AcidBase Pro — Titration Curves

Simulation Controls

Titration Type

Acid Conc. (mol/L)

Base Conc. (mol/L)

Acid Volume (mL)

Speed (×10)

Add Base:

Titration Flask

Current pH

1.00

Readings

Base Added

0.0mL

Total Vol.

25.0mL

Equiv. Vol.

25.0mL

½ Equiv.

12.5mL

HCl (strong acid) + NaOH (strong base)

Titration Curve Hover for pH · Click to capture · —

↑ Hover the titration curve to see region-specific calculations

pH Calculator

Complete step-by-step pH calculation for any point on any titration curve.

Parameters

Titration Type

Acid Conc. (mol/L)

Acid Volume (mL)

Base Conc. (mol/L)

Base Volume Added (mL)

Constants used:
Kₐ(CH₃COOH) = 1.8×10⁻⁵ | pKₐ = 4.745
Kb(NH₃) = 1.8×10⁻⁵ | pKb = 4.745
Ka(NH₄⁺) = Kw/Kb = 5.56×10⁻¹⁰ | pKa = 9.255
Kw = 1.0×10⁻¹⁴ at 25°C

Step-by-Step Solution

Enter values and click Calculate pH

What is Titration?

A titration is a quantitative analytical technique used to determine the unknown concentration of a solution (the analyte) by reacting it with a standard solution of precisely known concentration (the titrant), delivered from a graduated glass cylinder called a burette.

In an acid-base titration, the acid and base react until the stoichiometric equivalence point is reached — the point at which the moles of titrant exactly equal the moles of analyte. An indicator or pH electrode identifies this endpoint.

Key Principle: At the equivalence point, moles of acid = moles of base (for a 1:1 reaction).
n_acid = C_acid×V_acid = C_base×V_base = n_base

Acid-Base Chemistry (Brønsted-Lowry)

According to the Brønsted-Lowry theory, an acid is a proton (H⁺) donor and a base is a proton acceptor. Every acid has a conjugate base and every base has a conjugate acid.

For any acid HA dissociating: HA + H₂O ⇌ H₃O⁺ + A⁻

Acid Dissociation Constant

Ka = [H₃O⁺][A⁻] / [HA] pKa = −log₁₀(Ka) Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ (at 25°C) Ka × Kb = Kw (conjugate pair) pKa + pKb = 14

Strong vs Weak Acids and Bases

Strong Acids and Bases

Strong acids (HCl, HNO₃, H₂SO₄) and strong bases (NaOH, KOH) dissociate completely in water. For HCl: [H⁺] = C_HCl directly. For NaOH: [OH⁻] = C_NaOH directly.

Weak Acids and Bases

Weak acids (CH₃COOH, Ka = 1.8×10⁻⁵) and weak bases (NH₃, Kb = 1.8×10⁻⁵) only partially dissociate. Their degree of dissociation depends on Ka/Kb and concentration.

Initial pH of a Weak Acid (simplified, valid when Ka ≪ Ca)

[H⁺] ≈ √(Ka × Ca) pH = ½(pKa − log Ca)

The Titration Curve

System	Initial pH	Equiv. pH	Curve Feature
Strong acid + Strong base	~1 (0.1M HCl)	7.00	Sharp S-curve; steep inflection at equivalence
Weak acid + Strong base	~2.87 (0.1M CH₃COOH)	>7 (~8.72)	Buffer plateau; gradual rise; inflection above 7
Strong acid + Weak base	~1	<7 (~5.28)	Sharp initially; buffer plateau after equivalence

Henderson-Hasselbalch Equation

Henderson-Hasselbalch

pH = pKa + log₁₀([A⁻] / [HA]) [A⁻] = conjugate base concentration (or moles, same V) [HA] = weak acid remaining pKa = −log₁₀(Ka)

Since volume is equal for both species in the same flask, moles can be used directly in the ratio: pH = pKa + log(n_A⁻/n_HA).

Buffer Solutions

A buffer resists pH change upon addition of small amounts of acid or base. It consists of a weak acid and its conjugate base in comparable concentrations. Maximum buffering capacity occurs when [A⁻] = [HA], i.e., at the half-equivalence point where pH = pKa.

