Titration Problems
1973
A sample of 40.0 milliliters of a 0.100 molar HC2H3O2 solution is
titrated with a 0.150 molar NaOH solution. K a for acetic acid =
1.8x10-5
= MbVb = (0.0274 L)(0.135 M) = 3.70x10-3 mol
HA
molec .wt . 
mass HA

mol HA
0 . 68 2 g
3 . 70 1 0
3
 18 4
g/mol
mol
(b)
(c) HA + OH- --> A- + H2O
(a) What volume of NaOH is used in the titration in
initial:
0.00370 mol
order to reach the equivalence point?
added:
(0.0106L)(0.135M) = 0.00143 mole
(b) What is the molar concentration of C2H3O2- at the
remaining: (0.00370 - 0.00143) = 0.00227 mol
equivalence point?
(c) What is the pH of the solution at the equivalence (d) pH = -log[H3O+]; [H3O+] = 10-pH = 10-5.65
point?
= 2.2x10-6M
Answer:


6
[ H 3 O ][ A ]
( 2 . 2 10
) ( 0 . 00 143 / v )
K a 

(a) MaVa=MbVb
[ HA ]
( 0 . 00 227 / v )
(e)
(0.100M)(40.0 mL) = (0.150M)(Vb)
= 1.4x10-6
Vb = 26.7 mL
1979 B
(b) acetate ion is a weak base with
A solution of hydrochloric acid has a density of 1.15 grams per
milliliter and is 30.0% by weight HCl.
Kb=Kw/Ka = 1.010-14/1.810-5 = 5.610-10
4 . 00 mm ol

[ CH 3 COO ] o 
 0 . 06 00 M (a) What is the molarity of this solution of HCl?
( 40 . 0 mL  26 . 7 mL )
(b) What volume of this solution should be taken in
[CH3COO-]eq = 0.600M -X
order to prepare 5.0 liters of 0.20 molar
hydrochloric acid by dilution with water?
[OH-] = [CH3COOH] = X
2
(c) In order to obtain a precise concentration, the 0.20
X
10
5
5 . 6 1 0

; X  9 . 66 1 0
M
molar hydrochloric acid is standardized against
( 0 . 06 00  X )
pure HgO (molecular weight = 216.59) by
0.0600M-9.6610-5M = 0.0599M [CH3COO-]eq
titrating the OH- produced according to the
14
Kw
1 . 0 1 0
following quantitative reaction.

10
[ H ] 
 1 . 04 1 0
M
 
5
[ OH ]
9 . 66 1 0
HgO(s) + 4 I- + H2O --> HgI42- + 2 OH(c)
In a typical experiment 0.7147 grams of HgO
pH = -log [H+] = -log(1.0410-10) = 9.98
required 31.67 milliliters of the hydrochloric acid
1978 A
solution for titration. Based on these data what is
A 0.682 gram sample of an unknown weak monoprotic organic
the molarity of the HCl solution expressed to four
acid, HA was dissolved in sufficient water to make 50 milliliters
of solution and was titrated with a 0.135 molar NaOH solution.
significant figures.
After the addition of 10.6 milliliters of base, a pH of 5.65 was Answer:
recorded. The equivalence point (end point) was reached after the
addition of 27.4 milliliters of the 0.135 molar NaOH.
1 . 15 g
100 0 mL


