Equilibrium Previous Year Questions — NEET Chemistry

4 past-year Equilibrium questions from NEET (Chemistry).

Q48 (2024)

In a qualitative analysis, $Bi^{3+}$ is detected by appearance of precipitate of $BiO(OH)(s)$. Calculate pH when the following equilibrium exists at 298 K: $$BiO(OH)(s) \rightleftharpoons BiO^+(aq) + OH^-(aq), \quad K = 4 \times 10^{-10}$$ (Given: $\log 2 = 0.3010$)
  1. (1) 4.699
  2. (2) 8.714
  3. (3) 9.301
  4. (4) 5.286
### Related Formula $$K_{sp} = s^2 \implies s = \sqrt{K_{sp}}$$ $$[H^+] = \frac{K_w}{[OH^-]}$$ $$pH = -\log_{10} [H^+]$$ ### Core Logic $$K = [BiO^+][OH^-] = s^2 = 4 \times 10^{-10}$$ $$s = [OH^-] = 2 \times 10^{-5} \text{ M}$$ $$[H^+] = \frac{10^{-14}}{2 \times 10^{-5}} = 0.5 \times 10^{-9} = 5 \times 10^{-10} \text{ M}$$ $$pH = -\log (5 \times 10^{-10}) = 10 - \log 5 = 9.301$$ ### Step 1: Final Calculation $$pH = 9 + \log 2 = 9 + 0.3010 = 9.301$$ ### Pattern Recognition For 1:1 salt equilibrium, $s = \sqrt{K}$. Then $pOH = -\log s$, $pH = 14 - pOH$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium

Q68 (2024)

Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at an alkaline pH close to the equivalence point during this titration is:
  1. (1) pinkish red to yellow
  2. (2) yellow to pinkish red
  3. (3) colourless to pink
  4. (4) pink to colourless
### Related Formula $$\text{Phenolphthalein: } pH < 8.3 \text{ (Colourless)}, \quad pH > 10.0 \text{ (Pink)}$$ ### Core Logic In acidic/neutral medium before endpoint, phenolphthalein is colourless. Near the equivalence point in alkaline medium, it turns pink. ### Step 1: Conclusion Observed colour change is colourless to pink (Option 3). ### Pattern Recognition Phenolphthalein indicator transition in alkali titrations: Colourless $\rightarrow$ Pink. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

Q78 (2024)

At 298 K, a certain buffer solution contains equal concentrations of $X^-$ and $HX$. $K_b$ for $X^-$ is $10^{-10}$. What is the pH of this buffer solution?
  1. (1) 2
  2. (2) 10
  3. (3) 4
  4. (4) 6
### Related Formula $$K_a \times K_b = K_w = 10^{-14}$$ $$pH = pK_a + \log \frac{[X^-]}{[HX]}$$ ### Core Logic $$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4} \implies pK_a = 4$$ Given $[X^-] = [HX]$: $$pH = 4 + \log (1) = 4 + 0 = 4$$ ### Step 1: Conclusion pH of the buffer solution is 4. ### Pattern Recognition When conjugate acid and base concentrations are equal, $pH = pK_a = 14 - pK_b$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium

Q87 (2024)

Given below are certain reactions. Identify the reaction for which $K_P \neq K_C$.
  1. (1) $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$
  2. (2) $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$
  3. (3) $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
  4. (4) $H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g)$
### Related Formula $$K_P = K_C(RT)^{\Delta n_g}$$ ### Core Logic $K_P \neq K_C$ whenever $\Delta n_g \neq 0$. (1) $\Delta n_g = 2 - 2 = 0 \implies K_P = K_C$ (2) $\Delta n_g = 2 - 2 = 0 \implies K_P = K_C$ (3) $\Delta n_g = 2 - (1 + 3) = -2 \neq 0 \implies K_P \neq K_C$ (4) $\Delta n_g = 2 - 2 = 0 \implies K_P = K_C$ ### Step 1: Conclusion Reaction (3) has $K_P \neq K_C$. ### Pattern Recognition If sum of gaseous moles on reactant side equals product side, $K_P = K_C$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

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