Buffer Range: Effective within ±1 pH unit of pKa. For acetic acid: pH 3.74 – 5.74

The Equivalence Point

Strong acid + Strong base → NaCl (neutral salt); pH = 7.00
Weak acid + Strong base → CH₃COONa (alkaline salt); pH > 7 (acetate hydrolyses)
Strong acid + Weak base → NH₄Cl (acidic salt); pH < 7 (ammonium hydrolyses)

Equivalence Point pH via Hydrolysis

Weak acid + Strong base: [OH⁻] = √(Kb(A⁻) × cA⁻) where Kb = Kw/Ka Strong acid + Weak base: [H⁺] = √(Ka(BH⁺) × cBH⁺) where Ka = Kw/Kb

Indicators

Indicator	pH Range	Colour Change	Best for
Methyl orange	3.1 – 4.4	Red → Yellow	Strong acid + Strong/weak base
Methyl red	4.4 – 6.2	Red → Yellow	Strong acid + Strong base
Bromothymol blue	6.0 – 7.6	Yellow → Blue	Strong acid + Strong base (precise)
Phenolphthalein	8.2 – 10.0	Colourless → Pink	Weak acid + Strong base
Thymol blue	8.0 – 9.6	Yellow → Blue	Weak acid + Strong base

Key Formulas Quick Reference

Essential Equations

pH = −log[H⁺] pOH = −log[OH⁻] pH + pOH = 14 (25°C) Strong acid: [H⁺] = Cacid Strong base: [OH⁻] = Cbase Weak acid initial: [H⁺] = √(Ka×Ca), pH = ½(pKa−logCa) Weak base initial: [OH⁻] = √(Kb×Cb), pOH = ½(pKb−logCb) Buffer (H-H): pH = pKa + log([A⁻]/[HA]) Half-equiv pt: pH = pKa (when [A⁻] = [HA]) Equivalence (HA+NaOH): [OH⁻] = √(Kw/Ka × cA⁻) → pH = 14+log[OH⁻] Equivalence (HCl+NH₃): [H⁺] = √(Kw/Kb × cBH⁺) → pH = −log[H⁺] After equiv (excess NaOH): pH = 14 + log[OH⁻]excess After equiv (excess HCl): pH = −log[H⁺]excess

Sample Calculations

Fully worked examples for every region of each titration curve. Default: Ca = Cb = 0.1 mol/L, Va = 25.0 mL → Veq = 25.0 mL.

Reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) | Strong acid, complete dissociation, no buffer region

Region 1 Initial pH (Vb = 0 mL)

No base added — pure strong acid. HCl dissociates 100%, so [H⁺] = Cacid directly.

Worked Example — 0.1 mol/L HCl, 25 mL

HCl → H⁺ + Cl⁻ (complete dissociation)
[H⁺] = 0.1 mol/L

pH = −log(0.1) = −log(10⁻¹) = 1.00

pH = 1.00

Region 2 Before Equivalence — Excess HCl (Vb = 10.0 mL)

NaOH is added but HCl remains in excess. The unreacted H⁺ determines pH.

Worked Example — 10.0 mL NaOH added

Calculate moles:n(HCl) = 0.025 L × 0.1 mol/L = 2.50×10⁻³ mol n(NaOH) = 0.010 L × 0.1 mol/L = 1.00×10⁻³ mol

Excess H⁺:n(H⁺)excess = (2.50 − 1.00)×10⁻³ = 1.50×10⁻³ mol

Total volume:V = 25.0 + 10.0 = 35.0 mL = 0.0350 L [H⁺] = 1.50×10⁻³ / 0.0350 = 0.04286 mol/L

pH = −log(0.04286) = 1.37

pH = 1.37

Region 3 Equivalence Point (Vb = 25.0 mL)

All HCl neutralised → NaCl(aq). NaCl is a neutral salt; neither Na⁺ nor Cl⁻ hydrolyse. Solution is essentially pure water.