30 . 0 gHCl

1 mol
 9 . 5 M
1L
100 g
35 . 5 g
(a) Calculate the number of moles of acid in the (a) 1 mL
original sample.
(b) MfVf = MiVi
(b) Calculate the molecular weight of the acid HA.
(0.20M)(5.0L) = (9.5M)(V)
(c) Calculate the number of moles of unreacted HA
V = 0.11 L
remaining in solution when the pH was 5.65.
0 . 714 7 g
 0 . 003 300 mol Hg O
216 . 59 g mol
(d) Calculate the [H3O+] at pH = 5.65
(c)
(e) Calculate the value of the ionization constant, Ka,
mol OH- prod. = 2 (mol HgO) = 0.006600 mol
of the acid HA.
mol HCl req. = mol OH- prod. = 0.006600 mol
Answer:
0 . 00 6600 m o l
(a) at equivalence point, moles HA = moles NaOH
M HCl 
 0 . 20 84 M
0 . 03 167 L
The molecular weight of a monoprotic acid HX was to be
A solution of barium hydroxide is titrated with 0.1-M sulfuric determined. A sample of 15.126 grams of HX was dissolved in
acid and the electrical conductivity of the solution is measured as distilled water and the volume brought to exactly 250.00
the titration proceeds. The data obtained are plotted on the graph millilitres in a volumetric flask. Several 50.00 millilitre portions
of this solution were titrated against NaOH solution, requiring an
below.
average of 38.21 millilitres of NaOH.
1982 D
The NaOH solution was standardized against oxalic acid
dihydrate, H2C2O4.2H2O (molecular weight: 126.066 gram mol-1).
The volume of NaOH solution required to neutralize 1.2596
grams of oxalic acid dihydrate was 41.24 millilitres.
C ond uc tiv ity,

10
20
30
40
50
60
70
80
(a) Calculate the molarity of the NaOH solution.
(b) Calculate the number of moles of HX in a 50.00
millilitre portion used for titration.
(c) Calculate the molecular weight of HX.
(d) Discuss the effect of the calculated molecular
weight of HX if the sample of oxalic acid
dihydrate contained a nonacidic impurity.
Answer:
1.2 596 g
Millilitres of 0.1-M H2SO4
(a) For the reaction that occurs during the titration
described above, write a balanced net ionic
equation.
(b) Explain why the conductivity decreases, passes
through a minimum, and then increases as the
volume of H2SO4 added to the barium hydroxide is
increased.
(c) Calculate the number of moles of barium
hydroxide originally present in the solution that is
titrated.
(d) Explain why the conductivity does not fall to zero
at the equivalence point of this titration.
Answer:
(a) Ba2+ + 2 OH- + 2 H+ + SO42- --> BaSO4(s) + 2 H2O
(b) The initial conductivity is high because of the
presence of Ba2+ and OH- ions. The conductivity
decreases because Ba2+ forms insoluble BaSO4
with the addition of SO42-. The conductivity also
decreases because OH- combines with the addition
of H+ ions by forming H2O.
Beyond the equivalence point conductivity
increases as H+ and SO42- ions are added.
(c) # mol Ba(OH)2 = # mol H2SO4
12 6.066
(a) mol H2C2O4.2H2O =
=
-3
9.9916x10 mol
H2C2O4 + 2 NaOH --> Na2C2O4 + 2 H2O
9 . 99 16 1 0
M NaOH
3
2 m o lNa OH
2
m o l 
 1 . 99 83 1 0 m o l
1 m o lH 2 C 2 O 4

1. 99 83 1 0
2
mo l
 0 . 48 46 M
0 . 04 124 L
(b) mol HX = mol NaOH
0.03821 L x 0.4846 M = 0.01852 mol HX
0 . 01 852 m o l
(c)
50 . 00 m L
M W 
25 0 . 00 m L  0 . 09 260 m o l HX
15 . 12 6 g
0 . 09 260 m o l
 16 3 . 3
g
mo l
(d) The calculated molecular weight is smaller than
true value, because:
measured g H2C2O4 is larger than true value,
calculated mol H2C2O4 is larger than true value,
calculated mol NaOH is larger than true value,
calculated M NaOH is larger than true value
calculated mol HX is larger than true value,
therefore,
g HX ( tr ue v alu e )
MW 
=0.1M  0.04L = 0.004 mol
mol HX ( c a lc ulate d , an d too lar ge )
(d) BaSO4(s) dissociates slightly to form Ba2+ and
C
SO42-, while the water ionizes slightly to form H+ 1984
Discuss
the roles of indicators in the titration of acids and bases.
and OH-.
Explain the basis of their operation and the factors to be
1983 B
considered in selecting an appropriate indicator for a particular
titration.
28 . 35 m LN aO H
0 . 27 64 m o l
1 m o lHNO 3
Answer:



1L
1 m o lNa OH
An indicator signals the end point of a titration by (b) 25 . 00 m LHN O 3
changing color.
= 0.3134M HNO3
An indicator is a weak acid or weak base where the
MfVf=MiVi;(0.3134M)(500mL) = (M)(10.00mL)
acid form and basic form of the indicators are of
different colors.
M = 15.67M HNO3
g L HNO 3
An indicator changes color when the pH of the
% HNO 3 in c onc . s ol ’n 
10 0 %
solution equals the pKa of the indicator. In selecting an
g L s ol ’ n
(c)
indicator, the pH at which the indicator changes color
grams HNO3 in 1 L conc. solÆn =
should be equal to (or bracket) the pH of the solution
15 .67 mo lHNO 3
63 . 02 g
g
at the equivalence point.

