Overview of Acid-Base Equilibria Questions -- KEY
The following questions are designed to give you an overview of the topics in Acids and Bases
and Equilibria in Acid-Base Solutions. Although not comprehensive, these few questions do a
good job of covering the majority of the concepts and calculations related to this subject area.
Please do your best to answer the questions completely. Keys will be posted.
Acid-Base Equilibria
1.
a. A typical weak acid, HA, is prepared in such a way as to have an initial concentration
in water of 0.100 M. At equilibrium the solution has pH = 2.87, what is Ka for this acid?
The molarity of the conjugate base and the molarity of the hydronium ion are known since the
pH is given (pH = 2.87 gives [A-] = [H+] = 10-pH = 1.35x10-3 M.
The reaction information is summarized as . . .
H3O+ (aq)
+ A- (aq)
----
~0M
0M
- 1.35x10-3 M
----
+ 1.35x10-3 M
+ 1.35x10-3 M
0.10 - 1.35x10-3
M
----
1.35x10-3 M
1.35x10-3 M
initial
change
equilibrium
HA (aq)
+ H2O (l)
0.10 M
<===>
As you can see, all the concentrations are known and it is a simple matter to solve for Ka.
Ka =
[H O ][A ] = (1.35x10 )(1.35x10 ) = 1.8x10
[HA]
(0.10 − 1.35x10 )
+
3
−
−3
−3
−5
−3
As you can plainly see the reaction only proceeded to a small extent in the forward
direction in order to reach equilibrium . . . so this is a WEAK acid!
b. The conjugate base of the above acid exists as a sodium salt, NaA. What is Kb for this
conjugate base.
Remember for conjugate acid base pairs (and only for pairs): Kw = Ka*Kb
So Kb = Kw/Ka = (1.00x10-14)/( 1.8x10-5) = 5.6x10-10
c. For a 0.100 M solution of the conjugate base in part b, what is the [OH-] at
equilibrium AND what is the pH?
NOTE: in this problem I referred to the base as B- . . . this is an inconsequential change.
This is a straightforward equilibrium problem. The changes in concentrations and the final
equilibrium concentrations will be expressed in terms of a single variable . . .
112
B- (aq)
initial
change
+ H2O (l)
0.10 M
-xM
equilibrium 0.10 - x M
<===> OH- (aq)
+ HB
(aq)
----
~0M
0M
----
+xM
+xM
----
xM
xM
The variable expressions of the concentrations can be plugged into the equilibrium constant
expression and this expression can be solved. This should all look very familiar by now!
Kb =
[OH ][HB] = (x)(x)
[B ] (0.10 − x)
−
−
5.6 x10 −10 ≈
x2
0.10
Giving x = [OH-] = [HB] = 7.5x10-6 M and the following equilibrium conditions
B- (aq)
+ H2O (l)
equilibrium ~0.10 M
<===>
OH- (aq)
-6
----
7.5x10 M
+ HB (aq)
7.5x10-6 M
And pOH = 5.12 along with pH = 14.00 - pOH = 8.88
d. For the same pair of weak acid and conjugate base in the questions above, what is the
pH of a buffer in which [HA] = [A-] = 0.25 M?
For a buffer system, pH = pK a
with pKa = -logKa = 4.74
[A ]
+ log
−
[HA]
When [HA] = [A-], then pH = pKa = 4.74
e. For the same pair of weak acid and conjugate base to form a buffer with pH = 5.00
when [HA] = 0.25, what does [A-] have to be?
In a buffer system, pH = pK a
[A ]
+ log
−
[HA]
and [HA] = 0.25 M
with pKa = -logKa = 4.74
Then for pH = 5.00, we find
which then solves to -0.342 = log [A-]
5.00 - 4.74 = log[A-]- log[0.25]
[A ] = 0.455 M
This makes good sense, b/c for pH > pKa, [A-] should be > [HA]
113
so . . .
f. When 4.00 L of the buffer in part e (the pH = 5.00 buffer) has 0.100 mole of HCl (g)
added, what is the new pH? HINT: should it be higher or lower than 5.00? Why?
The simplest solution is, pH = 4.74 + log
[4 * 0.455 − 0.100] = 4.93
[4 * 0.25 + 0.100]
Note: moles of A- were used instead of [A-] and moles of HA instead of [HA]. This is perfectly
acceptable since both species exist in the same 4.00 L volume. Also note that the answer
makes good sense b/c an addition of acid should decrease the pH
g. When 0.100 mole of HCl (g) is added to 4.0 L of pure water (pH = 7.00), what is the
new pH? What are [H+] and [OH-] and pOH? How does the pH change of the water
compare to that of the above buffer?
