College Preparatory Program • Saudi Aramco
Acid-Base Titration Solution
Acid-Base Titration Solution Key
CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH-(aq) Kb = 4.38 x 10-4
In aqueous solution of methylamine at 25C, the hydroxide ion concentration is 1.50 x 10-3
M. In answering the following, assume that temperature is constant and that volumes are
additive.
(a) Write the equilibrium constant expression for the reaction above.
𝐶𝐻3 𝑁𝐻3+ [𝑂𝐻−]
𝐾𝑏 =
[𝐶𝐻3 𝑁𝐻2 ]
(b) Determine the initial concentration of methylamine.
Since the solution is a weak base, construct an ICE table
[𝑪𝑯𝟑 𝑵𝑯𝟐 ]
𝑪𝑯𝟑 𝑵𝑯+
𝟑
[𝑶𝑯−]
Initial
[𝐶𝐻3 𝑁𝐻2 ]𝑖
0
0
Change
−𝑥
+𝑥
+𝑥
[𝐶𝐻3 𝑁𝐻2 ]𝑖 − 𝑥
𝑥
𝑥
Equilibrium
Substitute the expression of the equilibrium concentrations for each chemical species into
the equilibrium constant expression for 𝐾𝑏 .
𝐶𝐻3 𝑁𝐻3+ 𝑂𝐻 −
𝑥 (𝑥)
𝐾𝑏 =
=
= 4.38 × 10−4
𝐶𝐻3 𝑁𝐻2
[𝐶𝐻3 𝑁𝐻2 ]𝑖 − 𝑥
Since we are looking for the initial concentration of methylamine, [𝐶𝐻3 𝑁𝐻2 ]𝑖 , the value of 𝑥
must be known. In the problem, the [𝑂𝐻 −] is given as 1.50 × 10−3 𝑀. This is equal to 𝑥.
Hence, we can now calculate [𝐶𝐻3 𝑁𝐻2 ]𝑖 from the equation.
1.50 × 10−3 2
= 4.38 × 10−4
[𝐶𝐻3 𝑁𝐻2 ]𝑖 − 1.50 × 10−3
Solving the equation,
[𝐶𝐻3 𝑁𝐻2 ]𝑖 = 𝟔. 𝟔𝟎 × 𝟏𝟎−𝟑 𝑴
College Preparatory Program • Saudi Aramco
Acid-Base Titration Solution
(c) Determine the percent ionization of methylamine in the above solution.
Percent ionization is the percentage of the initial concentration of the base that dissociates to
form 𝐶𝐻3 𝑁𝐻3+ and 𝑂𝐻 − given by the following formula:
% 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑
× 100%
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡
𝑥
1.50 × 10−3
=
× 100% =
× 100%
[𝐶𝐻3 𝑁𝐻2 ]𝑖
6.60 × 10−3
% 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 = 𝟐𝟐. 𝟕%
(d) What is the number of moles of HCl that should be added to 100. mL of 0.100 M CH3NH2
to produce a solution buffered at pH = 10.125.
A base buffer solution must contain both the weak base (CH3NH2) and its conjugate acid
(CH3NH3+). Adding HCl into the weak base solution will produce the conjugate acid as
shown by the following net ionic equation.
𝑪𝑯𝟑 𝑵𝑯𝟐 + 𝑯+ → 𝑪𝑯𝟑 𝑵𝑯+
𝟑
The amount of HCl added must not exceed the initial concentration of the CH3NH2,
otherwise, no buffer will be formed. The ratio of the weak base and its conjugate acid will
determine the pH. In other words, the problem basically asks for the amount of HCl that
must be added so that the ratio is just right to give a pH=10.125.