 98 7 . 5 L
1L
1 m ol
For example, when a strong acid is titrated with a
strong base, the pH at the equivalence point is 7, so we
grams sol’n in 1 L conc. Sol’n
would choose an indicator that changes color at a pH =
1 . 42 g s ol’n
100 0 mL
g

142 0 L
7. {Many other examples possible.}
1 mL
1L
1987 B
The percentage by weight of nitric acid, HNO 3, in a sample of
concentrated nitric acid is to be determined.
(a) Initially a NaOH solution was standardized by
titration with a sample of potassium hydrogen
phthalate, KHC8H4O4, a monoprotic acid often used
as a primary standard. A sample of pure KHC8H4O4
weighing 1.518 grams was dissolved in water and
titrated with the NaOH solution. To reach the
equivalence point, 26.90 millilitres of base was
required. Calculate the molarity of the NaOH
solution. (Molecular weight: KHC8H4O4 = 204.2)
(b) A 10.00 millilitre sample of the concentrated nitric
acid was diluted with water to a total volume of
500.00 millilitres. Then 25.00 millilitres of the
diluted acid solution was titrated with the
standardized NaOH solution prepared in part (a).
The equivalence point was reached after 28.35
millilitres of the base had been added. Calculate
the molarity of the concentrated nitric acid.
(c) The density of the concentrated nitric acid used in
this experiment was determined to be 1.42 grams
per millilitre. Determine the percentage by weight
of HNO3 in the original sample of concentrated
nitric acid.
Answer:
1 . 51 8 g 
(a)
1 mo l
20 4 . 2 g
 7 . 43 4 1 0
3
mo l ac id
= mol NaOH required to neut.
7 . 43 4 1 0
3
0 . 02 690 L
mo l
 0 . 27 64 M N aO H
( 0 . 01 52 3 L ) ( 0 . 22 11 mol L )
[ HA ] 
 0 . 04 810 M
0 . 07 00 0 L
K 
[A

[O H

[A
] 
] [ HA ]

