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Homework Help: Chemistry Lab Titration

  1. Oct 30, 2009 #1
    Hello all,

    We did a titration lab today. We used Ca(OH)2 in a burret and titrated a 10mL solution of vinegar. The solution of Vinegar was mixed with water, in a 1:20 ratio(inside a flask). So for every 10mL of solution there was 0.5mL of pure vinegar. We used phenothalen as the indicator.

    Here is the data we collected:

    Volume(Base)= Average(3 trials) = 35.1 mL Ca(OH)2 into vinegar and water solution.
    Conecntration(Base) = 0.025 mol/L

    Volume(Acid) = 0.5 mL (1/20 ratio at 10 mL per trial)
    Conecntration(Acid) = UNKNOWN (I keep getting 3.51 mol/L which is apparently way too high)

    Chemical equation:
    2 CH3COOH + Ca(OH)2 = Ca(CH3COO)2 + 2 H2O

    http://img251.imageshack.us/img251/6696/lastscanm.jpg [Broken]

    Notice i keep getting around 21.06% we were told we should see about 2% I been at this all day and am new to chemistry and cannot for the life of me figure this out.

    Please help me.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 30, 2009 #2


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    Homework Helper
    Education Advisor
    Gold Member

    One possibility: Look for any decimal error. Without actually checking your work, the similarity in numeric results suggests a possible decimal errorl

    One curiosity: Is calcium hydroxide actually a strong base? If not, then the reason for titrating a weak acid (acetic in vinegar) with a weak base is unusual.
  4. Oct 30, 2009 #3
    Thanks for your reply. We titrated using a weak base because our teacher ran out of sodium hydroxide. So calcium hydroxide was the substitute. I have checked over the calculation several times and cannot find where i have gone wrong with the decimals. Im beginning to wonder if the method is correct at this point that formula was created by the teacher.
  5. Oct 30, 2009 #4


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    Staff: Mentor

    Calcium hydroxide is a strong base (it is almost completely dissociated), although of low solubility.

    Edit: and this low solubility is an obvious problem here, saturated solution of calcium hydroxide is about 0.011M, it can't be 0.025M.

    Last edited: Oct 30, 2009
  6. Oct 30, 2009 #5
    ok thanks for your reply i dont know what to say the teacher gave us the 0.025 figure i also saw another one online for 0.027 for calcium hydroxide. at this point im very confused i think i will take a late penlty for this lab and ask her on monday. im very new to chemistry sorry.
  7. Oct 30, 2009 #6
    I just looked at the vinegar bottle and it says 5% acetic acid by volume.

    My average volume was 35.1 for Ca(OH)2. Ill try to start this from scratch. Ill need a proper value for concentration for limewater in mol/L if possible. This should really come off the bottle but the only thing i got to work with is 0.025 mol/L.
  8. Oct 31, 2009 #7


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    Staff: Mentor

    Hm, that's interesting - I have checked solubility tables and they suggest 22 mmol/L, while Ksp suggests lower concentration. I have recalculated - and taking ionic strength of the solution into account I got 16.6 mmol/L.

    Edit: one more take and it seems 22 mmol/L is OK, there is also CaOH+ complex present - about 5 mmol/L.

    Last edited: Oct 31, 2009
  9. Oct 31, 2009 #8
    Ok thanks for confirming that. Ultimately what i dont understand and what i cannot explain is why i get 21 percent while the vinegar bottle says only 5%. i did get similar volumes to my peers this is what im confused about.
  10. Oct 31, 2009 #9
    Bump please.
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