Read the particles first. Calculate the pH second.
JEE Main · JEE Advanced · NEET UG32 min study guide9 indexed sections
BY THE END, YOU CAN
Distinguish acid strength, concentration and pH.
Calculate weak-acid and weak-base equilibria while checking approximations.
Identify a genuine buffer and explain its response to added acid or base.
Use neutralization stoichiometry before a buffer equation.
Recognize buffer exhaustion, salt hydrolysis and temperature-dependent neutrality.
BEFORE YOU START
Moles, molarity and dilution: n = CV.
Equilibrium constants and the common-ion effect.
Base-10 logarithms and balancing ionic equations.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2015. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Answer checked against the official final key. Worked solution written for this lesson.
Mix 100 mL of 2.0 M acetic acid with 100 mL of 1.0 M NaOH. For Kₐ = 2.0 × 10⁻⁵, estimate the mixture’s pH. This isolates the acid–base part of the original shared experimental passage.
The common final volume makes [acetate]/[acid] = 1. Therefore pH = pKₐ + log₁₀1 = −log₁₀(2.0 × 10⁻⁵) ≈ 4.70.
Source: JEE (Advanced) 2015, Paper 2, Chemistry Q40; official post-feedback answer-key paper. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. The particle story behind pH
A beaker labeled “acid” does not tell you enough to calculate its pH. You need to know how much solute is present, whether it transfers protons almost completely or only partly, what else is dissolved, and the temperature. Begin every problem with a particle inventory rather than with a memorized formula.
A Brønsted–Lowry acid donates a proton; a base accepts one. When HA gives a proton to water, the products are H₃O⁺ and A⁻. The pair HA/A⁻ differs by one proton and is a conjugate acid–base pair. H⁺ is convenient shorthand for the hydrated proton in aqueous calculations. A strong acid is extensively ionized in dilute water. A weak acid establishes an equilibrium with substantial un-ionized acid remaining.
“Strong” describes extent of ionization in the solvent; “concentrated” describes amount per volume. A very dilute strong acid can have a higher pH than a more concentrated weak acid. Likewise, a salt is not automatically neutral: its ions may react with water. Sodium acetate contains the basic ion CH₃COO⁻, whereas ammonium chloride contains the acidic ion NH₄⁺.
HA + H₂O ⇌ H₃O⁺ + A⁻
A⁻ is the conjugate base of HA.
B + H₂O ⇌ BH⁺ + OH⁻
BH⁺ is the conjugate acid of the weak base B.
02 / UNDERSTAND & APPLY
2. pH, water and the temperature condition
For the dilute-solution model used here, pH ≈ −log₁₀[H₃O⁺] and pOH ≈ −log₁₀[OH⁻], with concentrations expressed numerically in mol L⁻¹. Strictly, pH is defined using hydrogen-ion activity. Concentration calculations assume activity effects are small. This matters in concentrated or high-ionic-strength solutions.
All numerical examples below assume 25 °C, K_w = 1.0 × 10⁻¹⁴, and ideal dilute behavior unless stated otherwise. A neutral solution has equal hydronium and hydroxide concentrations. Its pH equals pK_w/2, which is 7.00 at 25 °C but need not be 7 at another temperature. A pH below 7 does not by itself establish acidity if the temperature is unspecified.
The logarithm compresses multiplicative changes: lowering pH by one unit increases hydronium activity tenfold. Do not average pH values when mixing solutions. Add or subtract reacting moles, divide by the combined volume, and then take the logarithm.
K_w = [H₃O⁺][OH⁻]; pH + pOH = pK_w
The familiar sum 14.00 applies at 25 °C in this model.
At neutrality: [H₃O⁺] = [OH⁻] = √K_w
Neutrality is equality of the two ion concentrations, not a universal pH number.
ORIGINAL WORKED EXAMPLE
Mixing strong reagents
Mix 30.0 mL of 0.100 M HCl with 20.0 mL of 0.100 M NaOH. Find pH.
Answer pH = 1.70. The pH values of the starting solutions were never averaged.
03 / UNDERSTAND & APPLY
3. Weak acids and bases: derive before approximating
For a monoprotic weak acid of initial concentration C with no added conjugate base, let x be the concentration that ionizes. If water contributes negligible hydronium, equilibrium concentrations are x, x and C − x for H₃O⁺, A⁻ and HA. Substitution gives K_a = x²/(C − x). This equation is the starting point; the square-root expression is a conditional shortcut.
If x is much smaller than C, replace C − x with C and obtain x ≈ √(K_aC). Check the approximation afterward. A common classroom criterion is x/C ≤ 0.05, but this is an error-control convention rather than a law of nature. If it fails, solve the quadratic. If the answer approaches the 10⁻⁷ M scale at 25 °C, water autoionization may also matter; the simple acid-only quadratic is then insufficient.
