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Ionic equilibrium is where chemistry and logarithms meet, and it is easy to grab the wrong formula. The fix is to decide what is in the beaker first: a weak acid alone, a weak acid with its salt, or a salt alone. Each case has one tool. This post works through all of them with numbers, then lists the traps.
What does the syllabus ask for?
The JEE (Main) 2026 syllabus lists ionisation of acids and bases, ionisation of water, the pH scale, the common ion effect, "hydrolysis of salts and pH of their solutions", solubility products and buffer solutions. The NEET (UG)-2026 syllabus has the same topics. The values below come from the NCERT Class 11 chapter on Equilibrium: Kw = 1.0 × 10⁻¹⁴ at 298 K, Ka of acetic acid 1.74 × 10⁻⁵ and its pKa 4.76.
How do you find the pH of a weak acid?
Example 1. pH of 0.10 M acetic acid.
HA ⇌ H⁺ + A⁻, with Ka = x² ÷ (0.10 − x).
Assume x is small compared with 0.10: x = √(1.74 × 10⁻⁵ × 0.10) = √(1.74 × 10⁻⁶) ≈ 1.32 × 10⁻³ M.
pH = −log(1.32 × 10⁻³) ≈ 2.88.
Check the assumption: 1.32 × 10⁻³ ÷ 0.10 ≈ 1.3% ionised. Under about 5%, the approximation is fine. If it were larger, you would solve the quadratic.
How does a buffer hold its pH?
A buffer is a weak acid with its conjugate base (or a weak base with its conjugate acid). NCERT derives the working equation: pH = pKa + log([salt] ÷ [acid]).
Example 2. pH of a solution 0.10 M in acetic acid and 0.20 M in sodium acetate.
pH = 4.76 + log(0.20 ÷ 0.10) = 4.76 + 0.30 = 5.06.
With equal concentrations, log 1 = 0 and pH = pKa, which is why NCERT says to choose an acid whose pKa is close to the pH you want.
Example 3. To 1.0 L of a buffer that is 0.10 M in both acetic acid and acetate, add 0.010 mol of HCl.
Work in moles first. The H⁺ converts acetate to acetic acid: acetate 0.10 − 0.010 = 0.090 mol, acid 0.10 + 0.010 = 0.11 mol.
pH = 4.76 + log(0.090 ÷ 0.11) = 4.76 − 0.087 ≈ 4.67.
Compare: 0.010 mol of HCl in 1.0 L of pure water would give pH 2.0. The buffer moved by less than 0.1.
Why is a salt solution not neutral?
A salt of a weak acid and a strong base, such as sodium acetate, gives a basic solution because acetate takes H⁺ from water. NCERT explains each case: weak acid with strong base gives pH above 7, strong acid with weak base below 7, and for a weak acid with a weak base, pH = 7 + ½(pKa − pKb).
Example 4. pH of 0.10 M sodium acetate.
Kb of acetate = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 1.74 × 10⁻⁵ ≈ 5.75 × 10⁻¹⁰.
[OH⁻] = √(5.75 × 10⁻¹⁰ × 0.10) ≈ 7.6 × 10⁻⁶ M, so pOH ≈ 5.12.
pH = 14 − 5.12 = 8.88.
The shortcut pH = 7 + ½(pKa + log C) = 7 + ½(4.76 − 1) = 8.88 gives the same answer, and deriving it once from step 1 is better than memorising it.
What is the 10⁻⁸ M HCl trap?
A strong acid at 1.0 × 10⁻⁸ M seems to give pH 8, which would make an acid basic. NCERT works it properly: at this dilution water's own H⁺ is larger than the acid's, so you must include both. Solving x² + 10⁻⁸x − 10⁻¹⁴ = 0 gives pH ≈ 6.98, slightly acidic, as it must be.
Why do buffers confuse even advanced students?
Orgill and Sutherland (2008) interviewed general chemistry, analytical chemistry and biochemistry students, and asked them to think aloud while solving buffer problems. Students at all levels struggled to explain how buffers work and held misconceptions about them, including students who had met buffers in several courses. Being able to use the Henderson equation is not the same as knowing why the pH barely moves. Explain it in words: the added H⁺ is mostly used up converting A⁻ to HA, so the ratio changes only a little.
Which mistakes should you avoid?
Using the Henderson equation for a weak acid alone. With no salt, use the square root.
Skipping the 5% check after an approximation.
Adding strong acid without the moles step (Example 3).
Using Kw = 10⁻¹⁴ at temperatures other than 298 K.
Forgetting water's ions in very dilute solutions.
Many of these are not knowledge gaps but choice-of-method gaps; our post on why you understand theory but fail numericals explains the fix. The same idea of counting particles links to colligative properties and the van 't Hoff factor.
A practice routine
Day 1: five weak acid and weak base pH problems, with the 5% check written out.
Day 2: five buffer pH problems, including one weak base buffer using pOH.
Day 3: four "add acid or base to a buffer" problems, moles first.
Day 4: four salt hydrolysis problems, deriving the formula each time.
Day 5: a mixed set of ten; before each, write which of the four cases it is.
When you cannot tell which case a problem is, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, find the pH after adding 0.010 mol of NaOH to the buffer in Example 3.
