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In the lab, a titration is a burette and a colour change. On paper it is a curve, and exam questions ask you to calculate points on it and choose an indicator from it. This post computes a full weak acid–strong base curve and the steep part of a strong acid–strong base curve, then shows how the shape decides the indicator.
What does the syllabus ask for?
Both the JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list "the chemistry involved in the titrimetric exercises – acids, bases and the use of indicators" in their practical chemistry units, alongside ionic equilibrium and buffers in the physical chemistry units. The NCERT Class 11 lab manual unit on titrimetric analysis gives the indicator rules: phenolphthalein for a weak acid against a strong alkali, methyl orange for a strong acid against a weak base, either for strong against strong, and no suitable indicator for weak against weak. This post explains why. Check your own year's syllabus.
Why do students find titrations hard?
Sheppard studied sixteen American high school students' understanding of titrations using qualitative and computer-based tasks. Most could not describe concepts such as pH, neutralisation and strength accurately, and most could not relate them to actual solutions. The study pointed to, among other factors, instruction that emphasises solving numerical problems. So as you calculate each point below, say in words what is in the flask.
Worked example: acetic acid against NaOH
Titrate 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH. NCERT's Equilibrium chapter gives Ka = 1.74 × 10⁻⁵, so pKa = 4.76. The acid is 2.50 mmol, so equivalence is at 25.0 mL.
0 mL: weak acid alone. [H⁺] = √(1.74 × 10⁻⁵ × 0.100) = 1.32 × 10⁻³ M, pH = 2.88.
10.0 mL: buffer region. 1.00 mmol OH⁻ converts acid to acetate: acid 1.50 mmol, acetate 1.00 mmol. pH = 4.76 + log(1.00 ÷ 1.50) = 4.58. Volumes cancel in the ratio.
12.5 mL: half-equivalence. Acid and acetate are equal, so pH = pKa = 4.76. This is how pKa is read off a measured curve.
25.0 mL: equivalence. The flask holds 2.50 mmol acetate in 50.0 mL, which is 0.0500 M. Acetate is a base: Kb = 1.0 × 10⁻¹⁴ ÷ 1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰. [OH⁻] = √(5.75 × 10⁻¹⁰ × 0.0500) = 5.36 × 10⁻⁶ M, pOH = 5.27, pH = 8.73.
26.0 mL: excess base. 0.100 mmol OH⁻ in 51.0 mL is 1.96 × 10⁻³ M, pOH = 2.71, pH = 11.29.
The equivalence point is basic because acetate takes protons from water. "Equivalence" means equal moles, not neutral pH.
How steep is the jump near equivalence?
Compare 0.1 mL either side of equivalence.
Strong acid (25.0 mL of 0.100 M HCl) with 0.100 M NaOH. At 24.9 mL, 0.010 mmol H⁺ remains in 49.9 mL: pH = 3.70. At 25.1 mL, 0.010 mmol OH⁻ is in excess in 50.1 mL: pH = 10.30. The pH jumps by more than six units for 0.2 mL.
Acetic acid with NaOH. At 24.9 mL, acid 0.010 mmol and acetate 2.49 mmol: pH = 4.76 + log 249 = 7.16. At 25.1 mL the excess OH⁻ dominates: pH ≈ 10.30. The jump is smaller, from about 7.2 to 10.3, and it sits entirely above 7.
Which indicator should you choose?
The Royal Society of Chemistry's Education in Chemistry worksheet on indicators gives methyl orange's range as pH 3.1 to 4.4 and phenolphthalein's as 8.3 to 10.0.
Strong acid, strong base. The jump from 3.7 to 10.3 covers both ranges, so either works, as the lab manual says.
Weak acid, strong base. The jump from 7.2 to 10.3 contains phenolphthalein's range but not methyl orange's. Methyl orange would change colour while the flask is still a buffer, far before equivalence.
Strong acid, weak base. The mirror case: the equivalence point is acidic, because the conjugate acid of the base gives H⁺, so methyl orange is the right choice.
Weak acid, weak base. The change at equivalence is gradual, so no indicator gives a sharp end point.
Which mistakes should you avoid?
Using the starting volume after base is added. Total volume matters at equivalence and beyond.
Applying the Henderson equation at 0 mL or at equivalence, where there is no buffer pair.
Assuming pH 7 at equivalence for every titration.
Forgetting that half-equivalence gives pH = pKa directly.
The buffer region is the Henderson-Hasselbalch equation at work; our buffer calculations guide goes deeper, and our ionic equilibrium guide covers the weak acid and salt pH formulas used at the start and at equivalence. A whole curve is easier to understand when you can watch it build; the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Day 1: compute the five points above for 0.100 M ammonia titrated with 0.100 M HCl.
Day 2: sketch four curve types side by side and mark where each indicator changes colour.
Day 3: four "which indicator?" questions with a one-line reason each.
Day 4: a timed set of previous-year titration questions.
Today, plot the five points from the worked example on graph paper and draw the curve through them.
