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posted by  col33 on 7/11/2008 4:54:52 PM  |  status: Closed  

titration

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Question Details:
What volume of 1.00 M NaOH is required to titrate 100. mL of vinegar (5.0% acetic acid by mass, 1.31 g/mL)?

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Sage
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posted by SamiLiLOnE on 7/11/2008 9:47:18 PM  |  status: Live
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col33's comment:
"Very helpful, thank you!"
Response Details:
First use the density of vinegar to find its mass
1.31 g x 100 mL = 131 g vinegar
   mL

Then from the mass, multiply by the mass percent of acetic acid to find out the grams of acetic acid
131 g vinegar x 5.0% = 6.55 g acetic acid

Now find mol acetic acid using molar mass
6.55 g HC2H3O2 x mol HC2H3O2 = 0.109 mol acetic acid
                                     60 g

Convert mol to molarity using mol calculated and volume given
0.109 mol/0.100 L = 1.09 M

M1V1 = M2V2
(1.00 M)V1 = (1.09 M)(0.100 L)
V = 0.109 L = 109 mL NaOH

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