Calculating pH of buffer solutions is easier with the use of the Henderson-Hasselbach
equation:
𝑝𝐻 = 𝑝𝐾𝑎 + log
[𝑏𝑎𝑠𝑒]
[𝐶𝐻3 𝑁𝐻2 ]
= 𝑝𝐾𝑎 + log
[𝑎𝑐𝑖𝑑]
[𝐶𝐻3 𝑁𝐻3+]
𝑝𝐾𝑎 = − 𝑙𝑜𝑔 𝐾𝑎 . Since 𝐾𝑎 is not given, the value can be calculated using the following
equation: 𝐾𝑎 =
𝑝𝐾𝑎 = 10.641
𝐾𝑤
𝐾𝑏
where 𝐾𝑤 = 1 × 10−14 while 𝐾𝑏 = 4.38 × 10−4 . Therefore
College Preparatory Program • Saudi Aramco
Acid-Base Titration Solution
If we let 𝑥 = 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝑙 𝑎𝑑𝑑𝑒𝑑, the stoichiometry of the reaction is shown by the table
below:
𝒎𝒐𝒍𝒆𝒔 𝑪𝑯𝟑 𝑵𝑯𝟐
𝒎𝒐𝒍𝒆𝒔 𝑯𝑪𝒍 →
𝒎𝒐𝒍𝒆𝒔 𝑪𝑯𝟑 𝑵𝑯+
𝟑
Initial
(0.100 L)(0.100 M)
=0.0100 moles
𝑥
0
Change
−𝑥
−𝑥
0
completely used
up (Limiting
Reactant)
-𝑥
Final
0.0100 − 𝑥
𝑥
Remember that in the log ratio, the moles of the species can be used instead of a
concentration because the volume units will cancel out anyway.
𝑝𝐻 = 𝑝𝐾𝑎 + log
[𝐶𝐻3 𝑁𝐻2 ]
[𝐶𝐻3 𝑁𝐻3+]
10.125 = 10.641 + log
0.0100 − 𝑥
𝑥
𝒙 = 𝟎. 𝟎𝟎𝟕𝟔𝟔 𝒎𝒐𝒍𝒆𝒔 𝑯𝑪𝒍
College Preparatory Program • Saudi Aramco
Acid-Base Titration Solution
(e) 125 .0 ml of 0.200 M CH3NH3Cl is titrated with 0.100 M NaOH to the equivalence point
and beyond the equivalence point.
We have to identify the acid-base reaction involved in this particular titration. Notice that
−
+
the acid in the problem is a salt. 𝑪𝑯𝟑 𝑵𝑯+
𝟑 is the one that will react with the𝑶𝑯 , Na and
Cl- are weaker conjugates.
−
𝑪𝑯𝟑 𝑵𝑯+
𝟑 + 𝑶𝑯 → 𝑪𝑯𝟑 𝑵𝑯𝟐 + 𝑯𝟐 𝑶
(i) Find the pH of a 0.200 M CH3NH3Cl solution.
Initially the flask only contains the acid salt solution (𝑪𝑯𝟑 𝑵𝑯+
𝟑 ). Hence, the pH can
be solved from the dissociation of the acid salt in H2O using the ICE table below.
+
𝑪𝑯𝟑 𝑵𝑯+
𝟑 + 𝑯𝟐 𝑶 → 𝑪𝑯𝟑 𝑵𝑯𝟐 + 𝑯𝟑 𝑶
𝑪𝑯𝟑 𝑵𝑯+
𝟑
[𝑪𝑯𝟑 𝑵𝑯𝟐 ]
𝑯𝟑 𝑶+
Initial
0.200 M
0
0
Change
−𝑥
+𝑥
+𝑥
0.200 − 𝑥
𝑥
𝑥
Equilibrium
𝐾𝑤
𝑥2
1 × 10−14
𝐾𝑎 =
=
=
𝐾𝑏
0.200 − 𝑥 4.38 × 10−4
𝑥 = 2.14 × 10−6 𝑀
𝑝𝐻 = − log[ 𝑯𝟑 𝑶+] = −𝒍𝒐𝒈 2.14 × 10−6 = 𝟓. 𝟔𝟕
College Preparatory Program • Saudi Aramco
Acid-Base Titration Solution
(ii) Find the volume of NaOH needed to reach the equivalence point.
−
𝑪𝑯𝟑 𝑵𝑯+
𝟑 + 𝑶𝑯 → 𝑪𝑯𝟑 𝑵𝑯𝟐 + 𝑯𝟐 𝑶
At the equivalence point, there is no limiting or excess reactant, both the 𝑪𝑯𝟑 𝑵𝑯+
𝟑
and 𝑶𝑯− completely used up in the reaction.
Hence, to calculate the volume needed to reach equivalent point, the following
equation can be used because the ratio of acid to base is 1:1
−
𝑚𝑜𝑙𝑒𝑠 𝑪𝑯𝟑 𝑵𝑯+
𝟑 = 𝑚𝑜𝑙𝑒𝑠 𝑶𝑯
(𝐶 × 𝑉)𝑪𝑯𝟑𝑵𝑯+𝟑 = (𝐶 × 𝑉)𝑂𝐻 −
0.200 𝑀 125.0 𝑚𝐿 = 0.100 𝑀 𝑉𝑂𝐻 −
𝑽𝑶𝑯− = 𝟐𝟓𝟎. 𝟎 𝒎𝑳
(iii) Find the pH at the equivalence point.
