Here’s why you should know this. 38.
collected
the
of aof20.0
38. Data
Data collected
during during
the titration
of atitration
20.0 mL sample
a 0.10mL
M solution of
a monoprotic acid with a solution of NaOH of unknown concentration are plotted
sample of a 0.10 M solution of a monoprotic acid
in the graph above. Based on the data, which of the following are the approximate
pKawith
of the acid
and the molar
the NaOH?
a solution
of concentration
NaOH ofofunknown
concentration
areyou
plotted
the graph
pKa [NaOH]
First,
know thein[NaOH]
= 0.10 Mabove.
because at the
point, the same volume (20.0 mL) of NaOH
onM the equivalence
data,
which
of ofthe
following
A Based
4.7 0.050
is used as the volume
acid,
which you’veare
beenthe
told is
0.10
it’s important
to remember
pKaM.ofSecond,
the acid
and the
molar that at the
B approximate
4.7 0.10 M
halfway point of the titration, the pH = pKa of the weak
of the
acid.
At theNaOH?
halfway point, [HA] = [A-]. Plugging that
C concentration
9.3 0.050 M
D 9.3
0.10 M
pK
(A)
(B)
(C)
a
4.7
4.7
9.3
in to the Ka formula, you’ll see that [H+] = Ka, therefore
at the halfway
point, the pKa = pH
[NaOH]
0.050 M
0.10 M
0.050 M