Why is the answer 4.92 in one key and 4.94 in another?

It depends on the acid constant you are given. Ka = 1.8 × 10^-5 gives pKa 4.74 and pH 4.92. Some textbooks give pKa = 4.76 directly, which gives pH = 4.76 + 0.176 = 4.94. If you are given Ka = 1.75 × 10^-5 instead, pKa = 4.757 and the pH is 4.93. Use the value printed in your problem.

When can I use Henderson-Hasselbalch instead of an ICE table?

When both the weak acid and its conjugate base are present in significant amounts (both well above Ka, and their ratio between about 0.1 and 10). For a weak acid alone, use an ICE table.

Chemistry 2: pH of a Buffer of 0.10 M Acetic Acid and 0.15 M Sodium Acetate (Henderson-Hasselbalch)

The Problem

A buffer is made so that it contains 0.10 M acetic acid ($\text{CH}_3\text{COOH}$) and 0.15 M sodium acetate ($\text{CH}_3\text{COONa}$). For acetic acid, $K_a = 1.8\times10^{-5}$.

(a) Calculate the pH of the buffer.

(b) Calculate the pH after 0.010 mol of HCl is added to 1.00 L of this buffer (assume no volume change).

Part (a): pH of the Buffer

Step 1: Identify the pair. Weak acid $\text{HA} = \text{CH}_3\text{COOH}$ (0.10 M). Conjugate base $\text{A}^- = \text{CH}_3\text{COO}^-$ from sodium acetate (0.15 M). Na⁺ is a spectator.

Step 2: Find pKa.

$$ \text{p}K_a = -\log(1.8\times10^{-5}) = 4.74 $$

Step 3: Apply Henderson-Hasselbalch.

$$ \text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} = 4.74 + \log\frac{0.15}{0.10} $$

$$ \text{pH} = 4.74 + \log(1.5) = 4.74 + 0.176 = \mathbf{4.92} $$

Sense check: there is more base than acid, so the pH should be a little above the pKa. 4.92 > 4.74, as expected.

Part (b): Add 0.010 mol HCl to 1.00 L

Step 1: Moles before. In 1.00 L: HA = 0.10 mol, A⁻ = 0.15 mol.

Step 2: The strong acid reacts completely with the base:

$$ \text{CH}_3\text{COO}^- + \text{H}^+ \longrightarrow \text{CH}_3\text{COOH} $$

Species	Before (mol)	Change (mol)	After (mol)
CH₃COO⁻	0.15	−0.010	0.14
H⁺ (from HCl)	0.010	−0.010	0
CH₃COOH	0.10	+0.010	0.11

Step 3: New pH. The volume is the same for both species, so moles can go straight into the ratio:

$$ \text{pH} = 4.74 + \log\frac{0.14}{0.11} = 4.74 + 0.105 = \mathbf{4.85} $$

The pH drops by only about 0.07 units. The same 0.010 mol HCl in 1.00 L of pure water would give pH 2.00. That contrast is the whole point of a buffer.

Bonus (NaOH instead): 0.010 mol NaOH turns 0.010 mol HA into A⁻, giving $\text{pH} = 4.74 + \log\frac{0.16}{0.09} = 4.99$.

Traps That Cost Points

Flipping the ratio. It is base over acid: $\log\frac{[\text{A}^-]}{[\text{HA}]}$. Acid over base gives 4.57, a common wrong choice.
Doing an equilibrium on the strong acid. Added HCl or NaOH reacts to completion first (stoichiometry table), and only then do you use Henderson-Hasselbalch.
Using concentrations of HCl in a different volume. If the added acid changes the total volume, convert everything to moles first; the volume cancels in the ratio.
Using pKb. For an ammonia/ammonium buffer you need the pKa of NH₄⁺ (14.00 − pKb), not pKb.

When This Shortcut Fails

Henderson-Hasselbalch assumes the acid and base concentrations barely change from their starting values. That is true here (both 0.10 M or more, Ka is tiny). For a weak acid with no added conjugate base, set up an ICE table instead, like the 0.150 M acetic acid ICE table pH problem.