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Mixing questions give you two solutions and ask for one number. The trap is that each mixture needs a different tool: subtraction for strong with strong, Henderson's equation for a partly neutralised weak acid, hydrolysis for a fully neutralised one. Pick the wrong tool and the answer is off by several pH units. This post gives one decision method and works every case with numbers.
What does the syllabus include?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list the pH scale, the common ion effect, "hydrolysis of salts and pH of their solutions" and buffer solutions. The constants here come from the NCERT Class 11 chapter on Equilibrium: Kw = 1.0 × 10⁻¹⁴ at 298 K and, from its Table 6.6, Ka of formic acid = 1.8 × 10⁻⁴, so pKa = 3.74. Check the syllabus for your exam year.
Why do students get mixing problems wrong?
Watters and Watters (2006) studied first-year biochemistry students and found that most "juggled" the numbers in a problem without attending to its structure. Most striking, all but one could not do a simple calculation on logs without a calculator. Both habits hurt here. The fix is to describe the mixture in words before using any log.
The method in four steps
Convert everything to millimoles (mL × M = mmol).
React the strong species completely with whatever neutralises it.
Name what is left: excess strong acid or base, weak acid with its salt, or salt alone.
Divide by the total volume and use the tool for that case.
Case 1: strong acid with strong base
Example 1 (NCERT exercise 6.66a): 10 mL of 0.2 M Ca(OH)₂ with 25 mL of 0.1 M HCl.
OH⁻: 10 × 0.2 × 2 = 4.0 mmol, because each Ca(OH)₂ gives two OH⁻.
H⁺: 25 × 0.1 = 2.5 mmol.
Excess OH⁻: 1.5 mmol in 35 mL, so [OH⁻] = 0.0429 M.
pOH = 1.37, pH = 12.63.
Example 2 (exercise 6.66c): 10 mL of 0.1 M H₂SO₄ with 10 mL of 0.1 M KOH. Treating H₂SO₄ as fully dissociated, H⁺ = 2.0 mmol and OH⁻ = 1.0 mmol. Excess H⁺ = 1.0 mmol in 20 mL = 0.050 M, so pH = 1.30.
Case 2: two strong acids mixed
Example 3: equal volumes of HCl at pH 1 and pH 3. Average the concentrations, not the pH values.
[H⁺] = (0.1 + 0.001) ÷ 2 = 0.0505 M, so pH = 1.30.
The answer is not 2. The stronger solution dominates because pH is a log scale.
Case 3: weak acid partly neutralised by strong base
Example 4: 20 mL of 0.10 M HCOOH with 5 mL of 0.10 M NaOH.
Acid: 2.0 mmol. OH⁻: 0.5 mmol.
After reaction: 1.5 mmol HCOOH and 0.5 mmol HCOO⁻. This is a buffer.
pH = pKa + log(salt ÷ acid) = 3.74 + log(0.5 ÷ 1.5) = 3.74 − 0.48 = 3.27.
The volumes cancel in the ratio, so you can use millimoles directly. An exact quadratic gives 3.28; the Henderson equation is close enough here.
With 10 mL of NaOH instead, exactly half the acid is converted, salt equals acid and pH = pKa = 3.74. NCERT's Problem 6.24 does the base version: 50.0 mL of 0.10 M NH₃ with 25.0 mL of 0.10 M HCl leaves equal NH₃ and NH₄⁺, and the pH is 9.24.
Case 4: weak acid exactly neutralised
Example 5: 20 mL of 0.10 M HCOOH with 20 mL of 0.10 M NaOH. Now only sodium formate is left: 2.0 mmol in 40 mL, or 0.050 M. Formate is a weak base.
Kb = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁴ = 5.6 × 10⁻¹¹.
[OH⁻] = √(5.6 × 10⁻¹¹ × 0.050) = 1.7 × 10⁻⁶ M.
pOH = 5.78, pH = 8.22.
"Neutralised" does not mean pH 7. A weak acid with a strong base gives a basic solution.
Case 5: strong acid with weak acid
Example 6: a solution 0.010 M in HCl and 0.10 M in acetic acid (NCERT Ka 1.74 × 10⁻⁵). The H⁺ from HCl suppresses the weak acid. Taking [H⁺] ≈ 0.010, [acetate] = Ka × 0.10 ÷ 0.010 = 1.7 × 10⁻⁴ M, a tiny extra. [H⁺] ≈ 0.01017 M and the pH is 1.99, against 2.00 for the HCl alone. This is the common ion effect.
Common mistakes
Forgetting that Ca(OH)₂ and H₂SO₄ give two ions each.
Dividing by one solution's volume instead of the total.
Using Henderson's equation when one component has run out.
Calling the equivalence point of a weak acid neutral.
Putting added strong acid straight into a log without reacting it first.
For buffer capacity and hydrolysis in more depth, see our guide to pH, buffers and salt hydrolysis, and for a full curve, acid-base titration curves. When you cannot tell which case a mixture falls into, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning.
A practice routine
Do NCERT exercise 6.66b: 10 mL of 0.01 M H₂SO₄ with 10 mL of 0.01 M Ca(OH)₂. Ignoring any precipitate, the acid and base cancel exactly and the pH is 7.
Redo Examples 4 and 5 with acetic acid (pKa 4.76). Your answers should be 4.28 and 8.73.
For each problem this week, write step 3 of the method in words before calculating.
Practise logs without a calculator: log 2 = 0.30, log 3 = 0.48, log 5 = 0.70.
Pick five mixing questions from your last test and, before solving, label each as Case 1 to 5.
