Calculate the pH in the flask at the following points in the titration?

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.

f. What would the pH be after 300 ml of 0.100 M NaOH was added?

g. How could the equivalence point be detected wthout the use of a visual indicator?

Answer:
a.
CH3COOH <> CH3COO- + H+
initial concentration
0.100
at equilibrium
0.100-x . . . . . . . . . .x . . . . .. . .x
1.7 x 10^-5 = x^2 / 0.1-x
x = [H+] = 0.00130 M
pH = 2.88

b.
moles CH3COOH = 100 x 0.1 /1000 = 0.01
moles NaOH = 25 x 0.1 /1000 = 0.0025
0.01 mole acetic acid + 0.0025 mole OH- >> 0.0075 mole acetic acid + 0.0025 mole acetate
Total volume = 125 mL = 0.125 L
[CH3COOH] = 0.0075 / 0.125 =0.06 M
[CH3COO-] = 0.0025 / 0.125 = 0.02 M
pK = - log 1.7 x10^-5 =4.77
pH = pK + log [CH3COO-]/ CH3COOH] =
= 4.77 + log 0.02/0.06 = 4.29

c.
moles acetic acid = 0.01
moles OH- = 50 x 0.1 /1000 = 0.005
Moles acetic acid = 0.01 - 0.005 = 0.005
Moles acetate = 0.005
Total volume = 150 mL = 0.150 L
[CH3COOH]= [CH3COO-] = 0.005 / 0.150 =0.0333
pH = 4.77 + log 0.0333/0.0333 = 4.77

d.
moles acetic acid = 0.01
moles NaOH = 75 x 0.1 /1000 = 0.0075
Moles acetic acid = 0.01 - 0.0075 = 0.0025
Moles acetate = 0.0075
Total volume = 175 mL = 0.175 L
[CH3COOH] = 0.0143 M
[CH3COO-] = 0.0428 M
pH = 4.77 + log 0.0428 / 0.0143 = 5.25

e.
moles acetic acid = 0.01
moles NaOH = 100 x 0.1 /1000 = 0.01
we are at equivalent point : in the solution there is only acetate. Total volume = 200 mL = 0.2 L
[CH3COO- ] = 0.01 /0.2 = 0.05 M
CH3COO- + H2O <> CH3COOH + OH-
for this equilibrium
K = Kw/Ka = 5.88 x 10^-10
5.88 x 10^-10 = x^2 / 0.05 -x
x =[OH-] = 5.42 x 10^-6 M
pOH = 5.26
pH = 8.74

f.
300 - 100 = 200 mL of OH- are in excess
moles OH- = 200 x 0.1/1000 = 0.02
total volume = 0.4 L
[OH-] = 0.02 / 0.4 = 0.05 M
pOH = 1.30
pH = 12.7
Stop trying to cheat on your homework and do the math.. haha

I hated inorganic chemistry... Organic was pretty fun though. If I had to suffer through it so do you!

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