Transcript Slide 1

Acids and bases
Revision of Year 12
• Acids are proton
donors (the reactant
loses an H in becoming
the product)
• Bases are proton
acceptors
• Acids an bases form
conjugate pairs
• Strong acids
dissociate completely
• Weak acids have
partial dissociation
pH = -log10[H3O+]
[H3O+] = 10-pH
[H3O+] = 1 x 10-14
[OH-]
[OH-] = 1 x 10-14
[H3O+]
An amphiprotic
substance can do
either!
Revision of Year 12
Identify the acids and bases by drawing a circle
around each acid and a square around each base.
a) HNO3 + H2O → H3O+ + NO3b) F- + H2O → HF + OHc) HS- + CH3COOH → CH3COO- + H2S
d) HCN + OH- → CN- + H2O
Revision of Year 12
Write equations to prove that the following are acids:
a) NH4+ + H2O ⇌ NH3 + H3O+
b) HCO3- + H2O ⇌ CO32- + H3O+
Write equations to prove that the following are bases:
a) CO32- + H2O ⇌ HCO3- + OHb) CH3COO- + H2O ⇌ CH3COOH + OH-
c) HCO3- + H2O ⇌ H2CO3 + OH-
a) Calculate the pH of a solution whose hydrogen ion
concentration is 8.85 x 10-2 molL-1.
pH = -log10[8.85 x 10-2]
pH = 1.06
b) What is the concentration of a hydrochloric acid
solution whose pH is 2.71?
[H3O+] = 10-2.71
[H3O+] = 1.95 x 10-3 molL-1
c) What is the [OH-] of a solution whose pH is 8.7?
[H3O+]
=
10-8.7
[H3O+] = 1.995 x 10-9 molL-1
[OH-] = 1 x 10-14
[H3O+]
[OH-] = 5.01 x 10-6 molL-1
d) Calculate the pH of a solution of KOH whose
concentration is 0.200 molL-1.
[H3O+] = 1 x 10-14
pH = -log10[5 x 10-14]
0.200
pH = 13.3
Place the substances in the correct boxes:
Strong acid
Strong base
Weak acid
Weak base
H2SO4
CH3COOH
NH4+
HCl
NaOH
NH3
Salts
Mixing an acid with a base produces a salt. The pH
of the salt is dependent on the relative strength of
the acid and base.
For weak acids
we
K becomes
ignore the
Ka.
water
Theasacid
it is
constant
dissociation
in
aqueous
constant
solutions
Ka and pKa
Write the equilibrium expression for the following
weak acid:
CH3COOH(aq) + H2O(l) ⇌ H3O+ + CH3COO-(aq)
+][CH
-]+][CHCOO
K
O
Ka==[H
[H
O
COO
]
33
33
[H[CH
2O][CH
3COOH]
3COOH]
The bigger the Ka,
the stronger the acid!
Strong acid
Weak acid
pH = -log10[H3O+]
pKa = -log10Ka
[H3O+] = 10-pH
Ka = 10-pKa
Dissociates completely
Equilibrium, so need Ka
The larger the pKa the weaker the acid
Calculating pH using Ka
Find the pH of 0.10 molL-1 acetic acid solution.
Ka(CH3COOH) = 1.74 x 10-5
1. Write equilibrium equation
CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)
2. Write expression
Ka = [CH3COO-][H3O+]
[CH3COOH]
3. Make assumptions
[H3O+] = [CH3COO-]
[CH3COOH] = 0.10molL-1
4. Put the assumptions into the equation and rearrange
to calculate [H3O+]
1.74 x 10-5 = [H3O+]2
√1.74 x 10-5 x 0.10 = [H3O+]
1.319 x 10-3 molL-1 = [H3O+]
0.10
+]
pH
=
-log
[H
O
5.Calculate pH
10
3
pH = 2.88
Kb and pKb
Kb is the base
dissociation constant
Ka x Kb = Kw
Therefore: Kb = 1 x10-14
Ka
In a question, decide
whether you write an
expression for Ka or
Kb. OH- as a product
means Kb. H3O+ means
Ka
Calculating pH using Ka
Find the pH of 2.00 molL-1 NH3 if Ka(NH4+) = 5.75 x
10-10
1. Write equilibrium equation
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
3. Calculate Kb Kb = 1 x10-14
5.75 x 10-10
2. Write expression
Kb = [NH4+][OH-]
[NH3]
= 1.74 x 10-5
3. Make assumptions
[OH-] = [NH4+]
[NH3] = 2.00molL-1
4. Put the assumptions into the equation and rearrange
to calculate [OH-]
1.74 x 10-5 = [OH-]2
√1.74 x 10-5 x 2.00 = [OH-]
2.00
5.90 x 10-3 molL-1 = [OH-]
5.Convert to [H3O+] then calculate pH
+]
pH
=
-log
[H
O
+
-14
10
3
[H3O ] = 1 x10
pH = 11.8
5.90 x 10-3 = 1.69 x 10-12 molL-1
Calculating the pH of acidic and basic salts
Everything is the same except you must show how the
salt dissolves into ions first before writing the
equation and expression.
e.g.
NH4NO3(s) → NH4+(aq) + NO3-(aq)
e.g.
CH3COONa(s) → CH3COO-(aq) + Na+(aq)
In summary
For acid (salt):
Write equation & expression
Make assumptions
Calculate [H3O+]
Calculate pH
For base (salt):
Write equation & expression
Calculate Kb from Ka
Make assumptions
Calculate [OH-]
Convert to [H3O+]
Calculate pH
Species in solution
Product of
[H3O+] and [OH-]
always equals
You may be asked to order the concentrations
of ions
10-14
within a solution
Ionic substances whose
ions do not react with
water
Solutions will be neutral
[H3O+] = [OH-] = 10-7
Strong base
Work out [OH-] from
formula in standard
form
[H3O+] = 10-14 - above
Acidic or basic salts
pH will be provided in question to work
out [H3O+]
Write dissolving eq.
