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Buffer Calculations
A buffer is a solution that has the ability to resist a change in pH upon the addition of a strong acid or a strong base. For a buffer to exist it must satisfy two conditions: (1) the solution must contain a weak acid and its conjugate weak base, and (2) the concentrations of the weak acid and the conjugate weak base must be reasonably similar. This essay discusses in more detail how to calculate the pH of a buffer, how to evaluate the change in a buffer’s pH after adding a strong acid or a strong base, and how to evaluate a buffer’s capacity to neutralize a strong acid or a strong base.
The Standard Equilibrium Approach to Calculating a Buffer’s pH
Thus far our standard approach to finding the pH of a solution is to write the equilibrium reaction responsible for controlling pH and use an ICE table to organize information. For example, to find the pH of a buffer that consists of 3_._ 39 × 10 −^2 M CH 3 COOH and 1_._ 00 × 10 −^2 M CH 3 COO–, we first write the acid dissociation reaction for acetic acid
CH 3 COOH( aq ) + H 2 O( l ) H 3 O+( aq ) + CH 3 COO–( aq )
for which the equilibrium constant is
K a =
[H 3 O+][CH 3 COO–]
[CH 3 COOH]
= 1. 8 × 10 −^5
Next, we use an ICE table to organize the problem
CH 3 COOH( aq ) + H 2 O( l ) H 3 O+( l ) + CH 3 COO–( aq ) initial 3_._ 39 × 10 −^2 — 0 1_._ 0 × 10 −^2 change − X — + X + X equilibrium 3_._ 39 × 10 −^2 − X — + X 1_._ 0 × 10 −^2 + X
Note that this table includes the initial concentrations for CH 3 COOH and for CH 3 COO–, which were given to us, and assumes that we can ignore the presence of H 3 O+^ from water (our standard assumption).^1 Defining the change in acetic acid’s concentration as X , we then fill in the remainder of the table. Next, we substitute the terms in the row labeled “equilibrium” into the equilibrium constant expression.
K a =
[H 3 O+][CH 3 COO–]
[CH 3 COOH]
( X )(1. 0 × 10 −^2 + X )
3. 39 × 10 −^2 − X
= 1. 8 × 10 −^5
To simplify the problem, we make the following two assumptions
3. 39 × 10 −^2 − X ≈ 3. 39 × 10 −^2
1. 0 × 10 −^2 + X ≈ 1. 0 × 10 −^2
Substituting these assumptions into the K a expression and solving gives
(^1) Note, as well, that we ignore water, H 2 O as it is not included in the equilibrium constant expression.
K a =
[H 3 O+][CH 3 COO–]
[CH 3 COOH]
( X )(1. 0 × 10 −^2 )
3. 39 × 10 −^2
= 1. 8 × 10 −^5
X = 6. 10 × 10 −^5
Before accepting this solution, we verify the assumptions by calculating the errors introduced into the calculation, finding that they are
% error =
3. 39 × 10 −^2 − (3. 39 × 10 −^2 − 6. 10 × 10 −^5 )
3. 39 × 10 −^2
× 100 = +0. 21%
% error =
1. 0 × 10 −^2 − (1. 0 × 10 −^2 − 6. 10 × 10 −^5 )
1. 0 × 10 −^2
× 100 = − 0. 6%
Clearly our assumptions are within reason; thus, we know that [H 3 O+] = X = 6_._ 10 × 10 −^5 and that the buffer’s pH is 4.21.
The Henderson-Hasselbalch Equation
Now, let’s look at buffers in a slightly different way. When we first introduced buffers in class we wrote the K a expression in a logarithmic form called the Henderson-Hasselbalch equation. For an acetic acid/acetate buffer this equation is
pH = p K a + log
[CH 3 COO–]
[CH 3 COOH]
There are a two important things to note about this equation. First, the equation shows us that the pH of an acetic acid/acetate buffer is always at a pH level similar to acetic acid’s p K a value. Given that K a for CH 3 COOH is 1_._ 8 × 10 −^5 , the pH of an acetic acid/acetate buffer is always near
pH ≈ p K a = −log(1_._ 8 × 10 −^5 ) = 4_._ 74
The actual pH will deviate from this, being more acidic when the concentration of CH 3 COOH is greater than that for CH 3 COO–, and more basic when CH 3 COO–^ is present at a concentration greater than that for CH 3 COOH.
