pH and pKa 1
K
a
= [H
3
O
+
] ×
[conjugate base]
[acid]
The solution in the titration flask is at equilibrium at all times during the titration.
The relationship of
K
a
to [H
3
O
+
] (and pH to p
K
a
) depends on the ratio [conjugate base]/[acid].
K
a
= [H
3
O
+
] ×
1
10
10
× K
a
= [H
3
O
+
]
pH = –log(10
×
K
a
)
pH = –log(10) – log
K
a
pH = pK
a
– 1
In the titration of a weak acid
at the beginning of the buffer region
[acid] is greater than [conjugate base]
when [acid] = 10 × [conjugate base]
Recall: log (a × b) = log a + log b
K
a
= [H
3
O
+
]
p
K
a
= pH
in the middle of the buffer region
[conjugate base] = [acid]
K
a
= [H
3
O
+
]
×
10
p
K
a
= pH –log(10)
pH = pK
a
+ 1
at end of the buffer region
[conjugate base] is greater than [acid]
when [conjugate base] = 10
×
[acid]
Dr Sheila Woodgate
© The University of Auckland 2002