Calculating Concentrations of Buffer Components
 2013, Sharmaine S. Cady
East Stroudsburg University
Buffers are solutions that resist large changes in pH when acid or base is added. Buffers are
typically made from a weak acid and the salt of its conjugate base. A solution of a weak base and salt of
its conjugate acid is less commonly used. Buffer capacity is an indicator of how much acid or base can
be added before the buffer can no longer resists changes in pH. Once the stoichiometric amount of the
acid component is neutralized by base, the pH of the buffer shows a dramatic increase in pH with
additional base. Similarly, the pH of the buffer shows a dramatic decrease in pH with additional acid
once the base component has been completely neutralized. The concentrations of the acid and base
components can therefore be determined by titrating a given volume of buffer with an acid or base of
known concentration and monitoring the pH. From the volume of acid needed for neutralization at the
equivalence point, the concentration of the base component can be calculated. Likewise, from the
volume of base needed for neutralization at the equivalence point, the concentration of the acid
component can be calculated.
The H2PO4-/HPO42- buffer system is analyzed for the concentrations of the two components.
H2PO4- is the acid component, which is titrated with a solution of NaOH of known concentration. The
balanced chemical reaction equation for the acid-base neutralization is
H2PO4-(aq) + OH-(aq)  HPO42-(aq) + H2O(l)
Na+ ions serve to maintain electrical neutrality. HPO4- is the base component, which is titrated with a
solution of HCl of known concentration. The balanced chemical reaction equation is
HPO42-(aq) + H+(aq)  H2PO4-(aq) + H2O(l)
In both reactions, there is a 1:1 molar stoichiometric ratio between the buffer component and the
titrant.
How to use pH to determine the equivalence point for the
acid component
The table of data shows the change in pH for 25.00 mL of the buffer when 0.50 mL of NaOH is added.
Table 1. Titration of H2PO4-/HPO42- Buffer System with 0.1000 M NaOH
Volume 0.1000 M NaOH
added, mL
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
pH
7.39
7.56
7.88
8.55
10.30
10.67
10.88
11.03
The pH vs. NaOH volume is plotted to determine the NaOH volume at the equivalence point.
Buffer pH vs. Volume 0.1000 M NaOH
12.00

10.00
equivalence point
8.00
pH 6.00
4.00
2.00
0.00
0.00
0.50
1.00
1.50
2.00
2.50
volume NaOH added, mL
3.00
3.50
4.00
The equivalence point on the graph is the point where the change in the slope of the line is the steepest.
This occurs between 1.50 and 2.00 mL. A vertical line is drawn from this point to the x-axis. The point at
which the line intersects the x-axis is the volume of 0.1000 M NaOH needed to neutralize the acid
component of the buffer. This value is 1.72 mL. The concentration of H2PO4- can now be determined by
dimensional analysis.
Solution:
Write down the given values with
their units and what they represent.
In this problem, two volumes and one concentration are known. The
reaction stoichiometry is 1 mol H2PO4- : 1 mol NaOH.
Vbuffer = 25.00 mL
VNaOH = 1.72 mL
[NaOH] = 0.1000 M
Write down the unknown and its
unit.
The molarity of H2PO4- in the buffer is unknown.
[H2PO4-] = ? M
Develop a strategy to solve the
problem.
The definition of molarity is
𝒎𝒐𝒍𝒂𝒓𝒊𝒕𝒚 (𝑴) =
𝒎𝒐𝒍 𝒔𝒐𝒍𝒖𝒕𝒆
𝑳 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
A dimensional analysis problem is set up to give mol H2PO4-/L buffer.
1 𝑚𝑜𝑙 𝐻2 𝑃𝑂4− 0.1000 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 1.72 𝑚𝐿 𝑁𝑎𝑂𝐻
∙
∙
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 1000 𝑚𝐿 𝑁𝑎𝑂𝐻 0.02500 𝐿 𝑏𝑢𝑓𝑓𝑒𝑟
𝑀 𝐻2 𝑃𝑂4− = 0.00688 𝑚𝑜𝑙 𝐻2 𝑃𝑂4− /𝐿 𝑏𝑢𝑓𝑓𝑒𝑟
Round off the answer to the correct
number of significant digits and give
the appropriate units.
The NaOH volume has 3 significant digits while the concentration and
volume of buffer have 4. The molarity is rounded off to 3 significant
digits. The answer is
[H2PO4-] = 0.00688 M
How to use pH to determine the equivalence point for the
base component
The table of data shows the change in pH for 25.00 mL of the buffer when 0.50 mL of HCl is added.
Table 2. Titration of H2PO4-/HPO42- Buffer System with 0.100 M HCl
Volume 0.100 M HCl added,
mL
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
4.00
4.50
5.00
5.50
6.00
6.50
7.00
7.50
8.00
8.50
9.00
9.50
10.00
10.50
11.00
11.50
12.00
pH
7.36
7.22
7.09
6.96
6.84
6.72
6.60
6.48
6.34
6.17
5.94
5.57
3.91
3.18
2.88
2.70
2.55
2.45
2.35
2.26
2.19
2.13
2.08
2.03
1.98
The pH vs. HCl volume is plotted to determine the HCl volume at the equivalence point.
Buffer pH vs. Volume 0.100 M HCl
8.00
7.00
6.00
5.00
equivalence point
pH 4.00

3.00
2.00
1.00
0.00
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 6.00 6.50 7.00 7.50 8.00 8.50 9.00 9.50 10.00 10.50 11.00 11.50 12.00 12.50 13.00
Volume HCl added, mL
The equivalence point on the graph is the point where the change in the slope of the line is the steepest.
This occurs between 5.50 and 6.00 mL. A vertical line is drawn from this point to the x-axis. The point at
which the line intersects the x-axis is the volume of 0.100 M HCl needed to neutralize the base
component of the buffer. This value is 5.80 mL. The concentration of HPO42- can now be determined by
dimensional analysis.
Solution:
Write down the given values with
their units and what they represent.
In this problem, two volumes and one concentration are known. The
reaction stoichiometry is 1 mol HPO42- : 1 mol HCl.
Vbuffer = 25.00 mL
VHCl = 5.80 mL
[HCl] = 0.100 M
Write down the unknown and its
unit.
The molarity of HPO42- in the buffer is unknown.
[HPO42-] = ? M
Develop a strategy to solve the
problem.
The definition of molarity is
𝒎𝒐𝒍𝒂𝒓𝒊𝒕𝒚 (𝑴) =
𝒎𝒐𝒍 𝒔𝒐𝒍𝒖𝒕𝒆
𝑳 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
A dimensional analysis problem is set up to give mol HPO42-/L buffer.
1 𝑚𝑜𝑙 𝐻𝑃𝑂42− 0.100 𝑚𝑜𝑙 𝐻𝐶𝑙
5.80 𝑚𝐿 𝐻𝐶𝑙
∙
∙
1 𝑚𝑜𝑙 𝐻𝐶𝑙
1000 𝑚𝐿 𝐻𝐶𝑙 0.02500 𝐿 𝑏𝑢𝑓𝑓𝑒𝑟
𝑀 𝐻𝑃𝑂42− = 0.0232 𝑚𝑜𝑙 𝐻𝑃𝑂42− /𝐿 𝑏𝑢𝑓𝑓𝑒𝑟
Round off the answer to the correct
number of significant digits and give
the appropriate units.
The HCl concentration and volume have 3 significant digits while the
volume of buffer has 4. The molarity is rounded off to 3 significant
digits. The answer is
[HPO42-] = 0.0232 M