titrating all day before the color would change. The base was too concentrated.
     
  4. May 18, 2018 #3

    Borek

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    Around 200 mL is what I got.

    Either 0.019587 or 0.0019587, but I bet it is just a typo.

    Much faster is to realize your 510 mL of the solution contains 10/50*1 g of NaOH.

    You never said what the burette volume was, they come in many sizes.

    Plenty of things that could go wrong, but planning a titration that requires adding titrant to burette is in general a faulty approach. Shouldn't you be titrating 10 mL of the KHP solution?
     
  5. May 18, 2018 #4

    symbolipoint

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    I only read this very quickly and not in great detail.

    The KHP is a STANDARD which is used for finding what is the concentration of the prepared NaOH titrant. The titrant was not prepared accurately but you must weigh the KHP accurately, even if you take an aloquot of its solution to use for the titration. Take care of all calculations in advance.
     
  6. May 18, 2018 #5

    symbolipoint

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    I took one more but still fast look at the original post. Yes, pay careful attention to all needed calculations ahead of time, and also check your computations and calculations to know and predict the effect of sample size between the KHP and amount of NaOH titrant.
     
  7. May 19, 2018 #6

    symbolipoint

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    This part of the procedure,
    if you handle the calculation correctly will predict a CONCENTRATION of 0.009803 M of NaOH titrant; and you would now have 510 ml. of this titrant.

    Now when you try to use KHP to check the actual concentration of NaOH titrant, you WILL NOT want to take all of the 510 ml. of this titrant. You would take smoe much smaller aloquot. (Formula weight of NaOH, 39.99 grams per mole. Formula weight of Potassium BiPhthalate, ?--------okay; KHP is 204.22 grams per mole.)

    Watch the quantity of KHP weighed very carefully and accurately. You did not weight exactly 0.4 grams KHP. You would or should weight to the nearest 0.0001 gram or better. If you were to weigh exactly 0.4000 grams of KHP, how many MOLES is this?
    You will then need to decide how much of your titrant you need to titrate this 0.4000 grams of KHP.
     
    Last edited: May 19, 2018
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