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Activity 3
Buffer solutions
Figure 1. The pH in the human body varies greatly from one part to the other. The blood maintains a pH of
about 7.4 to keep its processes normal. The pH in the blood is maintained through a buffer that resists pH
changes even if large amounts of acid or base is added to it.
Objectives
At the end of this activity, the students will have mastered the concepts in weak acid-base
equilibria and its applications in buffer solutions. More specifically, the students will have the
proficiency in
- Discussing the common-ion effect in acid-base equilibria
- Preparing buffer solutions
- Determining the buffer capacity and factors affecting it
Theory
In biochemical systems, weak acid-weak base reactions are the predominant acid-base
interactions. These chemicals dissociate poorly in water, creating an equilibrium of the non-ionized
species, the hydronium ion ( 𝐻 ), and its conjugate pair. These systems are sensitive to changes 3
in pH and such changes are crucial for most biochemical processes such as enzyme function and
balance in osmotic pressure. Therefore, there is a need to “protect” these processes from sudden
fluctuations in pH, by creating a “buffer” that resists these changes.
Common-ion effect
Consider a solution with equal concentrations of 𝐶𝐻 and. The ionization of 3
3
𝐶𝐻 produces and ions: 3
3
𝐶𝐻 3
3
2
3
−
.
The ionization of Na 𝐶𝐻 produces and ions. 3
𝐶𝐻 3
−
3
−
Because both acetate ions are produced in both ionization process, we say that the acetate is a
common ion. The presence of a common ion suppresses the ionization of a weak acid or a weak
base. According to Le Châtelier’s principle, the addition of acetate ions from sodium acetate into a
solution of acetic acid will suppress the ionization of acetic acid, i.e., the equilibrium shifts to the left,
thereby decreasing the hydronium concentration. Thus, a solution containing both 𝐶𝐻 and 3
𝑁𝑎𝐶𝐻 will be less acidic than a solution containing only at the same concentration. 3
3
The shift in equilibrium of the acetic acid ionization is caused by the acetate ions from the salt
(Figure 2).
Figure 2. Common-ion effect in acetic acid solution.
The common ion effect is the shift in equilibrium caused by the addition of a compound
having an ion in common with the dissolved substance. The common ion effect plays an important
role in determining the pH of a solution. To determine the extent of the effect of common ion on the
equilibrium pH, let us consider a solution of a hypothetical weak acid, HA, and a soluble salt of its
conjugate base, say XA. Consider the pH of a solution containing a theoretical weak acid, HA, and
a soluble salt of such weak acid, say XA. The equation of the dissociation of HA is written as
2
3
(𝑎𝑞 ) + 𝐴
− (𝑎𝑞 ).
The ionization constant 𝐾 is represented as 𝑎
𝑎
𝐻 3
𝑂
⎡ ⎢ ⎣
⎤ ⎥ ⎦
𝐴
−
[ ]
[ 𝐻𝐴 ]
or
3
⎡ ⎢ ⎣
𝐾 𝑎
𝐴
−
[ ]
[ 𝐻𝐴 ]
Taking the negative logarithm of both sides of the equation gives a p function for hydronium
concentration and ionization constant: 𝑝𝐻 =− log 𝑙𝑜𝑔 𝐻. 3
⎡ ⎢ ⎣
𝑎
=− log 𝑙𝑜𝑔 𝐾 𝑎
− log 𝑙𝑜𝑔 𝐻 3
⎡ ⎢ ⎣
=− log 𝑙𝑜𝑔 𝐾 𝑎
− log 𝑙𝑜𝑔
[𝐻𝐴]
[𝐴
− ]
𝑎
− log 𝑙𝑜𝑔
[𝐻𝐴]
[𝐴
− ]
This equation is called the Henderson-Hasselbalch equation , which is more familiarly expressed
as
𝑎
[𝐴
− ]
[𝐻𝐴]
where [𝐴 is the concentration of the conjugate base and is the concentration of the acid. The
−
] [𝐻𝐴]
Henderson-Hasselbalch equation is useful not only for finding the equilibrium pH in an acid-base
reaction but also for estimating the pH of a buffer solution. An extension of the
For a measurable amount of strong acid or strong base added to a 1-L buffer solution over a
measurable change in the pH of the solution, the buffer capacity, β,is
β =
𝑛 𝑏𝑎𝑠𝑒
𝑉 𝑠𝑜𝑙𝑛
Δ𝑝𝐻
𝑜𝑟 β =−
𝑛 𝑎𝑐𝑖𝑑
𝑉 𝑠𝑜𝑙𝑛
Δ𝑝𝐻
where: 𝑛 is the amount in moles of base or acid added to the buffer; change in pH,
Δ𝑝𝐻 = 𝑝𝐻 ; and is the sum of the volume of the buffer and the added base or 𝑓𝑖𝑛𝑎𝑙
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑠𝑜𝑙𝑛
acid.
