Transcript Calculating pH of strong acids and bases
Slide 1
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 2
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 3
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 4
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 5
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 6
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 7
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 8
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 9
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 2
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 3
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 4
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 5
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 6
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 7
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 8
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.
Slide 9
Calculating pH of strong acids
and bases
Strong acids or bases are those which dissociate
completely.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
So in a 0.5 mol L-1 solution of HCl there will be 0.5 mol L-1
of H3O+ and 0.5 mol L-1 of Cl-.
Similarly, NaOH is a strong base because 0.1 mol L-1 of
NaOH(aq) contains 0.1 mol L-1 of Na+ and 0.1 mol L-1 of
OH-.
There are only four strong acids that we normally deal
with: HCl, HBr, HNO3 and H2SO4. All other acids should
be considered as weak acids unless you are told
otherwise.
Likewise, the only strong bases you are likely to meet are
NaOH, KOH and Ca(OH)2.
Weak acids or bases differ from strong ones in that they
only partially dissociate. Their reactions with water are
equilibrium reactions.
Ethanoic acid (in vinegar) is a weak acid. Most ethanoic
acid molecules remain as molecules in water: only a few
react to form CH3COO- and H3O+.
Later on in this unit you will find out how to calculate
exactly how much that ‘most’ is.
Example 1: Calculating the pH of a strong acid.
Calculate the pH of a 0.0672 mol L-1 solution of nitric acid.
Since nitric acid is a strong acid:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
Remember that [x]
means ‘the
concentration of x’.
c (HNO3) = 0.0672 mol L-1 so [H3O+] = 0.0672 mol L-1
pH = -log[H3O+]
= -log(0.0672)
= 1.17
Do this problem on
your calculator and
check that you can
get this answer.
Example 2: Calculating the pH of a strong acid.
Calculate the pH of a 2.41 × 10-3 mol L-1 solution of sulfuric
acid.
Each molecule of H2SO4 will donate two protons:
H2SO4(aq) + H2O(l) → 2H3O+(aq) + SO42-(aq)
Therefore c (H2SO4) = 2.41 × 10-3 mol L-1
But [H3O+] = 2 × 2.41 × 10-3 mol L-1
= 4.82 × 10-3 mol L-1
pH = -log[H3O+]
= -log(4.82 × 10-3)
= 2.32
Do this problem on
your calculator and
check that you can
get this answer.
Example 3: Calculating the pH of a strong base.
Calculate the pH of a 0.00250 mol L-1 solution of potassium
hydroxide.
KOH is a strong base, therefore
c (KOH) = 0.00250 mol L-1 = [OH-]
Before we can calculate the pH we need to know [H3O+].
In any aqueous solution at room temperature
[H3O+][OH-] = 10-14
Therefore
So for our 0.00250 mol L-1 solution of potassium
hydroxide:
[OH-] = 0.00250 mol L-1
An alternative method for calculating pH from [OH–]
This method is shorter than converting [OH–] to [H3O+], but
may be harder to understand.
Just as we can write
pH = –log[H3O+]
so we can write
pOH = –log[OH–]
And since
1 10–14 = [H3O+][OH–]
–log(1 10–14 ) = (–log[H3O+]) + (–log[OH–])
14 = pH + pOH
If you don’t understand the mathematics of logs, don’t
worry – just learn the bottom formula.
Calculate the pH of a 0.00250 mol L-1 solution of
potassium hydroxide.