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Chemistry
Acids and Bases
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Cards (19)
Strong Acid
−
log
10
[
H
+
]
=
-\log_{10}\left[H^+\right]=
−
lo
g
10
[
H
+
]
=
p
H
pH
p
H
Weak Acid
H
A
⇔
H
+
HA\Leftrightarrow H^+
H
A
⇔
H
+
+
+
+
A
−
A^-
A
−
K
a
=
Ka\ =
K
a
=
[
H
+
]
[
A
−
]
[
H
A
]
\ \frac{\left[H^+\right]\left[A^-\right]}{\left[HA\right]}
[
H
A
]
[
H
+
]
[
A
−
]
[
H
+
]
=
\left[H^+\right]=
[
H
+
]
=
[
A
−
]
\left[A^-\right]
[
A
−
]
K
a
=
Ka\ =
K
a
=
[
H
+
]
2
[
H
A
]
\ \frac{\left[H^+\right]^2}{\left[HA\right]}
[
H
A
]
[
H
+
]
2
[
H
+
]
=
\left[H^+\right]=
[
H
+
]
=
K
a
[
H
A
]
\sqrt{Ka\left[HA\right]}
K
a
[
H
A
]
−
log
10
[
H
+
]
=
-\log_{10}\left[H^+\right]=
−
lo
g
10
[
H
+
]
=
p
H
pH
p
H
Strong base
K
w
=
Kw\ =
K
w
=
[
H
+
]
[
O
H
−
]
\ \left[H^+\right]\left[OH^-\right]
[
H
+
]
[
O
H
−
]
[
H
+
]
=
\left[H^+\right]=
[
H
+
]
=
K
w
[
O
H
−
]
\frac{Kw}{\left[OH^-\right]}
[
O
H
−
]
K
w
−
log
10
[
H
+
]
=
-\log_{10}\left[H^+\right]=
−
lo
g
10
[
H
+
]
=
p
H
pH
p
H
BUFFER
: Weak acid + Salt
K
a
=
Ka\ =
K
a
=
[
H
+
]
[
A
−
]
[
H
A
]
\ \frac{\left[H^+\right]\left[A^-\right]}{\left[HA\right]}
[
H
A
]
[
H
+
]
[
A
−
]
[
H
+
]
=
\left[H^+\right]=
[
H
+
]
=
K
a
[
H
A
]
[
A
−
]
\frac{Ka\left[HA\right]}{\left[A^-\right]}
[
A
−
]
K
a
[
H
A
]
−
log
10
[
H
+
]
=
-\log_{10}\left[H^+\right]=
−
lo
g
10
[
H
+
]
=
p
H
pH
p
H
BUFFER:
weak acid + strong base
H
A
+
HA+
H
A
+
O
H
−
→
H
2
O
+
OH^-\rightarrow H_2O+
O
H
−
→
H
2
O
+
A
−
A^-
A
−
K
a
=
Ka\ =
K
a
=
[
H
+
]
[
A
−
]
[
H
A
]
\ \frac{\left[H^+\right]\left[A^-\right]}{\left[HA\right]}
[
H
A
]
[
H
+
]
[
A
−
]
[
H
+
]
=
\left[H^+\right]=
[
H
+
]
=
K
a
[
H
A
]
[
A
−
]
\frac{Ka\left[HA\right]}{\left[A^-\right]}
[
A
−
]
K
a
[
H
A
]
−
log
10
[
H
+
]
=
-\log_{10}\left[H^+\right]=
−
lo
g
10
[
H
+
]
=
p
H
pH
p
H
acid - definition
proton
donor
base - definition
proton acceptor
strong acid
completely
dissociates
to ions in solution
pH
0-1
weak acid
slightly
dissociates
in solution
pH
3-5
pH =
−
log
10
[
H
+
]
-\log_{10}\left[H^+\right]
−
lo
g
10
[
H
+
]
[
H
+
]
\left[H^+\right]
[
H
+
]
=
1
0
−
p
H
10^{-pH}
1
0
−
p
H
pure water
H
+
H^+
H
+
+
+
+
O
H
−
⇔
H
2
O
OH^-\ \Leftrightarrow\ H_2O
O
H
−
⇔
H
2
O
equilibrium of water
forwards reaction is
endothermic
favoured when temperature of water is increased
more
H
+
H^+
H
+
produced - meaning water becomes more
acidic
as temperature increases
pKa =
−
log
10
K
a
-\log_{10}Ka
−
lo
g
10
K
a
Ka =
1
0
−
p
K
a
10^{-pKa}
1
0
−
p
K
a
methyl orange
used in reactions with a more
acidic
neutralisation
point
orange in acids - turns yellow at neutralisation point
Phenolphthalein
used in reaction with a more basic neutralisation point
pink in alkalis - turns colourless at neutralisation point
Buffer calculations - Acid+base
find number of
moles
in each species
calculate
concentration
at
equilibrium
using total volume
use
Ka
to find
[
H
+
]
\left[H^+\right]
[
H
+
]
and
pH
Buffer calculations - Acid+salt
find
moles
of salt
use
Ka
to find
pH
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