This property is the first derivative of the buffer titration curve. A titration curve is a graph of
the pH of the solution against the volume of titrant (viz. the strong acid or strong base) added to the
solution (Figure 5). A large capacity indicates high resistance to pH changes and therefore a flatter
slope of the titration curve. Nevertheless, capacity depends ultimately on component
concentrations, both the absolute and relative concentrations.
In terms of absolute concentrations, the more concentrated the buffer components, the
greater the capacity. The higher their concentrations, the more acid or base the solution can
accommodate before the relative acid-salt ratio is changed. Note that buffer pH is independent of
buffer capacity.
In terms of relative concentrations, the closer the component concentrations are to each
other, the greater the capacity. As a buffer functions, the concentration of one component increases
relative to the other. Because the concentration ratio determines the pH, the less the ratio changes,
the less the pH changes. The capacity is at a maximum when the ratio is unity, that is, when the
𝑝𝐻 = 𝑝𝐾. Therefore, a buffer whose pH is equal to or near the of its acid component has the 𝑎
𝑎
highest capacity for a given concentration.
Buffer range is the pH range over which the buffer is effective and is also related to the
relative buffer-component concentrations. The blue region in
Figure 5 represents the buffer range for acetic acid. This is
generally about 1 pH unit on either side of the 𝑝𝐾𝑎of the
conjugate acid. The midpoint of the buffering region
corresponds to the pH equal to the 𝑝𝐾𝑎, which is halfway to
the equivalence point in titration. This is most effective at
resisting large changes in pH when either strong acid or base
is added, i.e., at maximum capacity. However, once the curve
extends out of this range, the pH will change abruptly when a small amount of acid or base added
to the buffer system. In practice, if the ratio is greater than 10 or less than 0.1—that is, if one
𝐴
−
[ ]
[ 𝐻𝐴 ]
component concentration is more than 10 times the other—buffering action is poor. In general,
buffers have a usable range within a pH range of 𝑝𝐾. 𝑎
Examples
- Calculate the pH of a solution containing 0.200 𝑚𝑜𝑙 𝐿 propionic acid and
− 𝐶𝐻 3
2
( )
0.300 𝑚𝑜𝑙 𝐿 potassium propionate. The for propionic acid is
− 𝐾𝐶𝐻 3
2
( )
𝑎
1. 3183×.
−
SOLUTION:
Propionic acid dissociates poorly in water where hydronium and propionate ions coexist with
propionic acid in equilibrium.
3
2
2
3
2
− (𝑎𝑞 ) + 𝐻 3
(𝑎𝑞)
Potassium propionate readily dissolves in water forming propionate and potassium ions in
equimolar concentrations.
3
2
3
2
−
The common ion is the propionate ion. The initial concentrations, changes, and equilibrium
concentrations of the species involved are presented in the reaction table below.
𝐶𝐻 3
𝐶𝐻 2
𝐶𝑂𝑂𝐻
𝐻 2
𝑂 𝐶𝐻 3
2
−
𝑂
Initial, 𝑚𝑜𝑙 𝐿
− 0.200 0.300 0
Change, 𝑚𝑜𝑙 𝐿
− − 𝑥 + 𝑥 + 𝑥
Equilibrium,
𝑚𝑜𝑙 𝐿
−1 0. 200 − 𝑥^ 0. 300 + 𝑥^ + 𝑥
Determining the 𝐾 gives 𝑎
3
2
− ⎡ ⎢ ⎣
3
⎡ ⎢ ⎣
3
2
[ ]
𝑎
(0.300+𝑥 ) 𝑥( )
0.200−𝑥
= 1. 3183×
−
,
which is a quadratic equation. Solving for 𝑥, which is also 𝐻 , gives 3
⎡ ⎢ ⎣
3
⎡ ⎢ ⎣
= 8. 7880×
− .
