−
] = [H
2
PO
4
−
]
×
10
( pH  pK
a
)
= [ H
2
PO
4
−
]
×
10
(7.207.20 )
=
[ H
2
PO
4
−
]
And, since [HPO
4
2
−
]
+
[H
2
PO
4
−
]
=
0.05 M
[HPO
4
2
−
] = [H
2
PO
4
−
]
=
0.025 M
In 100ml we have 100ml
×
1L
1000 ml
×
0.025 M
=
0.0025
mole of each.
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Now, addition of 50 ml of 0.1M HCl accomplishes two things:
(1)
It dilutes the solution; and,
(2)
It introduces protons that will convert HPO
4
2
to H
2
PO
4

.
The moles of protons is given by:
50 ml
×
1L
1000ml
×
0.01M
= 0.0005 mole
Thus, 0.0005 mole of
HPO
4
2
−
will be converted to
H
2
PO
4
−
or
There will be 0.0025
 0.0005 mole HPO
4
2
−
and 0.0025
+
0.0005 mole H
2
PO
4
−
pH
=
pK
a
+
log
10
[HPO
4
2
−
]
[H
2
PO
4
−
]
= 7.2
+
log
10
0.0025
−
0.0005
0.0025
+
0.0005
=
7.02 (1)
And
[HPO
4
2
−
] =
(0.0025
−
0.0005) mole
100 ml
+ 50 ml
×
1000 ml
1 L
=
0.0133 M
[H
2
PO
4
−
] =
(0.0025
+
0.0005) mole
100 ml
+ 50 ml
×
1000
ml
1 L
=
0.0200 M
Note:
This amount of acid added to 100 ml of water gives pH = 2.5.
b.
If 50 ml of 0.01 M NaOH is added to 100 ml of 0.05 M phosphate buffer at pH 7.2,
what is the resultant pH?
What are the concentrations of H
2
P0
4

and HPO
4
2
in this
final solution?
Answer:
For NaOH, the same equations apply with one important difference: HPO
4
2
is
increased and H
2
PO
4

is decreased.
Thus, the fraction in the first equation (above) is inverted.
This gives pH = 7.38.
Note:
If added to water instead, this amount of NaOH would
result in a solution with pH = 11.5.