For a weak base, use K_b = [BH⁺][OH⁻]/[B] and solve for hydroxide first. Then calculate pOH and convert using pK_w. The acid and base constants connected by K_aK_b = K_w must belong to a conjugate pair. You cannot multiply the constants of unrelated acid and base species and apply this identity.
x = (−K_a + √(K_a² + 4K_aC))/2
Exact solution of the acid-only quadratic, still neglecting water contribution.
K_aK_b = K_w; pK_a + pK_b = pK_w
Applies to HA/A⁻ or BH⁺/B at the same temperature.
ORIGINAL WORKED EXAMPLE
Check a square-root estimate
A hypothetical monoprotic acid has K_a = 4.0 × 10⁻⁶ and C = 0.040 M. Estimate pH.
Set x ≈ √[(4.0 × 10⁻⁶)(0.040)] = 4.0 × 10⁻⁴ M.
x/C = 0.010, or 1.0%; the small-ionization approximation is reasonable.
pH ≈ −log(4.0 × 10⁻⁴) = 3.40. Water contribution is negligible on this scale.
Answer pH ≈ 3.40.
04 / UNDERSTAND & APPLY
4. Two partners make a buffer
A buffer contains appreciable amounts of a weak acid and its conjugate base, or a weak base and its conjugate acid. The two partners provide complementary responses. Added hydronium is consumed mainly by the basic partner. Added hydroxide is consumed mainly by the acidic partner. These reactions convert one partner into the other, so the composition changes and the pH changes a little.
For an acetate buffer, CH₃COO⁻ accepts a proton from added acid, while CH₃COOH donates a proton to neutralize added hydroxide. The sodium ion is a spectator. A mixture of HCl and NaCl is not an analogous buffer: chloride is too weak a base in water to supply useful acid-neutralizing capacity. An isolated weak acid also lacks the substantial conjugate-base reserve characteristic of a prepared buffer.
A buffer can be made directly by mixing a weak acid with a salt of its conjugate base, or indirectly by partially neutralizing a weak acid with strong base. The word “partially” is essential. At complete neutralization, no appreciable original acid remains, so a different calculation is needed.
A⁻ + H₃O⁺ → HA + H₂O
Response to added strong acid.
HA + OH⁻ → A⁻ + H₂O
Response to added strong base.
Both members of the conjugate pair are needed. A buffer resists pH change; it does not prevent it.
05 / UNDERSTAND & APPLY
5. Henderson–Hasselbalch, with its conditions
Rearrange K_a = [H₃O⁺][A⁻]/[HA] to isolate hydronium and take negative logarithms. The concentration form gives pH ≈ pK_a + log([A⁻]/[HA]). With activities, the corresponding relation is thermodynamic. In routine buffer problems, we make another approximation: replace equilibrium concentrations by the amounts present after stoichiometric neutralization divided by their common volume.
This shortcut is good when both partners remain appreciable, additional ionization changes their amounts only slightly, activity effects are acceptable, and water does not dominate. It is not valid merely because a question mentions a weak acid. It becomes unreliable near complete neutralization, at extreme dilution or for mixtures with other important equilibria. If the assumptions fail, use equilibrium, material balance and charge balance.
Because both partners occupy the same final volume, their concentration ratio equals their mole ratio. This is why moles are especially convenient after mixing. Equal partner amounts give pH ≈ pK_a. Ten times as much conjugate base gives pH ≈ pK_a + 1. A customary useful range is pK_a ± 1, corresponding to base/acid ratios from 0.1 to 10; this is a practical range, not an abrupt boundary.
pH ≈ pK_a + log₁₀(n(A⁻)/n(HA))
Use remaining amounts after neutralization, not blindly the starting amounts.
pOH ≈ pK_b + log₁₀(n(BH⁺)/n(B))
For a weak-base buffer, the conjugate acid is in the numerator of the pOH expression.
ORIGINAL WORKED EXAMPLE
Read a ratio
A buffer contains 0.030 mol HA and 0.060 mol A⁻. The acid has pK_a = 4.60. Find pH.
The base/acid ratio is 0.060/0.030 = 2.00.
pH ≈ 4.60 + log 2.00 = 4.60 + 0.301.
More base than acid means the answer should lie above pK_a.
Answer pH ≈ 4.90.
06 / UNDERSTAND & APPLY
6. The two-stage calculation: react, then equilibrate
When strong acid or base enters a weak conjugate-pair mixture, perform the essentially complete reaction first. Write initial moles, moles consumed and moles remaining. Only then ask whether the remaining mixture is a buffer. Trying to insert starting concentrations into Henderson–Hasselbalch ignores the chemical event that changed the composition.
