pH of strong & weak acids and bases, salt hydrolysis, strong acid–strong base titration curves, and [H+] ↔ pH / Ka ↔ pKa converters — all in one free chemistry calculator.
Strong acids and bases dissociate completely: pH = −log[H+], pOH = −log[OH−], pH + pOH = 14.00 (25 °C). For sulfuric acid the second proton is handled with Ka2 = 1.2×10−2.
Weak acid: HA ⇌ H+ + A−. Approximation x = √(Ka·C) is used when ionization is below 5%; otherwise the exact quadratic is solved.
Weak base: B + H₂O ⇌ BH+ + OH−. Same 5% rule as weak acids, then pH = 14.00 − pOH.
Salt hydrolysis. Choose the salt type, then enter the relevant Ka/Kb of the weak parent.
Strong acid–strong base titration. Enter concentrations and volumes; the curve shows pH from 0 to twice the equivalence volume.
Quick converters. Kw = 1.0×10−14 at 25 °C, so pH + pOH = 14.00.
Reference values compiled from standard general-chemistry data tables; small differences between textbooks are normal.
| Acid | Formula | Ka | pKa |
|---|---|---|---|
| Sulfuric acid (2nd proton) | H₂SO₄ | 1.2×10⁻² | 1.92 |
| Oxalic acid | H₂C₂O₄ | Ka1 5.9×10⁻², Ka2 6.4×10⁻⁵ | 1.23 / 4.19 |
| Phosphoric acid | H₃PO₄ | Ka1 7.5×10⁻³, Ka2 6.2×10⁻⁸, Ka3 4.2×10⁻¹³ | 2.12 / 7.21 / 12.38 |
| Citric acid | H₃C₆H₅O₇ | Ka1 7.4×10⁻⁴, Ka2 1.7×10⁻⁵, Ka3 4.0×10⁻⁷ | 3.13 / 4.77 / 6.40 |
| Hydrofluoric acid | HF | 6.8×10⁻⁴ | 3.17 |
| Nitrous acid | HNO₂ | 4.5×10⁻⁴ | 3.35 |
| Formic acid | HCOOH | 1.8×10⁻⁴ | 3.74 |
| Lactic acid | CH₃CH(OH)COOH | 1.4×10⁻⁴ | 3.85 |
| Benzoic acid | C₆H₅COOH | 6.3×10⁻⁵ | 4.20 |
| Acetic acid | CH₃COOH | 1.8×10⁻⁵ | 4.74 |
| Carbonic acid | H₂CO₃ | Ka1 4.3×10⁻⁷, Ka2 5.6×10⁻¹¹ | 6.37 / 10.25 |
| Hypochlorous acid | HClO | 3.0×10⁻⁸ | 7.52 |
| Hydrocyanic acid | HCN | 4.9×10⁻¹⁰ | 9.31 |
| Ammonium ion | NH₄⁺ | 5.6×10⁻¹⁰ | 9.25 |
| Phenol | C₆H₅OH | 1.3×10⁻¹⁰ | 9.89 |
| Base | Formula | Kb | pKb |
|---|---|---|---|
| Ethylamine | C₂H₅NH₂ | 5.6×10⁻⁴ | 3.25 |
| Dimethylamine | (CH₃)₂NH | 5.1×10⁻⁴ | 3.29 |
| Methylamine | CH₃NH₂ | 4.4×10⁻⁴ | 3.36 |
| Trimethylamine | (CH₃)₃N | 6.4×10⁻⁵ | 4.19 |
| Ammonia | NH₃ | 1.8×10⁻⁵ | 4.74 |
| Pyridine | C₅H₅N | 1.7×10⁻⁹ | 8.77 |
| Aniline | C₆H₅NH₂ | 4.3×10⁻¹⁰ | 9.37 |
| Problem | Result |
|---|---|
| pH of 0.10 M HCl (strong acid) | pH = 1.00 |
| pH of 0.10 M NaOH (strong base) | pOH = 1.00 → pH = 13.00 |
| pH of 0.10 M acetic acid (Ka = 1.8×10⁻⁵) | pH ≈ 2.87 (exact quadratic: 2.88) |
| pH of 0.10 M ammonia (Kb = 1.8×10⁻⁵) | pH ≈ 11.13 |
| pH of 0.10 M H₂SO₄ (diprotic) | [H⁺] ≈ 0.110 M → pH ≈ 0.96 |
| pH of 0.10 M NaHCO₃ (amphiprotic) | pH = (6.37 + 10.25)/2 ≈ 8.31 |
| 25.0 mL 0.100 M HCl + 25.0 mL 0.100 M NaOH | Equivalence point → pH = 7.00 |
Exactly 1.00. Hydrochloric acid is a strong acid, so it dissociates completely: [H+] = 0.10 mol/L and pH = -log(0.10) = 1.00. Enter any strong-acid concentration in the calculator's first tab to get the same result instantly.