Worked Example — 25.0 mL NaOH added

n(HCl) = n(NaOH) = 2.50×10⁻³ mol → complete neutralisationHCl + NaOH → NaCl + H₂O

NaCl does not hydrolyse. [H⁺] = [OH⁻] = √Kw = 10⁻⁷pH = 7.00

pH = 7.00 (neutral)

Region 4 After Equivalence — Excess NaOH (Vb = 30.0 mL)

Excess NaOH dominates. The excess OH⁻ ions control pH.

Worked Example — 30.0 mL NaOH added

n(OH⁻)excess = (3.00 − 2.50)×10⁻³ = 5.00×10⁻⁴ mol [OH⁻] = 5.00×10⁻⁴ / 0.055 = 9.09×10⁻³ mol/L

pOH = −log(9.09×10⁻³) = 2.04 pH = 14 − 2.04 = 11.96

pH = 11.96

Reaction: CH₃COOH + NaOH → CH₃COONa + H₂O | pKₐ = 4.745 | Ka = 1.8×10⁻⁵ | Equivalence point pH > 7

Region 1 Initial pH — Pure Weak Acid (Vb = 0)

Acetic acid partially dissociates. Use ICE table with approximation Ka ≪ Ca.

Worked Example — 0.1 mol/L CH₃COOH, 25 mL

ICE table for CH₃COOH ⇌ H⁺ + CH₃COO⁻:Ka = x²/(0.1 − x) ≈ x²/0.1 = 1.8×10⁻⁵ x² = 1.8×10⁻⁶

[H⁺] = √(1.8×10⁻⁶) = 1.342×10⁻³ mol/L pH = −log(1.342×10⁻³) = 2.87

pH = 2.87

Region 2 Buffer Region — Henderson-Hasselbalch (Vb = 10.0 mL)

NaOH converts CH₃COOH → CH₃COO⁻. Both weak acid and conjugate base coexist — classic buffer. Use Henderson-Hasselbalch.

Worked Example — 10.0 mL NaOH added

n(CH₃COOH)initial = 0.025 × 0.1 = 2.50×10⁻³ mol n(NaOH) = 0.010 × 0.1 = 1.00×10⁻³ mol

After neutralisation:n(CH₃COOH)remaining = (2.50 − 1.00)×10⁻³ = 1.50×10⁻³ mol n(CH₃COO⁻)formed = 1.00×10⁻³ mol

Henderson-Hasselbalch (moles, same volume):pH = pKa + log(n(A⁻)/n(HA)) pH = 4.745 + log(1.00/1.50) pH = 4.745 + log(0.6667) pH = 4.745 − 0.176 = 4.57

pH = 4.57

Region 3 Half-Equivalence Point (Vb = 12.5 mL)

Exactly half the acid is neutralised: n(A⁻) = n(HA). The ratio in H-H = 1. pH = pKa exactly.

Worked Example — 12.5 mL NaOH (= Veq/2)

n(NaOH) = 0.0125 × 0.1 = 1.25×10⁻³ mol = ½ × n(CH₃COOH)
∴ n(CH₃COO⁻) = n(CH₃COOH) = 1.25×10⁻³ mol

pH = pKa + log(1.25×10⁻³ / 1.25×10⁻³) pH = pKa + log(1) = pKa + 0 = 4.745

This is used to experimentally determine Ka: Ka = 10⁻ᵖᴴ at half-equiv. point.

pH = pKa = 4.745 (Key landmark on the curve!)

Region 4 Equivalence Point (Vb = 25.0 mL)

All acetic acid → sodium acetate. CH₃COO⁻ is the conjugate base of a weak acid — it hydrolyses to give a basic solution.

Worked Example — 25.0 mL NaOH added

n(CH₃COO⁻) = 2.50×10⁻³ mol in V = 50.0 mL[CH₃COO⁻] = 2.50×10⁻³ / 0.050 = 0.050 mol/L

Hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻Kb(CH₃COO⁻) = Kw/Ka = 1.0×10⁻¹⁴ / 1.8×10⁻⁵ = 5.56×10⁻¹⁰

[OH⁻] = √(Kb × c) = √(5.56×10⁻¹⁰ × 0.050) = √(2.78×10⁻¹¹) = 5.27×10⁻⁶ mol/L

pOH = −log(5.27×10⁻⁶) = 5.278 pH = 14 − 5.278 = 8.72

pH = 8.72 (basic — weak acid + strong base equivalence is always > 7)

Region 5 After Equivalence — Excess NaOH (Vb = 30.0 mL)

Excess NaOH dominates; acetate hydrolysis is negligible compared to strong base.