]
7 . 78 9 1 0

3
Kw
Ka
mo l
0 . 08 52 3 L

1 . 1 0
14
7 . 7 1 0
5
1 . 3 1 0
10
 0 . 09 14 M
1988 D
12
X
10
X
8
pH
X
X
6
X
X
X
4X
2
0
0
5
10
15
20
m illilitr es of Na OH
25
30
A 30.00 millilitre sample of a weak monoprotic acid was titrated
with a standardized solution of NaOH. A pH meter was used to
measure the pH after each increment of NaOH was added, and the
curve above was constructed.
(a) Explain how this curve could be used to determine
the molarity of the acid.
(b) Explain how this curve could be used to determine
the dissociation constant Ka of the weak
monoprotic acid.
(c) If you were to repeat the titration using a indicator
in the acid to signal the endpoint, which of the
following indicators should you select? Give the
reason for your choice.
Ka = 1x10-5
Ka = 1x10-8
Ka = 1x10-11
Methyl red
Cresol red
Alizarin yellow
(d) Sketch the titration curve that would result if the
weak monoprotic acid were replaced by a strong
monoprotic acid, such as HCl of the same
molarity. Identify differences between this
titration curve and the curve shown above.
Answer:
(a) The sharp vertical rise in pH on the pH-volume
curve appears at the equivalence point (about 23
mL). Because the acid is monoprotic, the number
of moles of acid equals the number of moles of
NaOH. That number is the product of the exact
volume and the molarity of the NaOH. The
molarity of the acid is the number of moles of the
acid divided by 0.30L, the volume of the acid.
(b) At the half-equivalence point (where the volume
of the base added is exactly half its volume at the
equivalence point), the concentration [HX] of the
weak acid equals the concentration [X-] of its
anion. Thus, in the equilibrium expression [H+][X]/[HX] = Ka, [H+] = Ka. Therefore, pH at the halfequivalence point equals pKa.
(c) Cresol red is the best indicator because its pKa
(about 8) appears midway in the steep equivalence
region. This insures that at the equivalence point
the maximum color change for the minimal
change in the volume of NaOH added is observed.
(d)
12
Lon ger e quiv ale nc e
r eg ion f or s tro ng a c id
10
8
pH
X
X
X
X
6
X
X
X
4 SXa m e v olum e of N aO H
a dde d f or equ iva len c e p oint
1989 A
In an experiment to determine the molecular weight and the
ionization constant for ascorbic acid (vitamin C), a student
dissolved 1.3717 grams of the acid in water to make 50.00
millilitres of solution. The entire solution was titrated with a
0.2211 molar NaOH solution. The pH was monitored throughout
the titration. The equivalence point was reached when 35.23
millilitres of the base has been added. Under the conditions of this
experiment, ascorbic acid acts as a monoprotic acid that can be
represented as HA.
(a) From the information above, calculate the
molecular weight of ascorbic acid.
(b) When 20.00 millilitres of NaOH had been added
during the titration, the pH of the solution was
4.23. Calculate the acid ionization constant for
ascorbic acid.
(c) Calculate the equilibrium constant for the reaction
of the ascorbate ion, A-, with water.
(d) Calculate the pH of the solution at the equivalence
point of the titration.
Answer:
(a) (0.2211M)(0.03523L) = 7.789x10-3 mol
1.3717g/7.789x10-3 mol = 176.1g/mol
(b) at pH 4.23, [H+] = 8.0x10-8M
[A

] 
[ HA ] 
K 

[H
][ A
( 0 . 02 000 L ) ( 0 . 22 11 mo l L
1
)
0 . 07 000 L
( 0 . 01 523 L ) ( 0 . 22 11 mo l L
1
0 . 07 000 L

]

[ HA ]
( 5 . 9 10
5
) ( 0 . 063 17 )
( 0 . 048 10 )
)
 0 . 06 317 M
 0 . 04 810 M
 7 . 7 10
5
(c) A- + H2O <=> HA + OHK 
[O H

[A
][ HA ]

]