Use the dissociation constant Ka. For 0.10 M acetic acid (Ka = 1.8 x 10^-5): x = sqrt(Ka x C) = sqrt(1.8 x 10^-6) = 1.34 x 10^-3 mol/L, so pH = -log(1.34 x 10^-3) = 2.87. The calculator applies the 5% rule automatically: if ionization exceeds 5% of C, it solves the exact quadratic x^2 + Ka*x - Ka*C = 0 instead (which gives pH 2.88 for this example).
13.00. Sodium hydroxide is a strong base: [OH-] = 0.10 mol/L, so pOH = -log(0.10) = 1.00, and pH = 14.00 - pOH = 13.00 at 25 C. The strong acid/base tab of the calculator handles both directions.
pH = -log[H+]. For example, [H+] = 1.0 x 10^-3 mol/L gives pH = 3.00, and [H+] = 1.34 x 10^-3 mol/L gives pH = 2.87. The Convert tab does this in both directions, plus [OH-] to pOH.
About 11.13. Ammonia is a weak base (Kb = 1.8 x 10^-5): [OH-] = sqrt(Kb x C) = 1.34 x 10^-3 mol/L, pOH = 2.87, and pH = 14.00 - 2.87 = 11.13. Select ammonia in the weak-base tab - its Kb is pre-filled.
About 5.13 - it is acidic. NH4Cl is the salt of a strong acid and a weak base, so the NH4+ ion hydrolyzes: Kh = Kw/Kb = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10. Then [H+] = sqrt(Kh x C) = 7.45 x 10^-6 mol/L and pH = 5.13. Use the Salt tab for this calculation.
Exactly 7.00. At equivalence, only water and a neutral salt remain, so neither H+ nor OH- is in excess. Before equivalence, excess H+ (or OH-) sets the pH; the Titration tab computes the pH at any added volume and draws the full curve with the equivalence point marked.
pKa = -log Ka. For acetic acid, Ka = 1.8 x 10^-5, so pKa = 4.74. The reverse is Ka = 10^-pKa. The Convert tab handles Ka to pKa and pKa to Ka instantly.
Treat the first proton as strong and the second with Ka2. For 0.10 M H2SO4: the first proton gives [H+] = 0.10 M, and the second proton (Ka2 = 1.2 x 10^-2) adds about 0.0098 M via the quadratic, for a total [H+] of about 0.110 M and pH about 0.96. Select 'Diprotic strong acid (H2SO4)' in the strong acid tab.
About 8.31. HCO3- is amphiprotic, so pH = (pKa1 + pKa2)/2 = (6.37 + 10.25)/2 = 8.31, independent of concentration. Choose 'Amphiprotic' in the Salt tab and enter the two pKa values.
pOH = -log[OH-], the base-side mirror of pH. At 25 C, Kw = 1.0 x 10^-14, so pH + pOH = 14.00 always. For 0.10 M NaOH, pOH = 1.00 and pH = 13.00.
No - buffer pH belongs to a different equation (Henderson-Hasselbalch: pH = pKa + log([base]/[acid])). Use the dedicated Henderson-Hasselbalch buffer calculator at nrtechstudio.com/tools/henderson-hasselbalch/ for buffer pH, base-to-acid ratios, and buffer preparation.
They are ideal-solution calculations at 25 C (activity approximated as concentration), which is the standard for classroom problems and lab planning. For very dilute solutions (below about 10^-6 M), water autoionization adds a small correction - the calculator flags this in the result.
The fraction of acid molecules that dissociate, times 100. For 0.10 M acetic acid, [H+] = 1.34 x 10^-3 M, so ionization = (1.34 x 10^-3 / 0.10) x 100% = 1.34%. The weak-acid tab reports this alongside pH, and uses the 5% rule to decide between the approximation and the exact quadratic.
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