Write how ion reacts with water
[OH-] = 10-14 – above
Species in solution
0.1 molL-1 CaCl2
[Ca2+] = 0.1
[Cl-] = 0.2
[H3O+] = [OH-] = 10-7
Use the formula
to determine the
[OH-]
Ions will not
react with water
so therefore
neutral
0.01 molL-1 NaOH
NaOH → Na+ + OH[Na+]=[OH-] = 0.01 or 10-2
Write
the
+
-12
So [H3O ] = 10
equation for
dissolving then
the ion reacting
with water
1.0 molL-1 NH4Cl, pH5
NH4Cl(s) → NH4+(aq) + Cl-(aq)
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
[Cl-] = 1.0
[H3O+] = 10-5 so [NH3] = 10-5
[OH-] = 10-9
[NH4+] = 1.0 – 10-5
HCl NH4Cl NaCl NH3
HCl
NH4Cl
Pick out the strong
acid and base.
Pick out the acidic
Remember these go
salt then the weak
to completion when
base
writing equations
NaOH
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
Strong acid, fully dissociated give a high conc H3O+ ions
resulting in a low pH
NH4Cl(s) → NH4+(aq) + Cl-(aq)
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Dissolves to give solution of NH4+ and Cl- ions. NH4+ ions are weakly acidic
and partially dissociate in water to give a small increase in H3O+ conc.
NaCl
Dissolves to give Na+ and Cl- that don’t react with water. Neutral
NH3
NH3(aq) + H2O⇌ OH-(aq) + NH4+(aq)
NaOH
NaOH(aq) → OH-(aq) + Na+(aq)
Weak base that reacts with water to give a small increase in OH- ions
Strong base, fully dissociated gives a high conc. OHions resulting in a high pH
Don’t be scared!
Use the formula as
given for the acid
and
write equation
+
HPab(aq) + H2O(l) ⇌ Pab-(aq) + H3O (aq)
[H3O+][Pab-]
[Hpab]
What equation links
H3O+ with pH?
[H3O+] = 10-pH
[H3O+] = 10-3.22
[H3O+] = 6.03 x 10-4 molL-1
HPab
Volume
Conc.
Moles
NaOH
20 x 10-3 L 12 x 10-3 L
0.0300
molL-1
0.05 molL-1
6 x 10-4 mol 6 x 10-4 mol
n = cv
n = 0.05 x (12 x 10-3)
n/v = c
6 x 10-4/20 x 10-3
= 0.03 molL-1
[H3O+] = 6.03 x 10-4 molL-1
[HPab] = 0.0300 molL-1
Ka = [H3O+][Pab-]
[HPab]
Ka = [H3O+] = [Pab-]
[HPab] = 0.0300 molL-1
Ka = (6.03 x 10-4)2
0.0300
= 1.21 x 10-5
Look back over the
question to see +
You have [H3O ]
what
You
Youyou
have
Make
have
have
[HPab]
the
yourKa
How
assumptions
expression
do we get
and
from
Ka
insert
to pKa?
pKa = -log10Ka
pKa = -log101.21 x 10-5
= 4.92
Amines are weak
bases...
CH3NH2(aq) + H2O⇌ OH-(aq) + CH3NH3+(aq)
NH4Cl(s) → NH4+(aq) + Cl-(aq)
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
[CH3NH2] > [CH3NH3+]
= [OH-] > [H3O+]
[NH4+] would
have been
[NH4+] would- be
theWeak
same bases
as [Cl only
] at +
In
greater
a
weak
acid,
than
its
[Has
Hmmmm...no
info.
3O
-] ]
[NH
the
same
3] is
Weak
Use
both
base
equations.
so
[OH
start,
but
at
+
+
dissociate
-]
[NH4Cl] = [Cl ] > [NH4 ] > [NH3] = partially
]have
>
[OH-]
+as
is[H
conjugate
greater
than
[OH
We’ll
toweak
use+ so
3O
[H
O
]
must
Use
[OH
first
be
] =3greater
[CH
formula
than
equilibrium,
some
would
3NH
3 ]to
it
must
be
greater
acids
partially
theonly
equations
predict
[H
] NH3
be changed
to+Cl3O
than
conjugate
dissociate
logically
Ions that move in
an electric
With water
field
Aminomethane is a weak base so only partially
dissociates in water. This means that there are
only a few ions free to move in solution, hence, a
weak electrolyte .
Greater than 7
Must be a weak
Salt
of athe
weak
acid
Show
equation
base
Show the equation
to show the salt
with water only one
dissolves
works!
Na3Cit(s) → 3Na+(aq) + Cit3-(aq)
Cit3-(aq) + H2O(l) ⇌ OH- + HCit2Cit3-(aq) + H2O(l) ⇌ H3O+ + ?!
The salt of a weak acid is a weak base. This
anion will accept a proton from water. The OHconcentration will increase causing an increase in
pH.
So the NH4+
What
Write happens
equilibrium
when
increases...what
What effect will
equation
NH4Cl dissolves?
of NH3
effect
this will
have
this
on have
the
Show +equation
on the
H3O
equilibrium
conc?
NH3(aq) + H2O(l) ⇌ OH- + NH4+
NH4Cl(s) → NH4+(aq) + Cl-(aq)
As the NH4Cl dissolves, the NH4+ concentration increases
This causes the equilibrium to shift to the left, which in
turn reduces the OH- concentration
This means that the H3O+ concentration will increase
causing the pH to decrease