Note to Students in Chemistry 32A lab 04dec08
See attached and updated extra credit titration exercises for use in your report.
This added item on titration is something new for Chem 32A Acid&Base lab, so
in using it we found the first edition needed some updating and correcting. Please
print out the 2nd and 3rd page for your report, using data from this first page (or
your own) as needed.
Part A (solid carbonate titrated with acid)
Many students did not get to this one, and we had some problems with the first
trial, so here is data you can use in your report:
1.2 grams of Na2CO3*H2O dissolved in 50 ml distilled water
4 drops phenolphthalein turns solution pink
38.0 ml of 0.5 Molar HCl required to neutralize, pink turns to clear
The idea here is to see if the predicted moles of acid based on known amount of
carbonate agree with the titration data. Don’t forget the acid to base mole ratio
Part B (titration of Vinegar)
A number of people participated, and some stayed late to go over calculations.
We found the formulas are correct, but a confusion factor will be a formula
illustration next to Vinegar, M2 =M1*(V1/V2) which applies to NaOH, versus the
intended M1 =M2*(V2/V1) for Vinegar. Here’s some data very similar to that
collected by one of the teams, which you may use for your calculations if you did
not get to or finish this titration.
V1 = 1.9 ml Vinegar
M2 = 0.1 molar NaOH (per the bottle label)
V2 = 16.2 ml NaOH consumed in titration
The label on the “4 monks” vinegar says 5% acetic acid for your data
comparison
Titration Practice, 4 items
Item 3.
Household Ammonia
I used 10% by weight ammonia (NH3) in water, the upper end of household
product. You need to know the molarity of ammonia to finish the calculation. The
easiest way to estimate molarity is to observe that 10% by weight (100 grams of
NH3 in 1000 grams or 1 liter of solution), can be used with the formula weight of
ammonia (17 grams/mole) to determine molarity of about 5.9 … but please show
your calculation. The rest of this item is the usual M1V1=M2V2
Acids & Bases, 10% Extra Credit attachment on Titration
(updated version from one used in Chem32 lab on 04 December 2008)
Part A -- Titration of a solid carbonate with dilute acid
Na2CO3*H2O+ 2 HCl  2 NaCl + CO2 + H2O
Weigh approx 1.2 g sodium carbonate (Na2CO3*H2O) actual = ________ gm
Dissolve sample in Erlenmeyer flask with approx 50 ml distilled water
Add a few drops of phenolphthalein to solution, note color
_________
Titrate with ___ M Hydrochloric acid, until color vanishes
_________ ml
Moles of Na2CO3*H2O grams sample/(124 grams per mole)
_________ mole
Expected moles acid per carbonate from reaction formula (2:1) _________ mole
Actual moles acid = ________ moles/liter * __________ liters = _________ mole
Experimental versus theoretical moles of Acid required
_________ mole
Acid % error (difference moles/theoretical moles)*100
_________ %
Part B– Titration of Vinegar, determination of vinegar strength
CH3-COOH + NaOH  CH3-COONa + HOH
Measure 10.0 ml of household vinegar, add to flask
Add 3-4 drops of phenolphthalein indicator solution
Titrate with dilute NaOH, molarity value provided is
V1 = _______ ml vinegar
Volume of NaOH consumed to produce red color
V2 = _______ ml NaOH
Calculated Molarity of Vinegar M1 = M2 (V2/V1)
M1 =_______ M Vinegar
Moles of acid in sample of vinegar (M1 * V1)
M 2 = _______ M NaOH
_________ moles Vinegar
Formula weight for Acetic Acid
Grams of acetic acid in sample (moles*gm/mole)
60.05
g/mole
________ grams acid
Percent Acetic acid in sample (gms acid / gms sample)
Assume that the density of Vinegar is 1.0 making ml=grams
_________ %
Comparison to product label, (error = difference/actual)
_________ %
Titration Practice, applications of M1V1=M2V2 and moles
Always convert ml to Liters when calculating with Molarity (moles/Liter)
1. Book Example page 229, known H2SO4 and unknown NaOH
M1 = 0.108 M/L V1 = 25.0 ml
M2 = _____
V2 = 33.48 ml
2. Diluting household Vinegar with water
M1 = 0.80 M/L V1 = 100.0 ml conc M2 = 0.45 M/L
V2 = ______ml Dilute
3. Neutralizing Household Ammonia (10% weight % in water) with HCl
(First calculate NH3 molarity using 100 gm NH3 per liter or 1000 gm solution)
100 grams NH3 / 17 grams per mole = M1 =______ moles of NH3 per liter
M1 = ________
V1 = 100.0 ml
M2 = 0.50 M HCl
V2 = _______ ml HCl
4. Titrating 10.0 ml of 0.50 M H3PO4 with 0.10M NaOH
M1 = 0.50 M/L V1 = 10.0 ml acid M2 = 0.10 M/L V2 = ______ml NaOH