Unformatted text preview: olution used in dilution in liters (1000 mL = 1L)
Moles of sodium acetate solution used in dilution (n = MV) 2. Finding the molarity of acetic acid (HC2H3O2) in solution B
Molarity of acetic acid solution on bottle
Volume of acetic acid solution used in dilution in milliliters
Volume of acetic acid solution used in dilution in liters
Moles of acetic acid solution used in dilution (n = MV)
Total volume of solution in dilution in milliliters
Total volume of solution in dilution in liters (1000 mL = 1 L)
Molarity of acetic acid in diluted solution (M = n/V) Department of Physical Sciences Kingsborough Community College The City University of New York Spring 2012
Experiment 6: Determination of the Equilibrium Constant for Bromocresol Green 9 1. Moles of Acetic Acid
Now we need to calculate the number of moles of acetic acid each time we added 2.00 mL of acetic acid
solution to the bromocresol green solution in Part 2. The volume of acetic acid solution added was 2.00
mL through 8.00 mL. Convert these volumes to liters. The molarity of the acetic acid solution was
calculated on the previous page and is the same for each addition. The moles of acetic acid added can be
found from n = MV.
Volume of
Acetic Acid
Solution Added
(mL) Volume of
Acetic Acid
Solution
Added
(L) Molarity of
Acetic Acid
Solution
(mol/L)*
M = n/V Moles of
Acetic Acid
Added
(mol) 2.00 mL
4.00 mL
6.00 mL
8.00 mL
*From the previous page. This amount should be the same for each addition.
2. Moles of Sodium Acetate (NaAc)
We also need to know the number of moles of sodium acetate used in the mixture for Part 2. This was
calculated on the previous page.
3: Calculating the pH of the acetate solutions
The following table shows the progression of the calculations needed to find the pH of each solution in
Part 2. We first use the moles of the acetic acid and the moles of sodium acetate (already calculated) to
find the ratio of sodium acetate to acetic acid. We then take the log (base 10) of this ratio. The pH of a mol NaAc . The pKa for mol HAc solution that contains a weak acid and its conjugate base is: pH = pKa + log acetic acid is 4.74. Find the pH of each solution.
Volume of
Acetic
Acid
Solution
Added
(mL)
2.00 mL Moles of
Acetic
Acid*
(mol) Moles of
Sodium
Ace...
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 Spring '13
 PatrickLloyd
 Chemistry, Equilibrium, Bases, pH

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