What this tool does
This calculator finds the new pH of a buffer after adding a small amount of strong
acid or strong base, using the Henderson–Hasselbalch (H–H) equation. It follows the two-part
logic shown in the figures:
stoichiometric part (neutralization + mixing) and then the
equilibrium part via H–H.
Buffer recap & Henderson–Hasselbalch
HA ⇌ A− + H3O+
For a weak acid, \(K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\).
Taking \(-\log_{10}\) gives
\[
\mathrm{pH} = \mathrm{p}K_a + \log_{10}\!\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right).
\]
For a weak base buffer \( \mathrm{B}/\mathrm{BH^+} \) one may write
\( \mathrm{pOH} = \mathrm{p}K_b + \log_{10}\!\big([\mathrm{BH^+}]/[\mathrm{B}]\big) \)
and then \( \mathrm{pH} = 14.00 - \mathrm{pOH} \).
Assumptions for H–H: (i) activities ≈ concentrations (dilute solutions),
(ii) both buffer components are present after neutralization,
(iii) the subsequent equilibrium shift is small compared with the stoichiometric amounts
(common-ion suppression).
Stoichiometric part (neutralization)
Add strong acid to a buffer → convert base to acid:
A− + H+ → HA.
Add strong base to a buffer → convert acid to base:
HA + OH− → A− + H2O.
Compute moles, subtract the limiting reagent, and account for the new total volume
\( V_{\text{tot}} \) after mixing.
If the strong reagent is small compared with the buffer components, the new ratio can be taken
with moles directly (volume cancels):
\[
\mathrm{pH} = \mathrm{p}K_a + \log_{10}\!\left(\frac{n_{\mathrm{A^-}}'}{n_{\mathrm{HA}}'}\right)
\qquad\text{or}\qquad
\mathrm{pOH} = \mathrm{p}K_b + \log_{10}\!\left(\frac{n_{\mathrm{BH^+}}'}{n_{\mathrm{B}}'}\right).
\]
If either numerator or denominator becomes ≤ 0 (one component entirely consumed), H–H is not valid.
Use an exact equilibrium/hydrolysis treatment or strong-acid/strong-base pH instead.
Equilibrium part (why H–H works)
After the stoichiometric step the solution is already close to its new equilibrium because a
common ion (either \( \mathrm{A^-} \) or \( \mathrm{BH^+} \)) is abundant. An ICE summary for the
weak-acid case illustrates the small subsequent shift \(x\):
| HA | A⁻ | H₃O⁺ |
|---|
| Initial (after mixing) | \(C_{\!HA}'\) | \(C_{\!A^-}'\) | 0 |
| Change | \(-x\) | \(+x\) | \(+x\) |
| Equilibrium | \(C_{\!HA}'-x\) | \(C_{\!A^-}'+x\) | \(x\) |
Using \(K_a=\dfrac{x(C_{\!A^-}'+x)}{C_{\!HA}'-x}\) and assuming \(x\ll C'\) recovers the H–H form
\( \mathrm{pH} = \mathrm{p}K_a + \log_{10}(C_{\!A^-}'/C_{\!HA}') \).
Procedure used by the calculator
- Read buffer type (acid buffer \( \mathrm{HA}/\mathrm{A^-} \) or base buffer \( \mathrm{B}/\mathrm{BH^+} \)),
initial concentrations and volumes, and the amount of strong acid/base added.
- Stoichiometry: convert the appropriate amount:
- Added acid: \( n_{\mathrm{A^-}}' = n_{\mathrm{A^-}} - n_{\text{added}} \), \( n_{\mathrm{HA}}' = n_{\mathrm{HA}} + n_{\text{added}} \).
- Added base: \( n_{\mathrm{HA}}' = n_{\mathrm{HA}} - n_{\text{added}} \), \( n_{\mathrm{A^-}}' = n_{\mathrm{A^-}} + n_{\text{added}} \).
Set \( V_{\text{tot}} = V_{\text{buffer}} + V_{\text{added}} \), and
\( C' = n'/V_{\text{tot}} \) if concentrations are needed.
- H–H step: compute
\[
\mathrm{pH} = \mathrm{p}K_a + \log_{10}\!\left(\frac{n_{\mathrm{A^-}}'}{n_{\mathrm{HA}}'}\right)
\]
for acid buffers, or
\[
\mathrm{pOH} = \mathrm{p}K_b + \log_{10}\!\left(\frac{n_{\mathrm{BH^+}}'}{n_{\mathrm{B}}'}\right),\quad
\mathrm{pH} = 14.00 - \mathrm{pOH}
\]
for base buffers.
- Report the final pH (and pOH) and show a compact ICE table to explain the small equilibrium shift.
Edge cases
- Component exhausted. If \(n_{\mathrm{A^-}}' \le 0\) or \(n_{\mathrm{HA}}' \le 0\) (or the analogous base-buffer
case), the H–H equation is not valid. The solution behaves as a salt-only hydrolysis or is dominated by
excess strong acid/base.
- Large additions. When the added strong reagent is comparable to or larger than the buffer amounts,
expect noticeable pH changes and increasing deviation from H–H.
Worked pattern (matches the figures)
Start with \(0.300\ \mathrm{L}\) of a buffer \( [\mathrm{HA}]=0.250\ \mathrm{M} \),
\( [\mathrm{A^-}]=0.560\ \mathrm{M} \). Add \(0.00600\ \mathrm{mol}\) strong acid:
\[
n_{\mathrm{HA}}' = 0.250\cdot0.300 + 0.00600 = 0.0810\ \mathrm{mol},\qquad
n_{\mathrm{A^-}}' = 0.560\cdot0.300 - 0.00600 = 0.174\ \mathrm{mol}
\]
\[
\mathrm{pH} = \mathrm{p}K_a + \log_{10}\!\left(\frac{0.174}{0.0810}\right)
= \mathrm{p}K_a + 0.340 \;(\text{≈ values in the figure}).
\]
Tip: When both buffer components end up divided by the same \(V_{\text{tot}}\), the H–H ratio can be formed
directly with moles, avoiding an extra dilution step.
