Save
...
Inorganic
Module 5
5.1.3~ acids, bases and buffers
Save
Share
Learn
Content
Leaderboard
Share
Learn
Created by
Vision
Visit profile
Cards (9)
pH of a buffer: weak acid + strong base
Work out the
moles
of
acid
and
base
identify if the
base
or
acid
is in
excess
limiting reagent
tells us the
mole
of the salt [A
−
^-
−
]
if the
base
is in
excess
work out the
moles unreacted
and the
new concentration
[
O
H
−
OH^-
O
H
−
]=
m
o
l
e
s
u
n
r
e
a
c
t
e
d
n
e
w
v
o
l
u
m
e
\frac{moles\ unreacted}{new\ volume\ }
n
e
w
v
o
l
u
m
e
m
o
l
es
u
n
re
a
c
t
e
d
[
H
+
]
\left[H^+\right]
[
H
+
]
=
K
w
[
O
H
−
]
\frac{Kw}{\left[OH^-\right]}
[
O
H
−
]
K
w
What do you then do if the acid is in excess
[
HA
]=
m
o
l
e
s
o
f
H
A
u
n
r
e
a
c
t
e
d
n
e
w
v
o
l
u
m
e
\frac{moles\ of\ HA\ unreacted\ }{new\ volume\ }
n
e
w
v
o
l
u
m
e
m
o
l
es
o
f
H
A
u
n
re
a
c
t
e
d
[A
−
^-
−
]=
m
o
l
e
s
o
f
s
a
l
t
f
o
r
m
e
d
[
A
−
]
n
e
w
v
o
l
u
m
e
\frac{moles\ of\ salt\ formed\ \left[A^-\right]}{new\ volume}
n
e
w
v
o
l
u
m
e
m
o
l
es
o
f
s
a
lt
f
or
m
e
d
[
A
−
]
then [
H
+
H^+
H
+
]=
K
a
[
H
A
]
[
A
−
]
\frac{Ka\left[HA\right]}{\left[A^-\right]}
[
A
−
]
K
a
[
H
A
]
then -
log
[
H
+
H^+
H
+
]
pH of a buffer: strong acid + strong base
work out the
moles
of the
acid
and the
base
identify if it is the
acid
or
base
that is in
excess
and work out the left over moles/moles
unreacted
if the base is in excess work out the new conc
new concentration=
m
o
l
e
s
u
n
r
e
a
c
t
e
d
t
o
t
a
l
v
o
l
u
m
e
\frac{moles\ unreacted}{total\ volume\ }
t
o
t
a
l
v
o
l
u
m
e
m
o
l
es
u
n
re
a
c
t
e
d
then
[
H
+
]
\left[H^+\right]
[
H
+
]
=
K
w
[
O
H
−
]
\frac{Kw}{\left[OH^-\right]}
[
O
H
−
]
K
w
What to you do when if the acid is in excess
new concentration=
m
o
l
e
s
u
n
r
e
a
c
t
e
d
t
o
t
a
l
v
o
l
u
m
e
\frac{moles\ unreacted}{total\ volume}
t
o
t
a
l
v
o
l
u
m
e
m
o
l
es
u
n
re
a
c
t
e
d
then -
log
[
H
+
]
\left[H^+\right]
[
H
+
]
calculating the pH of water
in
pure
water
[
H
+
]
=
\left[H^+\right]=
[
H
+
]
=
[
O
H
−
]
\left[OH^-\right]
[
O
H
−
]
therefore
Kw
=
[
H
+
]
2
\left[H^+\right]^2
[
H
+
]
2
therefore
[
H
+
]
\left[H^+\right]
[
H
+
]
=
K
w
\sqrt{Kw}
K
w
What is the effect of temperature on the pH of water and the neutrality of water
as the temperature
increase
the
equilibrium
moves to the right to lower the temperature
therefore
[
H
+
]
a
n
d
[
O
H
−
]
\left[H^+\right]and\left[OH^-\right]
[
H
+
]
an
d
[
O
H
−
]
increase
therefore Kw increases and the pH also
increase
however the water is still neutral as
[
H
+
]
=
\left[H^+\right]=
[
H
+
]
=
[
O
H
−
]
\left[OH^-\right]
[
O
H
−
]
How do you calculate the pH of a strong base
find
[
O
H
−
]
\left[OH^-\right]
[
O
H
−
]
concentration
then substitute into equation
[
H
+
]
\left[H^+\right]
[
H
+
]
=
K
w
[
O
H
−
]
\frac{Kw}{\left[OH^-\right]}
[
O
H
−
]
K
w
whereby the value of Kw at room temperature is
1
×
1
0
−
14
1\times10^{-14}
1
×
1
0
−
14
How do you calculate the pH of a diluted strong acid
[
H
+
]
\left[H^+\right]
[
H
+
]
=
[
H
+
]
o
l
d
\left[H^+\right]_{old}
[
H
+
]
o
l
d
×
\times
×
o
l
d
v
o
l
u
m
e
n
e
w
v
o
l
u
m
e
\frac{old\ volume}{new\ volume}
n
e
w
v
o
l
u
m
e
o
l
d
v
o
l
u
m
e
What is the relationship between the strength of an acid and its Ka and pKa value
the bigger the value of Ka the
stronger
the acid
the bigger the value of pKa the
weaker
the acid