For a buffer to exist the concentrations of the conjugate acid and the conjugate base cannot differ from each other by too much. As a guideline, we will adopt the following convention: a buffer exists when the relative amount of conjugate base-to-conjugate acid is within a range of 0.1–
conugate base conjugate acid
Under these conditions, a buffer’s pH falls within the limits
p K a − 1 ≤ pH ≤ p K a + 1
which simplifies to
pH = p K a + log
(moles CH 3 COO–)o (moles CH 3 COOH)o
This equation is nice because it allows us to calculate the buffer’s pH without having to first calculate the concentrations of the buffering species. As we will see shortly, this is useful when we determine how the addition of a strong acid or a strong base will change a buffer’s pH. Example 1. What is the pH of a buffer prepared by dissolving 25.0 g of K 2 HPO 4 and 25.0 g of KH 2 PO 4 in 250 mL of water? Solution. First, we determine which species is the weak acid and which is the conjugate weak base. Both compounds are ionic and will dissociate in water. The K+^ ions have no effect on the solution’s pH, so we ignore them. The remaining anions are amphiprotic, meaning they have both acidic and basic properties. Because dihydrogen phosphate has one more proton than monohydrogen phosphate, H 2 PO 4 – , is the weak acid. To find the pH of the resulting buffer, we find the initial moles of weak acid
25_._ 0 g KH 2 PO 4 ×
1 mol KH 2 PO 4 136_._ 1 g KH 2 PO 4
= 0_._ 1837 mol H 2 PO 4 –
and the initial moles of weak base
25_._ 0 g K 2 HPO 4 × 1 mol K 2 HPO 4 174_._ 2 g K 2 HPO 4
= 0_._ 1435 mol HPO 4 2–
Although the recipe contains information on the amount of water used to prepare the buffer, we don’t need to include this in our calculation (although we will use it in the next example). Knowing the moles of H 2 PO 4 – and HPO 4 2– , we complete the calculation using the p K a value for H 2 PO 4 – ; thus
pH = p K a + log
(mol HPO 4 2– )o (mol H 2 PO 4 – )o
= 7_._ 20 + log
Finding a Buffer’s pH After Adding a Strong Acid or Strong Base
When we add a strong base to a buffer it reacts with some of the conjugate weak acid and converts it into the conjugate weak base. For example, the following reaction takes place when we add a strong base, OH–, to the buffer from the Example 1
H 2 PO 4 – ( aq ) + OH–( aq ) H 2 O( l ) + HPO 4 2– ( aq )
Note we write this as a reaction that goes to completion. If you look at the reaction in reverse (from right-to-left), you will see that it is a weak base, HPO 4 2– , reacting with water to form its conjugate weak acid, H 2 PO 4 – , and hydroxide ion, which is the standard form for a base dissociation reaction. The equilibrium constant for the reverse reaction, therefore, is K b for HPO 4 2– , or 1_._ 59 × 10 −^7. This means the equilibrium constant for the reaction of H 2 PO 4 – and OH–^ is (1_._ 59 × 10 −^7 )−^1 or 6_._ 3 × 106. Clearly the reaction essentially goes to completion. Knowing that the reaction goes to completion makes it easy to find the moles of H 2 PO 4 – and the moles of HPO 4 2– after the addition of OH–; these are
moles H 2 PO 4 – = (moles H 2 PO 4 – )o − moles OH–
moles HPO 4 2– = (moles HPO 4 2– )o + moles OH–
Substituting into the Henderson-Hasselbalch equation gives
pH = p K a + log (HPO 4 2– )o + moles OH– (H 2 PO 4 – )o − moles OH–
If we add a strong acid instead of a strong base, the result is
pH = p K a + log
(HPO 4 2– )o − moles H 3 O+ (H 2 PO 4 – )o + moles H 3 O+
because adding H 3 O+^ converts some of the HPO 4 2– to H 2 PO 4 –.
Example 2. What is the pH if we add 10.00 mL of 1.00 M NaOH to the buffer from Example 1? Solution. The moles of NaOH added is
10_._ 00 mL NaOH ×
1 L
1000 mL
×
1 mol OH– 1 L NaOH = 0_._ 0100 mol OH–
Substituting into the Henderson-Hasselbalch equation gives
pH = p K a + log
(HPO 4 2– )o + moles OH– (H 2 PO 4 – )o − moles OH–^
= 7_._ 20 + log
Note that adding NaOH raises the pH by just 0.06 pH units, demonstrating that the buffer can neutralize a strong base with only a small change in pH. Adding a similar quantity of NaOH to an equivalent amount of water (250.0 mL) gives
[OH–] =
moles NaOH total volume
0_._ 0100 mol 0_._ 250 L + 0_._ 0100 L
= 0. 0385 M
or a pOH of 1.41 and a pH of 12.59. With the water initially at a pH of 7 (neutral), the change in pH for this unbuffered solution is 5.59 pH units.
Evaluating a Buffer’s Capacity
Adding a strong acid or a strong base slowly destroys a buffer’s ability to do its job because the ratio
[conjugate base] [conjugate acid]
in the Henderson-Hasselbalch equation changes. Once this ratio falls outside range of 0.1 to 10 a buffer ceases to exist. We define a buffer’s buffer capacity as the moles of strong acid or strong base the buffer can neutralize before it reaches the end of its range.^2 When we add a strong acid this means that the buffer capacity is defined by the equation
(moles weak base)o − moles H 3 O+ (moles weak acid)o + moles H 3 O+^
(^2) There is a more quantitatively accurate description of buffer’s maximum capacity; however, in the context of this course, this definition is sufficient.
0. 1435 − X
0. 1837 + X
0. 09207 + 0. 5012 X = 0. 1435 − X
1. 5012 X = 0. 05143
X = 3_._ 43 × 10 −^2 mol H 3 O+
Thus, the buffer capacity for this scenario is 3_._ 43 × 10 −^2 moles of strong acid.