The added HCl creates a [H+] = 0.100 mole/4.0 L = 0.025 M
This gives: pH = 1.60, pOH = 12.40, and [OH-] = 4.0x10-13 M (check [H+][OH-] = Kw) OK
Note: water is not a good buffer!!
2.
a. 0.100 M NaOH(aq) is used to titrate 25.00 mL of 0.200 M HCl(aq). Calculate the pH
during this titration at the following volumes of added NaOH: 0.00 mL, 25.00 mL, 49.00
mL, the equivalence point, 51.00 mL, and 60.00 mL.
The balanced reaction:
HCl (aq) + NaOH (aq) Æ
NaCl (aq) + H2O(l)
Note: M = mole per liter may also be mmole per mL . . . which for problems like this one can save a great deal of effort
initial mmole HCl = (0.200 mmole/mL)(25.00 mL) = 5.00 mmole HCl
mmole NaOH added = (0.100 mmole/mL)(volume added in mL) = 3rd column
mL NaOH need to reach equiv. pt. = 5.00 mmole needed / 0.100 M = 50.0 mL NaOH sol’n
NOTE: at the equivalence point only NaCl is present in the solution and pH = 7.00
mmole excess H+ = mmole initial HCl – mmole added NaOH = 4th column
mmole excess OH- = mmole added NaOH – mmole initial HCl = 5th column
[H+] = mmole excess H+ / total volume = 6th column before equiv. pt.
[OH-] = mmole excess H+ / total volume = 6th column after equiv. pt.
pH = 14.00 - pOH
pH = -log [H+], note after the equivlance pt. I used
mL
NaOH
0.00
25.00
49.00
50.00
mmole H+
initial
5.00
5.00
5.00
5.00
mmole
OH- added
0.00
2.50
4.90
5.00
mmole H+
excess
5.00
2.50
0.10
0.00
mmole OHexcess
0.00
51.00
60.00
5.00
5.00
5.10
6.00
-
0.10
1.00
[H+], M
pH
2.00E-01
5.00E-02
1.35E-03
1.00E-07
[OH-], M
1.32E-03
1.18E-02
0.70
1.30
2.87
7.00
11.12
12.07
Note: see “sketch” of titration curve below . . . due to limitations of Excel, the shape to the curve is a little odd
114
titration curve of HCl with NaOH
14.00
12.00
pH
10.00
8.00
6.00
4.00
2.00
0.00
0.00
20.00
40.00
60.00
mL 0.100 M NaOH
b. 0.100 M NaOH(aq) is used to titrate 25.00 mL of 0.200 M acetic acid(aq).
[for acetic acid, Ka = 1.8 x 10-5] Calculate the pH during this titration at the following
volumes of added NaOH: 0.00 mL, 25.00 mL, 49.00 mL, the equivalence point, 51.00
mL, and 60.00 mL.
HOAc (aq) + NaOH (aq) Æ NaOAc (aq) + H2O(l)
The balanced reaction:
Notes:
OAc- represents the acetate ion and HOAc represents acetic acid
M = mole per liter may also be mmole per mL . . . which for problems like this one can save a great deal of effort
initial mmole HOAc = (0.200 mmole/mL)(25.00 mL) = 5.00 mmole HOAc
mmole NaOH added = (0.100 mmole/mL)(volume added in mL) = 3rd column
mL NaOH need to reach equiv. pt. = 5.00 mmole needed / 0.100 M = 50.0 mL NaOH sol’n
pH at 0.00 mL is found using an ICE table for 0.20 M acetic acid [Ka = 1.8x10-5]
[H+] = square root of Ka*0.20 = 1.9x10-3 M, pH = -log[H+] = 2.72
mL
NaOH
mmole
HOAc
initial
mmole
OHadded
mmole
HOAc
excess
mmole
OAcformed
[H+], M
pH
0.00
25.00
49.00
5.00
5.00
5.00
0.00
2.50
4.90
5.00
2.50
0.10
2.50
4.90
2.72
4.74
6.43
50.00
51.00
60.00
5.00
5.00
5.00
5.00
5.10
6.00
0.00
-
5.00
5.00
5.00
1.90E-03
1.82E-05
1.35E-03
[OH-], M
6.11E-06
1.32E-03
1.18E-02
115
8.79
11.12
12.07
at 25.00 mL pH = pKa because [HOAc] =[OAc-] since one half of the acid has been converted
into the conjugate base and
pH = pKa + log [base]/[acid], where base means the basic buffer species and acid is the acidic
buffer species
at 49.00 mL added NaOH the remaining HOAc = 0.100 mmole and the OAc- = 4.90 mmole
since
pH = pKa + log [base]/[acid] we find pH = 4.74 + log (4.90/0.10) = 6.43
Note that we can still use this equation even though we are outside of the “buffer range” because
our assumptions about the mmole of basic and acidic buffer species are valid
At the equivalence pt. we need to solve an ICE table for the acetate ion with Kb = 5.6x10-10 since
Kb = Kw/Ka
we find [OH-] = square root of 5.6x10-10*(5.00 mmole/75.00 mL) = 6.1x10-6 M
Using what should now be “standard practice” we find pH = 14.00 – pOH = 8.79
Beyond the equivalence point the pH is controlled by the excess NaOH added and the
calculations are the same as found in part (a) at 51.00 and 60.00 mL added NaOH.