The pH of the solution is then
𝑝𝐻 =− log 𝑙𝑜𝑔 𝐻 3
⎡ ⎢ ⎣
=− log 𝑙𝑜𝑔 8. 7880×
−
( ) = 5. 06.
Alternatively, using the Henderson-Hasselbalch equation spares the calculation of 𝐻 3
⎡ ⎢ ⎣
and setting up a reaction table.
𝑝𝐻 =− log 𝑙𝑜𝑔 1. 3183×
−
( ) + log 𝑙𝑜𝑔^
( )
Comparing it with the pH of pure propanoic acid of the same concentration, which is 2.79,
indicates a shift to the left side of the equilibrium (i.e., favoring the production of propionic acid),
confirming the prediction from Le Châtelier’s principle.
- What is the change in the pH of the buffer system in the previous example after the addition
10.0 mL of 1.00 𝑚𝑜𝑙 𝐿 hydrochloric acid was added to 1.00 L of the buffered solution?
−
SOLUTION:
Hydrochloric acid is a strong acid which completely dissociates into 𝐻 and ions in 3
𝐶𝑙
−
equimolar concentrations, in this case 0.010 mol. 𝐶𝑙 is a spectator ion which does not participate
−
in the chemical reaction. The hydronium ions provided by HCl reacts completely with the propionate
ion, which we can quantify using a reaction table:
of HCOOH and NaOH. As 𝑂𝐻 reacts with HCOOH, neutralization of some of the HCOOH
−
produces the 𝐻𝐶𝑂𝑂 needed.
−
Example 1
Suppose that, for a biochemical experiment, you need to prepare a buffer whose pH is 3.90.
Select an appropriate acid and prepare a buffer with a solid sodium salt of its conjugate base and
from there determine the quantities needed from such acid and salt pair.
Conjugate acid-base pair. To maximize the capacity, the 𝑝𝐾 of the acid component should 𝑎
be close to 3.90. Candidates for the buffer are lactic acid (𝑝𝐾 , glycolic acid , 𝑎
𝑎
( = 3. 83)
and formic acid (𝑝𝐾. To avoid substances that are common in biochemical systems, you 𝑎
choose formic acid, HCOOH, and formate ion, 𝐻𝐶𝑂𝑂, supplied by a soluble salt, such as sodium
−
formate, HCOONa, as the basic component.
Ratio of component concentrations. The buffer must have formic acid-formate ratio
𝐴
−
[ ]
[ 𝐻𝐴 ]
𝑝𝐻−𝑝𝐾 𝑎 ( )
= 10
3.90−3. = 1. 4
Thus, in a known volume of the buffer solution, a mole of formic acid needs 1.4 mole of sodium
formate.
Amount of buffer components. Suppose you have a large stock of 0.40 𝑚𝑜𝑙 𝐿 HCOOH
−
and a reagent-grade solid sodium acetate and you need approximately 1.0 L of final buffer,
𝐻𝐶𝑂𝑂
𝐻𝐶𝑂𝑂𝐻
The amount of sodium formate that will give the needed 1.4/1.0 ratio from the molarity of formic
acid is
𝐻𝐶𝑂𝑂𝐻
= [𝐻𝐶𝑂𝑂𝐻 ]𝑉
𝐻𝐶𝑂𝑂𝐻
−
( )
𝐻𝐶𝑂𝑂
− = 1. 4^ (0. 40 𝑚𝑜𝑙^ ) = 0. 56 𝑚𝑜𝑙
Converting to mass,
𝑁𝑎𝐻𝐶𝑂𝑂
𝐻𝐶𝑂𝑂
𝑁𝑎𝐻𝐶𝑂𝑂
−
( ) = 38 𝑔.
Mixing and pH correction. Dissolve 38 g of HCOONa in the stock 0.40 𝑚𝑜𝑙 𝐿 HCOOH
−
and add more of the HCOOH solution to a total volume of 1.0 L. Because of nonideal behavior, the
buffer may vary slightly from the desired pH, so you add strong acid or strong base dropwise while
monitoring the solution with a pH meter.
