# Chemistry Lab Titration

1. Oct 30, 2009

### Matt1234

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

Solve
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.

Thanks,
Matt

Last edited by a moderator: May 4, 2017
2. Oct 30, 2009

### symbolipoint

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.

3. Oct 30, 2009

### Matt1234

Hi
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.

4. Oct 30, 2009

### 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.

--
methods

Last edited: Oct 30, 2009
5. Oct 30, 2009

### Matt1234

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.

6. Oct 30, 2009

### Matt1234

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.

7. Oct 31, 2009

### 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.

--
methods

Last edited: Oct 31, 2009
8. Oct 31, 2009

### Matt1234

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.

9. Oct 31, 2009