- #1
Matt1234
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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
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.
Thanks,
Matt
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
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.
Thanks,
Matt
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