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Mr Chiasson
Advanced Chemistry 12
1
Unit D: Acids and Bases – Part 2
Amphoteric Substances
Were you surprised in the last section to see water being described as an acid? In the
ammonia reaction, water acted as an acid because it donated a proton (hydrogen ion) to
ammonia:
NH3 (g) + H2O (l)
NH41+(aq) + OH1– (aq)
Compare this to another reaction we looked at earlier when we saw how hydrochloric acid
acted as an acid by donating a proton to water:
HCl (g) + H2O (l) → H3O1+ (aq) + Cl1- (aq)
In this reaction, water is acting as a base because it accepts a proton from HCl.
Substances that can act as
an acid is one reaction and
as a base in another are called
amphoteric substances
Just to confuse you, the term
amphiprotic means the same thing
Here is another example of an amphoteric substance. In the first reaction the bisulfate ion,
HSO41– acts as a base. In the second reaction it acts as an acid:
1.
HSO41– + H3O1+
2.
HSO41– + OH–
H2SO4 + H2O
H2O + SO42-
HSO41– accepts a proton from H3O1+
HSO41– gives (donates) a proton to OH1 –
Polyprotic Acids
We have seen examples of acids that contain more than one hydrogen ion that can be lost.
Sulfuric acid, H2SO4, for example, has two hydrogen ions that it can give up. The first
hydrogen ion is released as:
H2SO4 (aq) → H1+ (aq) + HSO41- (aq)
The second hydrogen will be more difficult to remove because it must now be removed from a
negative ion, HSO41-.
HSO41- (aq) → H1+ (aq) + SO42- (aq)
Mr Chiasson
Advanced Chemistry 12
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For acids that can donate more than one hydrogen, it will always be easier to donate the first
H+ than the second. If there are more hydrogens to release, each H1+ is more difficult to
remove because of the increasingly positive charge of the rest of the molecule.
Acids that can donate more than one
hydrogen ion are called polyprotic
How many H1+ can citric acid, H3C6H5O7, release?
What do you think the equations describing the release of the H1+ will be? Make your
predictions before you check . . .
Citric acid can release three H1+. The reactions would be:
1.
H3C6H5O7 (aq) → H1+ (aq) + H2C6H5O71- (aq)
2.
H2C6H5O71- (aq) → H1+ (aq) + HC6H5O72- (aq)
3.
HC6H5O72- (aq) → H1+ (aq) + C6H5O73- (aq)
Strong & Weak Acids & Bases
You have undoubtedly heard of the pH scale before and know that it has something to do with
indicating how strong or weak an acid is. In this part of the unit we will work towards defining
acid and base strength in terms of pH, but there are several important steps along the way. It
will be important that you understand each step.
•
•
Strong electrolytes conduct electricity well because the compound produces many ions
in solution.
Weak electrolytes conduct electricity poorly because they produce few ions in solution.
These terms apply equally well to acids and bases which are, of course, electrolytic solutions:
Strong acids produce many H1+ ions
(or H3O1+ ions)
weak acids produce few H1+ ions
The stronger the acid,
the more H1+ ions are produced
AND
Strong bases produce many OH1- ions
Mr Chiasson
Advanced Chemistry 12
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weak bases produce few OH1- ions
The stronger the base,
the more OH1- ions are produced.
Strong acids and bases are essentially one-way reactions - the acid or base breaks down
completely to produce ions. At equilibrium there are very few reactants left (very low
concentration); only products - the ions.
Strong Acids
Strong Bases
HCl (aq) → H1+ (aq) + Cl1- (aq)
NaOH (aq) → Na1+ (aq) + OH1- (aq)
H2SO4 (aq) → H1+ (aq) + HSO41- (aq)
Mg(OH)2 (aq) → Mg2+ (aq) + 2OH1- (aq)
Weak acids and bases, however, do not ionize completely. For weak electrolytes, equilibrium
lies to the left side of the equation (the reactant side) and there will be few ions present. The
double arrow is commonly used to indicate the partial ionization of the solution. Some
examples:
Weak Acids
HC2H3O2 (aq)
HCHO2 (aq)
H1+ (aq) + C2H3O21- (aq)
H1+ (aq) + CHO21- (aq)
Weak Bases
NH3 (aq) + H2O (l)
NH2CH3 (aq) + H2O (l)
NH41+ (aq) + OH1- (aq)
NH3CH31+ (aq) + OH1- (aq)
It is important that you don't confuse acid and base strength will dilution.
Dilute and concentrated, you should remember, refer to the relative amounts of solute and
solvent in a solution. By contrast, acid and base strength specially refer to the concentration
of ions in the solution.
A strong acid such as HCl is still a strong acid (completely ionizes to produce many ions)
even when it is dilute (lots of water and relatively small amounts of HCl). Ethanoic acid is still
a weak acid even when it is concentrated.
