Deep Dive: Examples #1, 2, and 3
1. Henderson-Hasselbalch
The Henderson-Hasselbalch (HH) equation itself is fine. After all, the HH equation is just an algebraic rearrangement of
the Law of Mass Action
 +
H · [Base]
10−pH · [Base][Base]Ka ==⇐⇒pH = pKa + log10[Acid][Acid][Acid]
Caution: In practice approximations to HH are then made. The purpose of this document is (i) to explain theseassumptions and (ii) to provide a way to check them.
2. Acetic Acid and Sodium Acetate Solutions
Examples #1, 2, and 3 involve combining acetic acid CH3COOH and sodium acetate CH3COONa
Example #1: A buffer is prepared containing 1.00 m acetic acid and 1.00 m sodium acetate. What is its pH?
Example #2: A buffer is prepared containing 0.800 m acetic acid and 1.00 m sodium acetate. What is its pH?
Example #3: A buffer is prepared containing 1.00 m acetic acid and 0.800 m sodium acetate. What is its pH?
The sodium acetate NaAc is assumed to dissociate completely (a very reasonable assumption). The sodium ion Na+
becomes a spectator ion.
NaAc(aq) GGA Na+(aq) + Ac–(aq)Ksp −→ ∞Of course, the acetic acid HAc is a weak acid and experiences this equilibrium reaction+–GGB
HAc(aq) F
GG H (aq) + Ac (aq)Ka = 1.77 × 10−5(2.1)

We use the HH equation to solve these three examples. The “Acid” is the acetic acid HAc and the “Base” is its conjugatebase the acetate ion Ac–
[Ac–][HAc] = equilibrium concentration of HAc
pH = pKa + log10
where[Ac–] = equilibrium concentration of Ac–[HAc]Stoichiometry from eq. (2.1) means we can writepH = pKa + log10[Ac–]◦ + x[HAc]◦ − x[HAc]◦ = initial concentration of HAc[Ac–]◦ = initial concentration of A–
x = change in concentration from HAc to Ac–

where(2.2)

Since x is unknown eq. (2.2) is not a quick way to get the pH.
3. Assumptions
So far eq. (2.2) is exact. It follows the same methods used with the Law of Mass Action previously. To make eq. (2.2)
practical we assume x is very small compared to both of the initial concentrationsx  [HAc]◦

Routine HH Assumption:andx  [Ac–]◦

The above is routinely assumed but rarely mentioned. In other words, the most commonly USED form of the HHequation is[Base]◦pH ≈ pKa + log10(3.1)[Acid]◦
Notice the approximation sign? Again, whether or not mentioned, eq. (3.1) is the most commonly USED form for HH.
Indeed, eq. (3.1) is what most people mean by “using the HH equation.”
Bottom Line:Exact:pH = pKa + log10[Base][Acid]

Appoximate:
1pH ≈ pKa + log10[Base]◦[Acid]◦4. Approximate Results
For these first three examples, this form of HH is what is used to get the numerical answers[Ac–]◦pH ≈ pKa + log10[HAc]◦
Specifically, with pKa = − log10 1.77 × 10−5 = 4.752 we get
1.00
Example #1:
pH ≈ 4.752 + log10−→
pH = 4.752
1.00


1.00
Example #2:
pH≈ 4.752 + log10−→
pH = 4.849
0.800


0.800
Example #3:
pH≈ 4.752 + log10−→
pH = 4.655
1.00(4.1)

Getting these answers for the pH was very fast.
5. Exact Results
If we do NOT assume that x is very small compared to the initial concentrations, then we need to use the Law of Mass
Action without any further approximations
Ka =x · ([Ac–]◦ + x)
[HAc]◦ − x(5.1)

Using the above we would get
Example #1:
Example #2:
Example #3:

x · (1 + x)
1−xx · (1 + x)
1.77 × 10−5 =0.8 − xx· (0.8 + x)
1.77 × 10−5 =1−x1.77 × 10−5 ==⇒x ≈ 1.770 × 10−5=⇒pH ≈ 4.752=⇒x ≈ 1.416 × 10−5=⇒pH ≈ 4.849=⇒x ≈ 2.212 × 10−5=⇒pH ≈ 4.655

XXX

The above took much more time to calculate. For these three examples, we see pH’s using the approximate HH equationeq. (4.1) match the exact pH’s using eq. (5.1) to at least three sig figs.
6. Five-Percent Rules
But what do we do if we only use the approximate HH equation eq. (4.1)? We still would like to check our answers. We
now define a 5% rule. Start with these assumptionsx< 5%[Acid]◦andx< 5%[Base]◦

Taking the negative logarithm of both sides of the Acid version we getx− log10> − log10 (0.050)=⇒log10 [Acid]◦ + pH > 1.3[Acid]◦Thus, a 5% rule requires for the initial concentration of both the Acid and Base that
log10 [Acid]◦ + pH > 1.3

andlog10 [Base]◦ + pH > 1.3(6.1)

We apply these checks to the examples. The initial concentrations were 1.00 m and 0.800 m so we havelog10 (1.00) = 0andlog10 (0.800) = −0.097

The approximate pH’s were 4.752, 4.655, and 4.849. We can see (without a calculator) that these pH’s are large enoughthat adding either 0 or −0.097 will still be greater than 1.3. This means these approximate pH’s from eq. (4.1) are good
to within 5%. (Of course, we knew this already because we calculated the exact results.)2

7. A Case That Fails
Here is an example where the approximations fail. Consider the case of [HAc] = 1.0 × 10−3 m and [NaAc] = 1.0 × 10−4 m.
The approximation form of HH gives[Ac–]◦pH ≈ pKa + log10≈ 3.74[HAc]◦
Next, we check with log10 [∗]◦ + pH > 1.3
log10 [Acid]◦ + pH = −3 + 3.74 = 0.74
log10 [Base]◦ + pH = −4 + 3.74 = −0.26
Since both 0.74 and −0.26 are less than 1.3, the approximate pH is off by more than 5%. So we need to use the Law ofMass Action. Doing this we find the exact result is pH ≈ 4.06. So for this example, the HH approximation would have anerror of about 8%. There is no way we would have known this unless we check using log10 [∗]◦ + pH > 1.3.

8. Summary
The following equation is exact but only useful if we already know the equilibrium concentrations of the acid and base
Exact:

pH = pKa + log10[Base][Acid]

(8.1)

The following equation is approximate and useful if we want to estimate the final pH using just the initial concentrations
Approximate:pH ≈ pKa + log10[Base]◦[Acid]◦

(8.2)

Important: People write eq. (8.1) when solving buffer problem but it is really eq. (8.2) that they use.
The pH estimated using eq. (8.2) has an error of less than 5% provided both initial concentrations [∗]◦ and estimated pHsatisfy
Equation (8.2) Has Less Than 5% Provided That:

log10 [∗]◦ + pH > 1.3

The above check indicates that small initial concentrations [∗] and/or small pH’s may cause more than 5% error. Theerror involved using eq. (8.2) instead of eq. (8.1) will increase as(i) acid strength increases
bigger Ka lead to bigger x so the initial concentrations may be far off from those at equilibrium(ii) initial concentrations [∗]◦ decreasesmaller starting concentrations means that percentage-wise changes are more significant

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