A student titrates 10.0mL of a 0.500 M solution of acetic acid (K a =1.76 x 10 -
ID: 508918 • Letter: A
Question
A student titrates 10.0mL of a 0.500 M solution of acetic acid (Ka =1.76 x 10-5) with 0.250 M NaOH.
a) 20.00 mL of NaOH were needed to reach the endpoint. What is the initial concentration of acetic acid ? [HA]0 =
b ) What is the initial pH of the acid? pH =
c ) What is the pH after 5.00 mL of NaOH have been added? pH =
d ) What is the pH after 10 .00 mL of NaOH have been added? pH=
e) What is the pH after 20.00 mL of NaOH have been added? pH=
f) What is the pH after 30.00 mL of NaOH have been added? pH=
g ) Which of the resulting solutions (b – f) , if any, are buffers? buffer ?
Explanation / Answer
a)
initial concentration of acetic acid = 0.500 M
b)
pKa = 4.75
pH = 1/2 (pKa - log C)
= 1/2 (4.75 - log 0.5)
pH = 2.53
c)
millimoles of acetic acid = 10 x 0.5 = 5
millimoles of NaOH = 5 x 0.250 = 1.25
CH3COOH + NaOH ----------------> CH3COO- + H2O
5 1.25 0
3.75 0 1.25
pH = pKa + log [salt / acid]
= 4.75 + log [1.25 / 3.75]
= 4.27
pH = 4.27
d)
millimoles of NaOH = 10 x 0.250 = 2.50
this is half equivalence point:
pH = pKa
pH = 4.75
e)
millimoles of NaOH = 20 x 0.250 = 5
this is equivalence point. here salt only remains.
salt concentration = 5 / (10 + 20) = 0.166 M
pH = 7 + 1/2 (pKa + log C)
= 7 + 1/2 (4.75 + log 0.166)
= 8.99
pH = 8.99
f)
millimoles of NaoH = 30 x 0.250 =7.5
[OH-] = 0.0625
pOH = 1.20
pH = 12.80
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