Note: see “sketch” of titration curve below . . . due to limitations of Excel, the shape to the curve is a little odd, adding more data
point could improve the shape.
titration of HOAc with NaOH
14.00
12.00
pH
10.00
8.00
6.00
4.00
2.00
0.00
0.00
10.00
20.00
30.00
40.00
mL 0.100 M NaOH
116
50.00
60.00
c. Make rough sketches of the above titrations. Note any significant differences and/or
similarities. [ex. look at initial pH, pH at equiv. pt., vol. of NaOH at equiv. pt., etc.]
See (a) and (b) above –
similarities: since the stoichiometry of both titrations are 1:1 and the concentration and volumes
used are the same, the equivalence point volume of NaOH is the same. Also, since the
concentration of NaOH and volumes are the same, the pH beyond the equivalence pt. is
the same for both curves
differences: HCl is a strong acid and acetic acid is a weak acid, therefore, the pH of the HCl
curve is lower at the start of the titration. Also, the acetic acid titration curve passes
through a buffer region [centered on pH = pKa at the ½ equivalence point]
d. Explain how you might select an indicator for the above titrations? HINT: pKIn.
Use the Henderson-Hasselbalch equation to understand that we want to choose an indicator that
will change color over a pH range of about 2 pH units [for the same reason that buffers
are effective over about a 2 pH unit range]. Then select an acid-base indicator with a pKa
that falls on the steep part of the titration curve . . . ideally pKa of the indicator will
perfectly match the pH at the equivalence point. When this is not true, there can be a
difference between the added titrant volume at which the color change of the indicator is
observed (known as the “end-point”) vs. the volume at which the “equivalence point” is
reached. Our goal is to choose an indicator that will minimize this “end-point” error.
See the table or figure of indicators in your text:
For the HCl titration we might use chlorphenol red pKa ~ 6, bromthymol blue pKa ~ 6.8, or
phenol red pKa ~ 7.2 . . . but since the equivalence pt. region is VERY steep for a strong
acid-strong base titration, we can even get away with using phenolphthalein pKa ~8.8
and only have a small “end-pt” error
For the HOAc titration we will want to avoid the chlorphenol red, bromthymol blue, and phenol
red options listed above. Sticking with phenolphthalein for this titration is an excellent
idea [phenolphthalein is a very common choice of indicator for a weak acid-strong base
titration indicator since the pH at the equivalence point will be controlled by a weak base
in these tirations the equivalence point pH will be > 7 but also < 11]
Note, that for the next curve [weak base-strong acid titration] we will select an indicator that
changes color at a pH < 7 but > 3 since a weak acid will be present at the equivalence
point.
e. Based on what you have learned from the above, make a sketch of the titration curve
of 25.00 mL 0.200 M ammonia [Kb = 1.8 x 10-5] with 0.100 M HCl. You needn’t make
calculations, but do understand how you would do so if asked.
See the next page: no calcs. discussed, but a table of values is included.
117
mL HCl
mmole
NH3
initial
mmole
H+
added
mmole
NH3
excess
mmole
NH4+
formed
[OH-], M
pH
0.00
25.00
49.00
5.00
5.00
5.00
0.00
2.50
4.90
5.00
2.50
0.10
2.50
4.90
11.28
9.26
7.57
50.00
51.00
60.00
5.00
5.00
5.00
5.00
5.10
6.00
0.00
-
5.00
5.00
5.00
1.90E-03
5.50E-10
1.35E-03
[H+], M
6.11E-06
1.32E-03
1.18E-02
5.21
2.88
1.93
Note: see “sketch” of titration curve below . . . due to limitations of Excel, the shape to the curve is a little odd, adding more data
point could improve the shape.
titration of NH3 with HCl
12.00
10.00
pH
8.00
6.00
4.00
2.00
0.00
0.00
10.00
20.00
30.00
40.00
50.00
60.00
mL 0.100 M HCl
3.
