What this calculator does
You mix a weak acid HA with a strong base (or a weak base B with a strong acid).
A stoichiometric neutralization forms the conjugate ion (the common ion) and the total volume changes.
Using the post-mix concentrations, the tool then solves the exact equilibrium to obtain
\([\text{H}_3\text{O}^+]\), \([\text{OH}^-]\), species concentrations, and the resulting pH/pOH.
Symbols
- \(C_{\text{weak}},\,V_{\text{weak}}\): initial concentration/volume of HA (or B).
- \(C_{\text{tit}},\,V_{\text{tit}}\): concentration/volume of the strong titrant (OH⁻ or H⁺).
- \(K_a\) (acid case) or \(K_b\) (base case); \(K_w\) at the given temperature.
- \(C_{\text{weak,mix}},\,C_{\text{conj,mix}}\): concentrations after mixing (before equilibrium shift).
- \(x\): equilibrium change; the produced \([\text{H}_3\text{O}^+]\) or \([\text{OH}^-]\) equals \(x\).
Step 1 — Stoichiometric neutralization (moles)
HA + OH⁻ → A⁻ + H₂O (acid mixed with strong base)
B + H⁺ → BH⁺ (base mixed with strong acid)
Compute moles:
\(n_{\text{weak},0}=C_{\text{weak}}V_{\text{weak}}\),\;
\(n_{\text{tit},0}=C_{\text{tit}}V_{\text{tit}}\).
The limiting reagent is consumed; total volume is \(V_{\text{tot}}=V_{\text{weak}}+V_{\text{tit}}\).
- If \(n_{\text{weak},0}>n_{\text{tit},0}\): leftover weak; conjugate formed \(=n_{\text{tit},0}\).
- If \(n_{\text{weak},0}=n_{\text{tit},0}\): equivalence; only conjugate remains.
- If \(n_{\text{weak},0}
Post-mix concentrations: \(C_{\text{weak,mix}}=\dfrac{n_{\text{weak}}}{V_{\text{tot}}}\),\;
\(C_{\text{conj,mix}}=\dfrac{n_{\text{conj}}}{V_{\text{tot}}}\).
Step 2 — Exact equilibrium relations
Acid path (HA with A⁻ present)
HA + H₂O ⇌ A⁻ + H₃O⁺
ICE table (after mixing)
| HA | A⁻ | H₃O⁺ |
|---|
| Initial | \(C_{\text{weak,mix}}\) | \(C_{\text{conj,mix}}\) | 0 |
| Change | −\(x\) | +\(x\) | +\(x\) |
| Equilibrium | \(C_{\text{weak,mix}}-x\) | \(C_{\text{conj,mix}}+x\) | \(x\) |
\(K_a=\dfrac{[\text{A}^-][\text{H}_3\text{O}^+]}{[\text{HA}]}=
\dfrac{x\,\big(C_{\text{conj,mix}}+x\big)}{C_{\text{weak,mix}}-x}\).
Rearranging:
\(x^2+\big(C_{\text{conj,mix}}+K_a\big)x-K_a\,C_{\text{weak,mix}}=0\),
with the positive root used for \(x=[\text{H}_3\text{O}^+]\).
Base path (B with BH⁺ present)
B + H₂O ⇌ BH⁺ + OH⁻
ICE table (after mixing)
| B | BH⁺ | OH⁻ |
|---|
| Initial | \(C_{\text{weak,mix}}\) | \(C_{\text{conj,mix}}\) | 0 |
| Change | −\(x\) | +\(x\) | +\(x\) |
| Equilibrium | \(C_{\text{weak,mix}}-x\) | \(C_{\text{conj,mix}}+x\) | \(x\) |
\(K_b=\dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}=
\dfrac{x\,\big(C_{\text{conj,mix}}+x\big)}{C_{\text{weak,mix}}-x}\).
Rearranging:
\(x^2+\big(C_{\text{conj,mix}}+K_b\big)x-K_b\,C_{\text{weak,mix}}=0\),
with \(x=[\text{OH}^-]\).
Step 3 — From \(x\) to pH/pOH
- Acid path: \([\text{H}_3\text{O}^+] = x\), \([\text{OH}^-] = K_w/[\text{H}_3\text{O}^+]\).
- Base path: \([\text{OH}^-] = x\), \([\text{H}_3\text{O}^+] = K_w/[\text{OH}^-]\).
- \(\mathrm{pH}=-\log_{10}[\text{H}_3\text{O}^+]\), \(\mathrm{pOH}=-\log_{10}[\text{OH}^-]\); at 25 °C, \(\mathrm{pH}+\mathrm{pOH}=14.00\).