Example 2
The paper by Zeitoun and Debevere (1992)
1 investigated lactic acid-sodium lactate buffer
solution to decontaminate poultry products which contributed to its increased shelf life. A buffer
solution with a pH of 3.00 was used in the study. Prepare a final volume of 250 mL of this solution
using 10.0 mL of 0.20- 𝑚𝑜𝑙 𝐿 lactic acid and determine the volume of 0.20- of sodium
− 𝑚𝑜𝑙 𝐿
−
lactate required. The 𝑝𝐾 of lactic acid is 3.86. 𝑎
Solution
Lactic acid dissociation is expressed as
3
2
3
− .
Using the rearranged Henderson-Hasselbalch equation, the concentration ratio of lactate to
lactic acid is
1 Zeitoun, A. A. & Debevere, J. M. (1992). Decontamination with lactic acid/sodium lactate buffer in combination with modified
atmosphere packaging effects on the shelf life of fresh poultry. Int. J. Food Microbiol., 16 (2), 89-98.
[ 𝑙𝑎𝑐𝑡𝑎𝑡𝑒 ]
[𝑙𝑎𝑐𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 ]
𝑝𝐻−𝑝𝐾 𝑎 ( )
= 10
3.00−3. = 0. 138,
meaning that in a volume of solution, a mole of lactic acid requires 0.138 moles of lactate.
To determine the actual amounts in 250 mL solution, we use the molarity formula 𝑐. 𝑀
𝑛
𝑉
Therefore, the amount of lactic acid in the 10 mL solution is
𝑙𝑎𝑐𝑡𝑖𝑐 𝑎𝑐𝑖𝑑
= [𝑙𝑎𝑐𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 ]𝑉
𝑙𝑎𝑐𝑡𝑖𝑐 𝑎𝑐𝑖𝑑
−
( ) (10 𝑚𝐿^ )^
1 𝐿
1000 𝑚𝐿 ( )
Since the lactate needed is 0.138 times the amount of lactic acid, the amount required for sodium
lactate is
𝑙𝑎𝑐𝑡𝑎𝑡𝑒
𝑙𝑎𝑐𝑡𝑖𝑐 𝑎𝑐𝑖𝑑
which has a volume
𝑙𝑎𝑐𝑡𝑎𝑡𝑒
𝑛 𝑙𝑎𝑐𝑡𝑎𝑡𝑒
[ 𝑙𝑎𝑐𝑡𝑎𝑡𝑒 ]
0.00276 𝑚𝑜𝑙
0.2 𝑚𝑜𝑙 𝐿
−
Generally, if the stock solutions of acid and conjugate base needed to prepare a buffer have
the same molar concentrations, the conjugate base can have a volume equal to 10 times the
𝑝𝐻−𝑝𝐾 𝑎
volume of acid.
Example 3
Spikevax, a Covid-19 vaccine developed by Moderna, contains
tris(hydroxymethyl)aminomethane (abbreviated tris, see Figure 6) to buffer the
mRNA viral material at a pH of 7.40. You are to prepare a 1000.0-mL buffer
solution at such pH with a maximum possible buffer capacity. You have a stock
solution of 0.10- 𝑚𝑜𝑙 𝐿 tris and 0.10- tris hydrochloride. The of tris
− 𝑚𝑜𝑙 𝐿
− 𝑝𝐾 𝑎
𝐻 is 8.072.
Solution
To prepare this buffer at maximum capacity, we need to prepare a solution with the largest
possible concentration of tris and tris 𝐻. We first determine the concentration ratio, followed by the
amounts, and then the volumes.
Tris hydrochloride dissociates into tris 𝐻 when dissolved in water and therefore the molarity
of tris 𝐻 is equal to that of tris hydrochloride. The chemical equilibrium for the dissociation of tris is
3
Using the rearranged Henderson-Hasselbalch equation, the concentration ratio of tris and
tris 𝐻 is
[ 𝑡𝑟𝑖𝑠 ]
𝑡𝑟𝑖𝑠 𝐻
[ ]
𝑝𝐻−𝑝𝐾 𝑎 ( )
= 10
7.40−8. = 0. 2128,
meaning that in a given volume of solution, a mole of tris 𝐻 requires 0.2128 moles of tris. This ratio
can then be translated into mole fractions. Tris 𝐻 will have a fraction
χ 𝑡𝑟𝑖𝑠𝐻
+ =^
1
1+0.
and for tris, χ If tris were the sole component in the buffer, it would have 𝑡𝑟𝑖𝑠
1+0.
an amount of 0. 1 𝑚𝑜𝑙 𝐿. However, the buffer solution can only have 0.
