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## Homework Statement

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

## Homework 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*)

## 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

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

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

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

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 ---------