How is the equation modified for a weak base buffer?

For a weak base (B) and its conjugate acid (BH+), use the form $pOH = pK_b + \log([BH^+] / [B])$, then find pH using $pH = 14.00 - pOH$.

What happens if a strong acid is added to the buffer?

Assume the strong acid reacts completely with the conjugate base. Subtract the moles of added strong acid from the conjugate base and add them to the weak acid, then recalculate the ratio.

When does the Henderson-Hasselbalch equation fail?

It fails for highly dilute buffers (concentrations below $10^{-3}$ M) or when the $K_a$ is relatively large (greater than $10^{-3}$), because the assumption that equilibrium concentrations equal initial concentrations breaks down.

How to use Henderson-Hasselbalch for a buffer

The Henderson-Hasselbalch equation computes the pH of a buffer solution from the acid dissociation constant and the ratio of conjugate base to weak acid concentrations. It applies when the x-is-small approximation is valid, requiring the initial concentrations of the acid and base to be at least 100 times larger than the dissociation constant.

The setup

Identify the weak acid (HA) and its conjugate base (A-) in the buffer system. Determine their initial molar concentrations, $[HA]$ and $[A^-]$ . Obtain the acid dissociation constant, $K_a$ , for the weak acid.

The steps

Convert the dissociation constant to its negative base-10 logarithm: $pK_a = -\log(K_a)$ . 2. Calculate the ratio of the conjugate base concentration to the weak acid concentration: $[A^-] / [HA]$ . 3. Substitute these values into the equation: $pH = pK_a + \log([A^-] / [HA])$ . 4. Evaluate the logarithm and add it to the $pK_a$ to find the pH.

Checking the result

Compare the calculated pH to the $pK_a$ . If $[A^-] > [HA]$ , the logarithmic term is positive, and the $pH$ must be greater than the $pK_a$ . If $[HA] > [A^-]$ , the logarithmic term is negative, and the $pH$ must be less than the $pK_a$ . If concentrations are equal, $pH = pK_a$ .

Common errors

A frequent error is inverting the ratio to $[HA] / [A^-]$ . Another is using the $pK_b$ of the conjugate base instead of the $pK_a$ of the weak acid. Finally, when mixing two solutions to form a buffer, failing to recalculate the new diluted concentrations or moles before using the equation will yield incorrect results.

Worked example

Calculate the pH of a buffer solution that is 0.150 M in acetic acid ( $CH_3COOH$ ) and 0.250 M in sodium acetate ( $CH_3COONa$ ). The $K_a$ for acetic acid is $1.8 imes 10^{-5}$ .

$[HA] = 0.150$ M $[A^-] = 0.250$ M $K_a = 1.8 imes 10^{-5}$ $pK_a = -\log(1.8 imes 10^{-5}) = 4.74$ $pH = 4.74 + \log(0.250 / 0.150)$ $pH = 4.74 + \log(1.667)$ $pH = 4.74 + 0.222$ $pH = 4.96$

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References: OpenStax Chemistry 2e, Chapter 14: Acid-Base Equilibria · Zumdahl Chemistry, Chapter 15: Applications of Aqueous Equilibria