If both partners survive, use their new ratio. If the original weak acid is just exhausted by added base, calculate the hydrolysis of its conjugate base at the new total concentration. If strong base is left over, its excess usually controls hydroxide. These are distinct chemical regions; one equation cannot cover all of them without checking what remains.
ORIGINAL WORKED EXAMPLE
Build a buffer by partial neutralization
Mix 50.0 mL of 0.200 M HA with 30.0 mL of 0.100 M NaOH. Given pK_a = 4.80, find pH.
Initial HA = 10.0 mmol; OH⁻ = 3.00 mmol.
HA + OH⁻ → A⁻ + H₂O leaves 7.00 mmol HA and creates 3.00 mmol A⁻.
Both partners are present. pH ≈ 4.80 + log(3.00/7.00).
Answer pH ≈ 4.43.
ORIGINAL WORKED EXAMPLE
Challenge an existing buffer
A 100 mL buffer initially contains 12.0 mmol HA and 8.00 mmol A⁻. Add 2.00 mmol HCl in a negligible volume. Let pK_a = 5.00.
H₃O⁺ consumes A⁻ and forms HA.
New amounts: A⁻ = 6.00 mmol and HA = 14.0 mmol.
pH ≈ 5.00 + log(6.00/14.0) = 4.63; initial pH was about 4.82.
Answer The pH falls by about 0.19, to 4.63.
07 / UNDERSTAND & APPLY
7. Capacity, dilution and the equivalence boundary
Two buffers can have the same pH but very different resistance to disturbance. Equal acid/base ratios fix the approximate pH; the absolute amounts determine how much added reagent can be consumed. A liter containing 0.10 mol of each partner withstands a specified small addition better than a liter containing 0.001 mol of each, even though their starting ratios match.
Moderate dilution leaves the ratio nearly unchanged, so the usual buffer pH estimate barely changes. It reduces capacity per unit volume. If the entire original sample is diluted without removing solute, its total inventory of neutralizing partners remains the same. At sufficiently extreme dilution, dissociation and water contributions become important and the usual “pH unchanged” approximation fails.
For a fixed total concentration of an ideal weak conjugate pair, its buffer contribution to capacity is greatest near equal partner concentrations. Acid-rich mixtures have a larger reserve against added base; base-rich mixtures have a larger reserve against added acid. Pick a pK_a close to the target pH instead of forcing an extreme ratio.
At the equivalence point of a weak-acid/strong-base titration, the conjugate base hydrolyzes and the solution is usually basic at 25 °C. Half-equivalence is different: half the acid has been converted to conjugate base, so pH ≈ pK_a under buffer assumptions. Never exchange those two landmarks.
ORIGINAL WORKED EXAMPLE
After the acid is exhausted
At equivalence, a hypothetical salt A⁻ has concentration 0.050 M. Its parent acid has K_a = 1.0 × 10⁻⁵. Estimate pH.
K_b(A⁻) = K_w/K_a = 1.0 × 10⁻⁹.
For A⁻ + H₂O ⇌ HA + OH⁻, [OH⁻] ≈ √[(1.0 × 10⁻⁹)(0.050)] = 7.07 × 10⁻⁶ M.
The hydrolyzed fraction is tiny. pOH ≈ 5.15, so pH ≈ 8.85.
Answer pH ≈ 8.85. A salt-only hydrolysis calculation replaces the buffer shortcut.
08 / UNDERSTAND & APPLY
8. Original practice: choose the chemistry first
Try each question before revealing the answer. These are original exam-style practice questions, not past-paper reproductions. Numerical answers use 25 °C and ideal dilute solutions. State the reaction and check what remains before reaching for a logarithm.
Suggested practice: allow yourself 14 minutes for these eight questions, then review every explanation. This is a study interval, not an official exam time limit. Mark each error as particle inventory, stoichiometry, equation choice or arithmetic; retry missed questions without the solution the next day.
ORIGINAL PRACTICE · 1
A solution has [H₃O⁺] = 2.5 × 10⁻⁴ M. Find pH and pOH.
Show a hint
Use the logarithm first, then pK_w.
Reveal answer & reasoning
pH = 3.60; pOH = 10.40.
−log(2.5 × 10⁻⁴) = 4 − log 2.5 = 3.602. At 25 °C pOH = 14.00 − 3.602 = 10.398. Round to two decimal places.
ORIGINAL PRACTICE · 2
Can 0.10 M HNO₃ plus 0.10 M KNO₃ form an effective acid buffer?
Show a hint
Does the anion supply useful proton-accepting capacity?
Reveal answer & reasoning
No.
HNO₃ is a strong acid in dilute water and nitrate is an extremely weak base. This is not an appreciable weak conjugate pair capable of consuming added acid and base.