At equivalence point, the major species that would affect the pH is only 𝑪𝑯𝟑 𝑵𝑯𝟐 .
Since this is a weak base and can dissociate in water, the pH is expected to be
greater than 7.
Since the moles of 𝑪𝑯𝟑 𝑵𝑯+
𝟑 that reacted is the same number of moles of 𝑪𝑯𝟑 𝑵𝑯𝟐
produced,
𝑚𝑜𝑙𝑒𝑠 𝑪𝑯𝟑 𝑵𝑯+
𝟑 = 0.200 𝑀 0.125 𝐿 = 0.0250 𝑚𝑜𝑙𝑒𝑠
𝑚𝑜𝑙𝑒𝑠 𝑪𝑯𝟑 𝑵𝑯𝟐 = 0.0250 𝑚𝑜𝑙𝑒𝑠
−
𝑪𝑯𝟑 𝑵𝑯𝟐 + 𝑯𝟐 𝑶 → 𝑪𝑯𝟑 𝑵𝑯+
𝟑 + 𝑶𝑯
The pH can be calculated from the ICE table of the dissociation of the weak base.
The initial concentration of 𝑪𝑯𝟑 𝑵𝑯𝟐 must be calculated based on the number of
moles produced and the total volume (original volume of the acid salt + volume of
the base added) of the solution at this point.
[𝑪𝑯𝟑 𝑵𝑯𝟐 ]𝑖 =
0.0250 𝑚𝑜𝑙𝑒𝑠
= 0.0667 𝑀
0.125 𝐿 + 0.250 𝐿
College Preparatory Program • Saudi Aramco
Acid-Base Titration Solution
[𝑪𝑯𝟑 𝑵𝑯𝟐 ] →
𝑪𝑯𝟑 𝑵𝑯+
𝟑
[𝑶𝑯−]
Initial
0.0667
0
0
Change
−𝑥
+𝑥
+𝑥
0.0667 − 𝑥
𝑥
𝑥
Equilibrium
𝐾𝑏 =
𝑥2
= 4.38 × 10−4
0.0667 − 𝑥
𝑥 = 0.00541 𝑀
𝑝𝑂𝐻 = − log 0.00541 = 2.27
𝑝𝐻 = 11.73
(iv) Sketch a rough curve for the above titration on the graph below and indicate the
initial pH, the pH at half-way to the equivalence point and the pH at the equivalence
point.
A titration curve can be sketched based on four points only.
 The initial point
The volume of NaOH added is zero.
The initial pH is calculated from part (e)(i), pH = 5.67

The equivalence point
The volume of NaOH added is 250.0 mL based on calculation in part (e)(ii)
The pH at equivalence point is calculated from part (e)(iii), pH=11.73

The half-equivalence point
The volume is half of the volume at equivalence point, 125.0 mL
Since the ratio of the base and acid is equal to 1 at this point, the HendersonHasselbach equation reduces to pH = pKa
Hence, the pH at half-equivalence point is 10.64

Beyond equivalence
Assume a volume greater than the equivalence volume. The difference between
the assumed volume and the equivalence volume is the excess NaOH unreacted
in the solution. This will determine the pH of the solution.
Let’s say, we have added 400.0 mL. The excess NaOH would then be 400.0 –
250.0 mL. Hence the excess base would be 150.0 mL.
College Preparatory Program • Saudi Aramco
Acid-Base Titration Solution
The concentration of the base must then be recalculated considering the total
volume of the solution at this point.
𝑂𝐻 − =
𝑚𝑜𝑙𝑒𝑠 𝑒𝑥𝑐𝑒𝑠𝑠 𝑁𝑎𝑂𝐻
0.150 𝐿 (0.100 𝑀)
=
= 0.0286 𝑀
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
(0.125 𝐿 + 0.400 𝐿)
𝑝𝑂𝐻 = − log 0.0286 = 1.54
𝒑𝑯 = 𝟏𝟒. 𝟎𝟎 − 𝟏. 𝟓𝟒 = 𝟏𝟐. 𝟒𝟔