Mr Chiasson
Advanced Chemistry 12
Ionization Constants: Ka, Kb, and Kw
As acid/base solutions are systems at equilibrium, we can write equilibrium constant
expressions for these systems. We'll return to the equilibrium constant, a concept we first
addressed in the ≡’m unit.
Ka and Kb
Here are several examples, including some that show water as a reactant. Pay attention to
the physical states and whether or not a particular substance is included in the equilibrium
expression. Also notice that we now identify the equilibrium constant as Ka for acids and Kb
for bases.
HCl (aq) → H+(aq) + Cl-(aq)
[H+] [Cl-]
Ka
=1.3×106
=
[HCl]
HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)
[H3O+] [Cl-]
Ka
=1.3×106
=
[HCl]
HCHO2 (aq) ↔ H+(aq) + CHO2-(aq)
[H+] [CHO2-]
Ka
=
= 1.8 × 10-4
[HCHO2]
Mg(OH)2 (aq) → Mg2+(aq) + 2 OH-(aq)
[Mg2+] [OH-]2
Kb
=
[Mg(OH)2]
NH3 (aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
[NH4+] [OH-]
Kb
=
= 1.8×10-5
[NH3]
Recall that the value of the equilibrium constant indicates which side of a reaction is favored.
When Keq is greater than 1 the product side is favored; when Keq is less than 1 the reactants
are favored.
4
Mr Chiasson
Advanced Chemistry 12
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So we now have a numerical way to indicate acid and base strength - Ka and Kb
A large value of Ka
means there are many
H+ ions in solution in other words, a strong acid
A large Kb indicates
many OH- ions a strong base
The Table of Acid and Base Strength gives Ka and Kb values for a number of acids and
bases. Ka values for very strong acids are typically not given as there is little use in
expressing them numerically. Be sure to print a copy of this table and keep it handy.
Kw
We usually do not think of water as producing ions, but that isn't the case. Water does ionize,
although not very well.
We can write the ionization equation for water in two ways (both mean essentially the same
thing). The equilibrium constant, Kw, can also be written for both equations as shown.
Notice that H2O does not appear in the Kw expression because it is a liquid, and you'll
remember that liquids and solids are not included in equilibrium constant expressions.
H2O(l) → H+(aq) + OH-(aq)
Kw
=
[H+] [OH-]
1.0 × 10-14
=
2 H2O(l) → H3O+(aq) + OH-(aq)
Kw
=
[H3O+] [OH-]
=
1.0 ×10-14
Values for Kw are given for 25°C.
Two more key items to note:
§
In pure water, the balanced equation tells us that the concentrations of H+ and
OH- will be equal to one another (this is what makes water neutral). Solving the
equations shown in the table above, we find that [H+] = [OH-] = 1.0×10-7
§
As long as temperature remains constant Kw is a constant - it's value will not
change.
Mr Chiasson
Advanced Chemistry 12
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The value of Kw is very small, meaning that very few ions are present. Most water remains
"intact" as H2O, and few ions form. Why, then, do we even mention it? For a very important
reason that we will examine in the next section.
Calculating [H+] and [OH–]
We can use Ka and Kb to calculate the concentrations of ions in acids and bases. This will be
crucial to determining pH later in this unit, so be sure to follow along carefully.
This is one of the most important sections of this unit mathematically, so be sure you
understand every step.
How you calculate ion concentrations
depends on whether you have a strong
acid (or base) or a weak acid (or base).
Calculating Ion Concentrations for Strong Acids & Bases
For strong acids and bases, the concentration of the ions can be readily calculated from the
balanced equation. Consider these examples carefully.
1.
Calculate the hydrogen ion concentration in a 0.050 M solution of hydrochloric acid.
Solution:
We know that HCl is a strong acid that ionizes completely in water (you should memorize
the list of strong acids). Begin by writing the balanced reaction:
HCl(aq) → H+(aq) + Cl-(aq)
From the balanced equation we see that 1 mole of HCl produces 1 mole of H+ (a 1:1
ratio), therefore the concentration of H+ will equal that of HCl.
Answer: [H+ ] = 0.050 M
2.
Also [Cl-] = 0.050 M
Calculate the hydroxide ion concentration in a 0.010 M solution of barium hydroxide,
Ba(OH)2. Barium hydroxide is a strong base.
Solution:
Always begin by writing a balanced equation:
Ba(OH)2 (aq) → Ba2+(aq) + 2 OH-(aq)
Mr Chiasson
Advanced Chemistry 12
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Since 2 moles of OH- are produced for every 1 mole of Ba(OH)2 , the concentration of OHwill be twice the concentration of Ba(OH)2 .
Also [Ba2+] = 0.010 M
Answer:
[OH-] = 2 × 0.010 = 0.020 M
Calculating Ion Concentrations for Weak Acids & Bases
Weak acids and bases require a much different approach to finding ion concentrations. Once
you know you have a weak acid or base, follow these steps in finding ion concentrations:
1. Write a balanced equation for the reaction
2. You will need to know the value of Ka or Kb - if it is not given in the question,
look it up.