Calculate the pH of the following two solutions: (i) 0.100 M H3PO4 and (ii) 0.100 M
Na3PO4. Also, find the concentrations of the following species in each of the two
solutions: H3PO4, H2PO4-, HPO4-2, and PO4-3. HINT: make sure you choose the
correct equilibrium constants at the correct times.
For H3PO4
Ka1 = 6.9 x 10-3
Ka2 = 6.2 x 10-8
Ka3 = 4.5 x 10-13
Acidic
species
H3PO4
H2PO41HPO42-
Relationship
Kw = Ka1*Kb3
Kw = Ka2*Kb2
Kw = Ka3*Kb1
Note: the best acid has the “least good” conjugate base
118
Basic
Conjugate
H2PO41HPO42PO43-
For Na3PO4
Kb3 = 6.9 x 10-3
Kb2 = 6.2 x 10-8
Kb1 = 4.5 x 1013
(i) First phosphoric acid, 0.100 M
H3PO4 (aq)
+ H2O (l)
<===>
H3O+ (aq)
+ H2PO41- (aq)
initial
0.10 M
----
~0M
0M
change
-xM
----
+xM
+xM
equilibrium
0.10 - x M
----
xM
xM
K a1
[H O ][H PO ] =
=
3
+
14
2
[HA]
[ x][ x]
= 6.6 x10 − 3
(0.10 − x )
Since this Ka is reasonably large we cannot simplify by the assumption that x is small . . . oh well
at least this is not on an exam! So we solve the quadratic equation and take the one root
and pH = 1.64
that is reasonable giving
x = [H3O+] = 2.3 x 10-2 M = [H2PO41-]
and [H3PO4] = 0.077 M
Note: this is 23% ionization so we certainly cannot avoid the quadratic!
Next we carry through these findings into the second equilibria . . .
H2PO41- (aq) + H2O (l)
<===>
H3O+ (aq)
+ HPO42- (aq)
initial
2.3 x 10-2 M
----
2.3 x 10-2 M
0M
change
-y M
----
+yM
+y M
----
2.3 x 10-2+ y M
yM
equilibrium 2.3 x 10-2 - y M
K a2 =
[H O ][HPO ] = [2.3x10
[H PO ] [2.3x10
3
+
4
−2
2-
14
2
+ y ][ y ]
−2
− y]
= 6.2 x10 − 8
This time we can assume that y is small because this Ka is small! When we do this we find that
y = [HPO42-] = Ka2 = 6.2 x 10-8 M and our assumption is valid and the pH = 1.64 is still correct
Next we carry through findings from steps one and two into the third equilibria . . .
HPO41- (aq)
+ PO43- (aq)
+ H2O (l)
<===>
H3O+ (aq)
initial
6.2 x 10-8 M
----
2.3 x 10-2 M
0M
change
-z M
----
+zM
+z M
----
2.3 x 10-2+ z M
zM
equilibrium 6.2 x 10-8 -z M
K a3 =
[H O ][PO ] = [2.3x10
[HPO ] [6.2x10
3
+
4
4
2-
3-
−2
+ z ][ z ]
−8
− z]
= 4.8 x10 −13
We can now VERY safely assume that z is small. This yields z = [PO43-] = 1.3 x 10-18 M and the
pH still is 1.64 !!!!!
119
(ii) For Na3PO4 0.100 M we have a very similar process starting with the phosphate and working
through the steps in the order Kb1, then Kb2, then Kb3 we will find . . .. .
from the first equilibrium
PO43- (aq)
+ H2O (l)
<===>
OH- (aq)
+ HPO42- (aq)
<===>
OH- (aq)
+ H2PO41- (aq)
solving the quadratic once again!!!!
[OH-] = [HPO42-] = x = 0.037 M
[PO43-] = 0.100 – x = 0.063 M
btw, this yields pOH = 1.43 giving pH = 12.57
from the second base ionization step we find
HPO42- (aq)
+ H2O (l)
[H2PO4-] = y = 1.6x10-7 M = Kb2
which is such a small change as to not alter the results found above and pH = 12.57 is still valid!
from the third base ionization step we find
H2PO41- (aq) + H2O (l)
<===>
OH- (aq)
+ H3PO4 (aq)
[H3PO4] = z = 6.5 x 10-18 M
which is once again such a tiny change that our prior assumptions are all OK and pH = 12.57 still
120