−
( ) (1. 000 𝐿^ ) = 0. 1 𝑚𝑜𝑙
times that amount. Therefore, in the buffer, tris 𝐻 can only have an amount of (0.1 mol) (0.8244) =
0.08244 mol. This translates to a tris hydrochloride volume of
𝑡𝑟𝑖𝑠𝐻
+ =^
0.08245 𝑚𝑜𝑙
0.10 𝑚𝑜𝑙 𝐿
−
Tris, on the other hand, will comprise
𝑡𝑟𝑖𝑠
Checking the buffer capacity
- Create two replicates of Buffer 1 [ + Duplicate] and rename each solution as
“Flask A” and “Flask B”. Record the pH in each solution. This will be the initial pH ( 𝑝𝐻) of 𝑖
the solutions.
- Create 0.5- 𝑚𝑜𝑙 𝐿 solutions of HCl and NaOH in separate flasks through dilution by
−
doing the following:
a. Get 2 new flasks from the stockroom. Into each flask pour 50 mL of distilled water.
(After pouring the water, remove it from the workbench.)
b. Rename one flask as 0.5 𝑚𝑜𝑙 𝐿 HCl and the other flask as 0.5 NaOH.
− 𝑚𝑜𝑙 𝐿
−
c. Pour 50 mL of 1 𝑚𝑜𝑙 𝐿 HCl into the flask labelled 0.5 HCl, and 50 mL of 1
−
𝑚𝑜𝑙 𝐿
−
NaOH into the flask labelled 0.5 NaOH. (After pouring the 1 mol L
𝑚𝑜𝑙 𝐿
− 𝑚𝑜𝑙 𝐿
−
HCl and 1 mol L
- NaOH, remove them from the workbench.)
- Into Flask A, add HCl in 1.0-mL increments until the pH changes from pH 5.00 to around
pH 4.00. Record the total volume of the added HCl and calculate the amount in moles (
𝑛 ) of the added HCl. Also record final total volume ( ) and the final pH ( ) of the 𝑎𝑐𝑖𝑑
𝑠𝑜𝑙𝑛
𝑓
solution, and calculate change in pH,
𝑓
𝑖
- Into Flask B, add NaOH in 1.0-mL increments until the pH changes from pH 5.00 to
around pH 5.60. Record the total volume of the added NaOH and calculate the amount in
moles ( 𝑛 ) of the added NaOH. Also record the final total volume ( ) and the final 𝑏𝑎𝑠𝑒
𝑠𝑜𝑙𝑛
pH of the solution, and calculate change in pH.
(Note: In case the pH changes significantly beyond 1.00 pH unit due to too much addition
of acid or base, right click and remove the flask and then create another duplicate from
Buffer 1.)
- Calculate separately the buffer capacity of Flask A and Flask B using the formulas for
determining the buffer capacity:
β 𝐹𝑙𝑎𝑠𝑘 𝐴
𝑛 𝑎𝑐𝑖𝑑
𝑉 𝑠𝑜𝑙𝑛
Δ𝑝𝐻
β 𝐹𝑙𝑎𝑠𝑘 𝐵
𝑛 𝑏𝑎𝑠𝑒
𝑉 𝑠𝑜𝑙𝑛
Δ𝑝𝐻
Activity 3A
Buffer Solutions
Individual Worksheet
Name:
Date:
Section
Code:
Schedule:
This part provides exercises for theoretical calculations. An online
version of this data sheet can be obtained by scanning the QR code in Figure
8 or through the URL: https://forms.gle/z325BPiBdLrYf2Wo8.