Kw
Ka

1 . 1 0
14
7 . 7 1 0
5
 1 . 3 1 0
10
(d) at equiv. pt.
[A

] 
7 . 789 1 0
3
mol
0 . 085 23 L
 0 . 091 4 M
[OH-]2 = (1.3x10-10)(9.14x10-2) = 1.2x10-11
[OH-] = 3.4x10-6M
pOH =-log(3.4x10-6) =5.47; pH = (14-5.47)= 8.53
1994 D
2
0
0
5
Ac id por tion at lo w er
pH f or str on g ac id
10
15
20
25
30
m illilitr es of Na OH
A chemical reaction occurs when 100. milliliters of 0.200-molar
HCl is added dropwise to 100. milliliters of 0.100-molar Na3P04
solution.
(a) Write the two net ionic equations for the
formation of the major products.
(b) Identify the species that acts as both a Bronsted
• Analytical balance
acid and as a Bronsted base in the equation in
• Phenolphthalein indicator solution
(a), Draw the Lewis electron-dot diagram for this
• Potassium hydrogen phthalate, KHP, a pure solid
species.
monoprotic acid (to be used as the primary
(c) Sketch a graph using the axes provided, showing
standard)
the shape of the titration curve that results when (a) Briefly describe the steps you would take, using
100. milliliters of the HCl solution is added slowly
the materials listed above, to standardize the
from a buret to the Na3PO4 solution. Account for
NaOH solution.
the shape of the curve.
(b) Describe (i.e., set up) the calculations necessary to
determine the concentration of the NaOH solution.
(c) After the NaOH solution has been standardized, it
is used to titrate a weak monoprotic acid, HX. The
equivalence point is reached when 25.0 mL of
pH
NaOH solution has been added. In the space
provided at the right, sketch the titration curve,
showing the pH changes that occur as the volume
0
of NaOH solution added increases from 0 to 35.0
mL HCl
mL. Clearly label the equivalence point on the
(d) Write the equation for the reaction that occurs if a
curve.
few additional milliliters of the HCl solution are
added to the solution resulting from the titration in
(c).
Answer:
(a) PO43- + H+ --> HPO42-; HPO42- + H+ --> H2PO4(b) HPO42. .–
:O
. .:
. .
. .
O
:
:
P
:
O
. . . . . .:H
:O
. .:
–
3–
+
2–
P O 4 + H  H P O 4
(d) Describe how the value of the acid-dissociation
constant, Ka, for the weak acid HX could be
determined from the titration curve in part (c).
(e) The graph below shows the results obtained by
0
mL HCl
titrating a different weak acid, H2Y, with the
(c)
standardized NaOH solution. Identify the negative
(d) H+ + H2PO4- --> H3PO4
ion that is present in the highest concentration at
1998 D (Required)
[repeated in lab procedures section]
the point in the titration represented by the letter A
An approximately 0.1-molar solution of NaOH is to be
on the curve.
standardized by titration. Assume that the following
materials are available.
• Clean, dry 50 mL buret
• 250 mL Erlenmeyer flask
• Wash bottle filled with distilled water
pH
–
2–
+
H P O 4 + H  H 2P O 4
In aqueous solution, ammonia reacts as represented
above. In 0.0180 M NH3(aq) at 25ºC, the hydroxide ion
concentration, [OH–] is 5.60x10–4 M. In answering the
following, assume that temperature is constant at 25ºC
and that volumes are additive.
(a) Write the equilibrium-constant expression for the
reaction represented above.
(b) Determine the pH of 0.0180 M NH3(aq).
Answer
(c) Determine the value of the base ionization
(a) • exactly mass a sample of KHP in the Erlenmeyer
constant, Kb, of NH3(aq).
flask and add distilled water to dissolve the solid.
(d) Determine the percent ionization of NH3 in
• add a few drops of phenolphthalein to the flask.
0.0180 M NH3(aq).
• rinse the buret with the NaOH solution and fill.
(e) In an experiment, a 20.0 mL sample of 0.0180 M
• record starting volume of base in buret.
NH3(aq) was placed in a flask and titrated to the
• with mixing, titrate the KHP with the NaOH
equivalence point and beyond using 0.0120 M
solution until it just turns slightly pink.
HCl(aq).
• record end volume of buret.
(i) Determine the volume of 0.0120 M HCl(aq)
• repeat to check your results.
that was added to reach the equivalence
mass of KHP
point.
(b) molar mass KHP = moles of KHP
(ii) Determine the pH of the solution in the flask
since KHP is monoprotic, this is the number of
after a total of 15.0 mL of 0.0120 M HCl(aq)
moles of NaOH
was added.
moles of NaOH
(iii) Determine the pH of the solution in the flask
L of titr an t
after a total of 40.0 mL of 0.0120 M HCl(aq)
= molarity of NaOH
was added.
Answer
[NH4+][OH–]
(a) Kb =
[NH3]
(b) pOH = -log(5.60x10–4) = 3.252
pH = 14 – pOH = 10.748
(c) Kb =
(d)
(5.6010–4)(5.6010–4)
= 1.80x10–5
(0.0180 – 5.6010–4)
5.6010–4
0.0180 x100% = 3.11%
(e) (i) NAVA = NBVB
(c)
(d) from the titration curve, at the 12.5 mL volume
point, the acid is half-neutralized and the pH =
pKa. Ka = 10pKa
(e) Y2- (could it be OH- ?)
1999 A Required
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
(0.0120 N)(VA) = (0.0180 N)(20.0 mL)
VA = 30.0 mL
(ii) at 15.0 mL it is half-titrated and [NH4+] =
[NH3], then the Kb = [OH–] = 1.80x10–5.
pOH = -log(1.80x10–5) = 4.745
pH = 14 – 4.75 = 9.255
(iii) at 40.0 mL, there is an excess of 10.0 mL of
HCl past equivalence point,
(10.0 mL)(0.0120 M) = 0.120 mmol H+ ions
0.120 mmol
+
60.0 mL = 0.00200M =[H ]
pH = -log(0.00200) = 2.70