100.0 ml of 0.100 M acetic acid is placed in a flask and istitrated with 0.100 M
ID: 822497 • Letter: 1
Question
100.0 ml of 0.100 M acetic acid is placed in a flask and istitrated 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 pointsin 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 be the appropriate indicator that was used ? consultfig 16.7 in BLB and explain your choice.
Explanation / 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
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