Exercises
- Can the following acid-conjugate base pairs be used as a buffer? Write the corresponding
chemical formula of the acid-conjugate base pairs and then check the box corresponding to
your answer.
a. Ammonia and ammonium chloride ☐ Yes ☐ No
b. Acetic acid and potassium acetate ☐ Yes ☐ No
c. Nitric acid and sodium nitrate ☐ Yes ☐ No
d. Phosphoric acid and monopotassium phosphate ☐ Yes ☐ No
e. Chlorous acid and tris ☐ Yes ☐ No
- Consider a glycolic acid ( 𝐻𝑂𝐶𝐻 ) and glycolate ( ) buffer system. Write the 2
2
−
chemical equation representing the following:
a. The glycolic acid-glycolate equilibrium
b. The buffer system titrated (added) with a strong base
c. The buffer system titrated (added) with a strong acid
- Prepare a 250-mL buffer with pH = 4.40 from solid reagents benzoic acid ( 𝐶 ; M = 6
5
122.123 𝑔 𝑚𝑜𝑙 ) and sodium benzoate ( ; M = 144.105 ). The for
−
𝐶 6
5
−
𝑝𝐾 𝑎
benzoic acid is 4.202.
a. Write the balanced equation showing the equilibrium of the buffer solution.
b. Calculate the concentration ratio of benzoate to benzoic acid using the
Henderson-Hasselbalch Equation.
☐ Buffer 1
☐ Buffer 2
☐ Buffer 3
☐ Their buffer capacities are equal
- From the stock solutions of 1.00- 𝑚𝑜𝑙 𝐿 acetic acid and 1.00- of sodium acetate, five
− 𝑚𝑜𝑙 𝐿
−
solutions were prepared with the acetic acid and acetate concentrations shown in the table
below. The 𝑝𝐾 of acetic acid is 4.756. 𝑎
Solutio
n
[Acetic acid],
𝑚𝑜𝑙 𝐿
−
[Acetate],
𝑚𝑜𝑙 𝐿
−
Final volume,
mL
A 0.0000351 0 100
B 0.10 0.10 100
C 0.010 0.010 100
D 0.10 0.15 100
E 0.15 0.10 100
a. Determine the volume of the stock solutions of 1.00- 𝑚𝑜𝑙 𝐿 acetic acid and 1.00-
− 𝑚𝑜𝑙 𝐿
−
of sodium acetate needed to prepare the solutions. Write your answer in the table below.
Show your solution.
Solution
[Acetic acid],
𝑚𝑜𝑙 𝐿
−
Vol. acetic acid,
mL
[Acetate],
𝑚𝑜𝑙 𝐿
−
Vol. of acetate,
mL
Final volume,
mL
A 0.0000351 0 100
B 0.10 0.10 100
C 0.010 0.010 100
D 0.10 0.15 100
E 0.15 0.10 100
b. Calculate for the initial pH of the solutions and write the pH in the table below. Show your
solution.
Solution (^) [Acetic acid], 𝑚𝑜𝑙 𝐿
− [Acetate], 𝑚𝑜𝑙 𝐿
− Initial pH
A* 0.0000351 0
B 0.10 0.
C 0.010 0.
D 0.10 0.
E 0.15 0.
https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php
ax
2
c. Which solution(s) have the same pH?
d. Which solution is closest to the 𝑝𝐾? 𝑎
e. Calculate the pH of the solutions and the change in pH after adding 5.00 mL of 0.10-
𝑚𝑜𝑙 𝐿 NaOH. Show your solution.
−
Solution Initial pH Final pH pH change
A
B
C
D
E
f. Calculate the pH of the solutions and the change in pH after adding 5.00 mL of 0.10-
𝑚𝑜𝑙 𝐿 HCl.
−
Show your solution.
Solution Initial pH Final pH pH change
A
B
C
D
E
g. Which solution is most resistant to pH change?
h. Which solution is least resistant to pH change?
Activity 3B
Buffer Solutions
Group Activity Worksheet
Group Number:
Group Members:
- Complete the data sheet for the determination of the capacity of Buffer 1. Show your solution
for the buffer capacity.
Flask A Flask B
Initial pH, 𝑝𝐻 𝑖
Total volume of added HCl, 𝑉 𝑎𝑐𝑖𝑑
n/a
Amount of acid added, 𝑛 𝑎𝑐𝑖𝑑 n/a
Total volume of added NaOH, 𝑉 𝑏𝑎𝑠𝑒
n/a
Amount of base added, 𝑛 𝑏𝑎𝑠𝑒 n/a
Total volume of solution, 𝑉 𝑠𝑜𝑙𝑛
Final pH, 𝑝𝐻 𝑓
ΔpH
Buffer capacity, β
- Are the capacities obtained in Flask A and Flask B close to each other? Why is it so?