Worked Example — 30.0 mL NaOH added

n(OH⁻)excess = (3.00 − 2.50)×10⁻³ = 5.00×10⁻⁴ mol [OH⁻] = 5.00×10⁻⁴ / 0.055 = 9.09×10⁻³ mol/L

pOH = 2.04, pH = 11.96

pH = 11.96

Reaction: HCl(aq) + NH₃(aq) → NH₄Cl(aq) | pKb(NH₃) = 4.745 | pKa(NH₄⁺) = 9.255 | Equivalence point pH < 7

Region 1 Initial pH — Pure Strong Acid (Vb = 0)

HCl in flask; no base added yet. Complete dissociation.

Worked Example — 0.1 mol/L HCl, 25 mL

[H⁺] = 0.100 mol/L → pH = −log(0.100) = 1.00

pH = 1.00

Region 2 Before Equivalence — Excess Strong Acid (Vb = 10.0 mL)

NH₃ reacts completely with H⁺: H⁺ + NH₃ → NH₄⁺. Remaining excess H⁺ determines pH.

Worked Example — 10.0 mL NH₃ added

n(HCl) = 2.50×10⁻³ mol; n(NH₃) = 1.00×10⁻³ mol n(H⁺)excess = (2.50 − 1.00)×10⁻³ = 1.50×10⁻³ mol

V = 35.0 mL = 0.035 L; [H⁺] = 1.50×10⁻³/0.035 = 0.04286 mol/L pH = 1.37

pH = 1.37

Region 3 Equivalence Point (Vb = 25.0 mL)

All HCl → NH₄Cl. NH₄⁺ is the conjugate acid of weak base NH₃ — it hydrolyses to give an acidic solution.

Worked Example — 25.0 mL NH₃ added

n(NH₄⁺) = 2.50×10⁻³ mol in 50.0 mL[NH₄⁺] = 0.050 mol/L

Hydrolysis: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺Ka(NH₄⁺) = Kw/Kb = 1.0×10⁻¹⁴/1.8×10⁻⁵ = 5.56×10⁻¹⁰

[H⁺] = √(Ka × c) = √(5.56×10⁻¹⁰ × 0.050) = √(2.78×10⁻¹¹) = 5.27×10⁻⁶ mol/L pH = −log(5.27×10⁻⁶) = 5.28

pH = 5.28 (acidic — strong acid + weak base equivalence is always < 7)

Region 4 After Equivalence — NH₄⁺/NH₃ Buffer (Vb = 30.0 mL)

Excess NH₃ coexists with NH₄⁺ — buffer region. Use H-H with pKa(NH₄⁺) = 9.255.

Worked Example — 30.0 mL NH₃ added

n(NH₃)excess = (3.00 − 2.50)×10⁻³ = 5.00×10⁻⁴ mol n(NH₄⁺) = n(HCl) = 2.50×10⁻³ mol

H-H for NH₄⁺/NH₃ system:pH = pKa(NH₄⁺) + log([NH₃]/[NH₄⁺]) pH = 9.255 + log(5.00×10⁻⁴ / 2.50×10⁻³) pH = 9.255 + log(0.200) = 9.255 − 0.699 = 8.56

pH = 8.56

Region 5 Half-Equivalence of NH₄⁺/NH₃ Buffer (Post-Equiv.)

When n(NH₃)excess = n(NH₄⁺) after the equivalence point, pH = pKa(NH₄⁺).

When [NH₃]excess = [NH₄⁺]

pH = pKa(NH₄⁺) + log(1) = 9.255

pH = pKa(NH₄⁺) = 9.255

Data Table

Capture data points during simulation or by clicking the graph. Export as CSV for analysis in Excel or Google Sheets.

#	Volume Base (mL)	pH	[H⁺] (mol/L)	Region	Titration Type	Timestamp
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