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Chemistry

Equilibrium

Both halves matter here: chemical equilibrium with Kp, Kc, and Le Chatelier's principle, then ionic equilibrium with pH, buffers, salt hydrolysis, and solubility product. Most NEET questions here are numericals, with pH, buffer, and Ksp calculations appearing year after year.

Exam rules change between cycles. Confirm dates, syllabus and pattern against NTA's official site. We are not NTA.

Weightage: what the paper actually did

Equilibrium carried 4 questions in the NEET 2026 paper. One paper is a data point, not a destiny: NTA moves weight between chapters every cycle, which is why the weightage tables deserve care. The stable facts are the paper's shape: 180 questions, 720 marks, +4 right, −1 wrong.

Sample PYQs from this chapter

Real questions NTA asked from Equilibrium, with the answer and the working. Try before you peek.

Question 1 · NEET 2026

In a qualitative analysis, Bi3+Bi^{3+} is detected by appearance of precipitate of BiO(OH)(s)BiO(OH)(s). Calculate pH when the following equilibrium exists at 298 K. BiO(OH)(s)⇌BiO+(aq)+OH−(aq)BiO(OH)(\text{s}) \rightleftharpoons BiO^{+}(\text{aq}) + OH^{-}(\text{aq}), K=4×10−10K = 4\times10^{-10}. (Given: log⁡2=0.3010\log 2 = 0.3010)
  1. A.8.714
  2. B.4.699
  3. C.5.286
  4. D.9.301
Answer and explanation

Answer: D

For the dissolution, K=s2K = s^2 since a solid gives equal amounts of BiO+BiO^{+} and OH−OH^{-}. So s=4×10−10=2×10−5 M=[OH−]s = \sqrt{4\times10^{-10}} = 2\times10^{-5}\ \text{M} = [OH^{-}]. Then pOH=5−log⁡2=4.699\text{pOH} = 5 - \log 2 = 4.699 and pH=14−4.699=9.301\text{pH} = 14 - 4.699 = 9.301. Why the other options are wrong: - (A) Uses an incorrect concentration or log manipulation. - (B) This is the pOH (4.699), not the requested pH. - (C) Arises from a mis-set solubility expression.

Question 2 · NEET 2026

At 298 K, a certain buffer solution contains equal concentrations of X−X^- and HX, KbK_b for X−X^- is 10−1010^{-10}. What is the pH of this buffer solution?
  1. A.2
  2. B.4
  3. C.6
  4. D.10
Answer and explanation

Answer: B

For the conjugate pair, Ka(HX)=KwKb=10−1410−10=10−4K_a(HX) = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-10}} = 10^{-4}, so pKa = 4. Using Henderson-Hasselbalch, pH=pKa+log⁡[X−][HX]pH = pK_a + \log\frac{[X^-]}{[HX]}; since [X−^{-}] = [HX], the log term is zero, giving pH = 4.

How to practise Equilibrium

Chapter by chapter, against real past-paper questions, on a screen: NEET is computer based from 2027. Read the chapter once, then attempt its PYQs until your accuracy stops climbing, and let the wrong answers tell you which lines to reread. If the chapter keeps refusing to stick, that has its own guide. There is also a free 12-question mock on the exam screen.

Nearby Chemistry chapters

Questions

How many questions did Equilibrium have in NEET 2026?

Equilibrium carried 4 questions in the NEET 2026 paper. Weightage shifts between cycles, so treat it as a probability, not a promise.

Where can I practise Equilibrium PYQs for NEET?

In the BrainIt app, free until the 2027 exam: its bank holds 50,000 past NTA questions tagged to all 83 chapters, including this one, with accuracy tracked per chapter. Sample questions from this chapter are on this page.

Is Equilibrium important for NEET 2027?

It is part of the Chemistry syllabus the paper draws from, and NEET 2027 is computer based, so practise it on a screen. Your own accuracy in the chapter matters more than any weightage table: a heavy chapter you already score well on needs less time than a light one you keep missing.

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