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mol NaOH= 31.43mL x 1L/1000mL x 0.108 mol/L = 0.00339 mol NaOH

*There is a 1:1 Ratio between AC (Acetic Acid) and NaOH therefore,

mol AC = 0.00339 mol NaOH x 1 mol AC/1 mol NaOH = 0.00339 mol AC

Mass of AC in Vinegar = 0.00339 mol AC x 60.05 g AC/1 mol AC = 0.204 g AC

% AC in Vinegar = (g AC/g Vinegar) = 0.204 g AC/4.441 g Vinegar) x 100% = 4.6%

Andrea

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1. The problem statement, all variables and given/known data

The problem is as follows:

100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide . An appropriate indicator is used. Ka for acetic acid is 1.7 x 10 -5. Calculate the pH in the flask at the following points in the titration.

a. when no NaOH has been added.

b. after 25.0 ml of NaOH is added

c. after 50.0 ml of NaOH is added

d. after 75.0 ml of NaOH is added

e. after 100.0 ml of NaOH is added

2. Relevant equations

According to my text:

(Liters of acetic acid solution)X(mol H+/1Liter of solution)=mol H+

(Liters of sodium hydroxide solution)X(mol OH-/1Liter of solution)=mol OH-

(Liters of acetic acid)+(Liters of hydroxide solution)=total Liters

Concentration of H+=(moles H+/total Liters)

pH=-log(concentration of H+ *above*)

3. The attempt at a solution

What I have so far is this, but my pH answers are nowhere near what they should be. What am I missing?

"100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide . An appropriate indicator is used. Ka for acetic acid is 1.7 x 10 -5Calculate the pH in the flask at the following points in the titration."

(0.1L soln)(0.100M H+/1L soln)=1.000 X 10-2

a. "when no NaOH has been added" pH=-log(0.100)=1.000

b. "after 25.0 ml of NaOH is added" (0.025 L soln)(0.100M OH-/1L soln)=2.5 X 10-3 mol OH-

100.0mL+25.0mL=125.0mL=0.125L [H+]=(7.5x10-3)/(0.125L)=6.000X10-2M pH=-log(6.000X10-2M)=1.222

c. "after 50.0 ml of NaOH is added" (0.050L soln)(0.100M H+/1L soln)=5.000 X 10-3 mol OH-

100.0mL+50.0mL=150.0mL=0.150mL [H+]=(5.000 X 10-3)/(0.150L)=3.333 X 10-2 pH=-log(3.333 X 10-2)=1.477

d. "after 75.0 ml of NaOH is added" (0.075 L soln)(0.100M H+/1L soln)=7.5X10-3 mol OH-

100.0mL+75.0mL=175.0mL=0.175L [H+]=(2.5 X 10-3)/(0.175L)=1.429X10-2 pH=-log(1.429X10-2)=1.845

e. "after 100.0 ml of NaOH is added" (0.1L soln)(0.100M H+/1L soln)=1.000 X 10-2 mol OH-

100.0mL+100.0mL=200.0mL=0.200L [H+]=pH=7.0

This is a table similar to that which is shown in my text, where I have filled in my answers.

H+(aq) + OH-(aq) --> H2O(l)

Before addition 1.000 X 10-2 0 ---------

Addition 25.0 ml 2.5 X 10-3

Addition 50.0 ml 5.000 X 10-3

Addition 75.0 ml 7.5X10-3

Addition100.0 ml 1.000 X 10-2

After Adn25.0 ml 7.5x10-3 0 --------

After Adn 50.0 ml 5.000 X 10-3 0 --------

After Adn 75.0 ml 2.5 X 10-3 0 --------

After And 100.0 ml 0 0 ---------

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Q: A 31.43 mL volume of 0.108 M NaOH is required to reach the phenolphthalein endpoint in the titration of a 4.441 g sample of vinegar Calculate the percent acetic acid in the vinegar?
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