100 mL of 0.02 M benzoic acid (pKa = 4.2) is titrated using 0.02 M NaOH. pH after 50 mL and 100 mL of NaOH have been added are
Answers
Answer:
Given 100ml of 0.02M C
6
H
5
COOH with pK
a
=4.2
So, K
a
=10
−4.2
=6.31×10
−5
It is titrated using 0.02M NaOH.
(i) After addition of 50ml of NaOH:
Number of moles of C
6
H
5
COOH=100×0.02=2mmol
Number of moles of NaOH=50×0.02=1mmol
Here after the addition of 50ml of NaOH,we get basic Buffer solution C
6
H
5
COONa and C
6
H
5
COOH.
Number of moles of C
6
H
5
COOH=2−1=1mmol
Number of moles of C
6
H
5
COONa=1mmol
pH=pK
a
+log
acid
salt
=4.2+log
1
1
pH=4.2
(ii) After addition of 100ml of NaOH:
Number of moles of C
6
H
5
COOH=100×0.02=2mmol
Number of moles of NaOH=100×0.02=2mmol
So, Number of moles of C
6
H
5
COONa=2mmol
Therefore the given point is equivalent point. The salt C
6
H
5
COONa is weak base, so we will find basic ionization constant.
Now,
K
b
=
K
a
K
w
=
6.31×10
−5
10
−14
=1.58×10
−10
Concentration of [C
6
H
5
COONa]=
Totalvolumeofthesolution
NumberofmolesofC
6
H
5
COONa
=
200
2
=0.01
[OH
−
]=
K
b
×[C
6
H
5
COONa]
=
(1.58×10
−10
)×0.01
=1.257×10
−7
M
pOH=−log[OH
−
]=−log(1.257×10
−7
)=6.9
pH=14−pOH=14−6.9≈7.0