ORIGINAL PRACTICE · 3
For HA, pK_a = 6.20. What A⁻/HA ratio gives pH 6.50?
Show a hint
Rearrange the logarithmic relation.
Reveal answer & reasoning
Approximately 2.0.
log([A⁻]/[HA]) = 6.50 − 6.20 = 0.30, so the ratio is 10^0.30 ≈ 2.00. The ratio is base over acid, consistent with pH above pK_a.
ORIGINAL PRACTICE · 4
A buffer contains 5.0 mmol HA and 5.0 mmol A⁻. Add 1.0 mmol NaOH. If pK_a = 4.00, estimate pH.
Show a hint
Hydroxide consumes HA.
Reveal answer & reasoning
pH ≈ 4.18.
After neutralization HA = 4.0 mmol and A⁻ = 6.0 mmol. pH ≈ 4.00 + log(6/4) = 4.176. The upward shift is consistent with adding base.
ORIGINAL PRACTICE · 5
A 100 mL mixture contains 2.0 mmol HA and 3.0 mmol A⁻. Add 4.0 mmol NaOH in negligible volume. The parent acid has K_a = 1.0 × 10⁻⁵. Estimate pH.
Show a hint
Is the weak acid reserve large enough?
Reveal answer & reasoning
pH ≈ 12.30.
All 2.0 mmol HA reacts, leaving 2.0 mmol excess OH⁻. In 0.100 L its concentration is 0.020 M, so pOH ≈ 1.70 and pH ≈ 12.30. Excess strong base dominates; a negative or zero acid amount must never enter a buffer logarithm. Here K_b(A⁻) = 1.0 × 10⁻⁹, and conjugate-base hydrolysis is suppressed by the excess hydroxide; its additional contribution is negligible.
ORIGINAL PRACTICE · 6
Two equal-volume buffers have the same HA/A⁻ ratio. One has ten times both concentrations. Which better resists the same small added amount of acid?
Show a hint
Separate ratio from inventory.
Reveal answer & reasoning
The more concentrated buffer.
Their approximate starting pH values match. The more concentrated sample has ten times the conjugate-base inventory, so the same acid addition changes its base/acid ratio by a smaller fraction.
ORIGINAL PRACTICE · 7
A weak acid has C = 1.0 × 10⁻³ M and K_a = 1.0 × 10⁻⁴. Is x ≈ √(K_aC) justified? Find a better x.
Show a hint
Check x/C; then use the quadratic.
Reveal answer & reasoning
No; x ≈ 2.70 × 10⁻⁴ M, giving pH ≈ 3.57.
The shortcut gives x = 3.16 × 10⁻⁴ M, a 31.6% change, far beyond the usual small-change criterion. The quadratic gives (−0.0001 + √0.00000041)/2 ≈ 0.000270 M. Water is negligible at this concentration.
ORIGINAL PRACTICE · 8
At a temperature where pK_w = 13.40, what is the neutral pH? Is pH 6.80 acidic or basic?
Show a hint
Neutrality occurs halfway through pK_w.
Reveal answer & reasoning
Neutral pH = 6.70; pH 6.80 is basic.
At neutrality pH = pOH = pK_w/2 = 6.70. For pH 6.80, pOH = 6.60, so hydroxide exceeds hydronium. Comparing only with 7 would give the wrong classification.
09 / UNDERSTAND & APPLY
9. A reliable exam route
When a problem feels unfamiliar, reduce it to three questions: What reacts almost completely? What species remain in appreciable amounts? Which equilibrium describes those remaining species? This sequence distinguishes a strong excess, a lone weak acid or base, a salt solution and a buffer without requiring a separate memorized trick for every wording.
Check the direction of every answer. Added acid should lower the pH of an ordinary buffer, added base should raise it, and equal HA/A⁻ amounts should put the pH near pK_a. A numerical result that contradicts this picture usually signals an inverted ratio, an incorrect neutralization sign or a missed volume.
Use moles during reactions and final volume when an absolute concentration is needed.
Keep pK_a and pK_b attached to the correct conjugate species.
At half-equivalence, use the conjugate-pair ratio; at equivalence, inspect salt hydrolysis.
State temperature and dilution assumptions.
The logarithmic relation is useful only after checking its chemical conditions.
KEEP THE BIG IDEAS
Quick revision
A buffer needs appreciable amounts of both members of a weak conjugate pair.
Stoichiometry comes before equilibrium whenever strong acid or base is added.
pH follows a ratio; resistance to change depends on available amounts.
Approximation checks are part of the solution, not optional decoration.
Neutral pH is pK_w/2, equal to 7.00 only under the usual 25 °C assumption.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.