3. Set up the equilibrium constant expression. You will know the value of Ka (or
Kb) and the concentration of the acid; you will be solving the equation for the
concentration of the ions.
Follow along with these examples very carefully!
1.
Calculate the hydrogen ion concentration in a 0.10 M acetic acid solution, HC2H3O2.
Ka for acetic acid, a weak acid, is 1.8 ×10-5.
Solution:
Begin by writing the balanced reaction:
HC2H3O2 (aq)
H+(aq) + C2H3O2-(aq)
The question gives us the concentration of the acid, HC2H3O2 (0.10 M).
We need to find the concentration of H+, which will also equal the concentration of
C2H3O2- (why?)
Because ionization is NOT complete because this is a weak acid , [H+] will NOT equal
[HC2H3O2]. Instead we must calculate it using the equilibrium constant expression.
Set up the
Ka equation:
[H+] [ C2H3O2-]
Ka
=
[HC2H3O2]
Mr Chiasson
Advanced Chemistry 12
Substitute values into the
equation. Let x equal the
unknowns
8
(x) (x)
1.8 ×10-5
=
0.10
x2
=
(1.8 ×10-5)(0.10)
x2
=
1.8 ×10-6
x
=
1.3×10-3
[H+]
=
1.3×10-3
[C2H3O2-]
=
1.3×10-3
Rearrange the equation
Take the square root
ANSWER
also
An important note - in our Ka equation we used a value of 0.10M for [HC2H3O2]. This isn't
quite correct. At equilibrium, the concentration of HC2H3O2 will actually be less:
initial [HC2H3O2]
=
subtract the lost [H+]
equilibrium [HC2H3O2]
0.100
- 0.013
=
0.087
Using this equilibrium concentration of [HC2H3O2] in our calculations above (instead of
0.10 M), however, does not change our answer significantly. Therefore for our
calculations we can safely ignore the negligible change in concentration of the acid.
2.
Calculate the hydroxide ion concentration, [OH-], in a 0.025 M solution of analine, C6H5NH2,
a weak base with Kb = 4.3×10-10
Solution:
This question will test some of your skills. Begin by writing a balanced equation. Since
analine is a base that doesn't contain the hydroxide ion, include H2O as a reactant. Also
remember that a base is a hydrogen acceptor and will gain an additional H+:
C6H5NH2 (aq) + H2O (l)
C6H5NH3+(aq) + OH-(aq)
As we did in the previous example, we now set up the Kb expression and solve for ion
concentrations. We see from the balanced equation that the ions have a 1:1 ratio, therefore
Mr Chiasson
Advanced Chemistry 12
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[OH-] will equal the [C6H5NH3+].
[C6H5NH3+] [OH- ]
Set up the Kb equation,
omitting liquid water:
Kb
=
[C6H5NH2]
Substitute values into the
equation. Let x equal the
unknowns
(x) (x)
4.3×10-10
=
0.025
x2
=
(4.3×10-10)(0.25)
x2
=
1.1×10-11
x
=
3.3×10-6
[OH-]
=
1.3×10-3 M
[C6H5NH3+]
=
1.3×10-3 M
Rearrange the equation
Take the square root
ANSWER
also
Finding [OH-] in Acids and [H+] in Bases
Remember Kw from the previous section? Now we learn why it is important.
When we need to determine ion concentrations of an acid, you should immediately realize
you will be finding the concentration of hydrogen ion (H+) and some anion (Cl- and C2H3O2- in
our acid examples above). The concentration of the anion normally doesn't concern us much.
When finding ion concentrations of bases, we determine the concentration of hydroxide ions,
OH-, and some cation (Ba2+ and C6H5NH3+ in our base examples above). Again, we aren't too
concerned with the cation concentrations.
Remember water?
H2O(l) → H+(aq) + OH-(aq)
Kw
=
[H+] [OH-]
=
1.0 × 10-14
When we make an acid solution, the hydrogen ion concentration will increase because we are
adding more H+ ions to those already present in water.
Mr Chiasson
Advanced Chemistry 12
10
Consider our strong acid example from above, in which [H+] = 0.05 M. If we add this to the
hydrogen ion concentration of pure water, 1.0×10-7, we get a total hydrogen ion concentration
of 0.0500001M. Clearly, the water adds little to the total and we can essentially ignore its
contribution.
[H+] from water:
1.0×10-7
[H+] from HCl
Total [H+]
=
0.000 000 1
+
0.05
0.050 001
Next, remember that equilibrium constants are A CONSTANT (as long as temperature does
not change). Thus the value of Kw will still have a value of 1.0 ×10-14 even though [H+] has
increased due to the presence of the acid. We can use this information to calculate the
concentration of hydroxide ions present in the aqueous solution:
Rearrange the equation
Kw
=
[OH- ]
=
[H+] [OH-]
Kw
[H+]
1.0 ×10-14
Substitute known values and solve for [OH- ]
-
[OH ]
=
0.05
[OH- ]
=
2.0 ×10-13