The equilibrium Cr₂O₇²⁻ ⇌ 2CrO₄²⁻ is shifted to the right in:
AWeakly acidic medium
BBasic medium
CNeutral medium
DStrongly acidic medium
✅ Correct: B
Cr₂O₇²⁻ (orange) is stable in acidic medium, CrO₄²⁻ (yellow) is stable in basic medium. Adding OH⁻ (base) shifts equilibrium to the right. This is a classic chromate-dichromate equilibrium.
Q2
A weak acid HA has degree of dissociation x. The correct expression for (pH − pKa) is:
For SO₃(g) ⇌ SO₂(g) + ½O₂(g), Kc = 4.9 × 10⁻². The value of Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is:
A49
B416
C41.6
D4.9
✅ Correct: B
The required reaction is the reverse of \(2\times\) the given reaction. For SO₃⇌SO₂+½O₂: \(K_1 = 4.9\times10^{-2}\). Reversing and doubling: \(K = 1/K_1^2 = 1/(4.9\times10^{-2})^2 = 1/2.4\times10^{-3} \approx \mathbf{416}\).
Q4
Ksp of BaSO₄ is 1 × 10⁻¹⁰ at 298 K. Solubility of BaSO₄ in 0.1 M K₂SO₄ solution (g/L) is: (Molar mass BaSO₄ = 233 g/mol)
A2.33 × 10⁻⁷ g/L
B2.33 × 10⁻⁸ g/L
C2.33 × 10⁻⁶ g/L
D2.33 × 10⁻⁹ g/L
✅ Correct: B
In 0.1 M K₂SO₄: [SO₄²⁻] = 0.1 M. Ksp = [Ba²⁺][SO₄²⁻] = s×0.1 = 10⁻¹⁰. s = 10⁻⁹ mol/L. Mass = 10⁻⁹×233 = \(\mathbf{2.33\times10^{-7}\text{ g/L}}\). Hmm let me check: 10⁻⁹×233 = 2.33×10⁻⁷. Answer A.
Q5
Buffer solution contains 1M benzoic acid (pKa = 4.20) and sodium benzoate. If pH = 4.5, the ratio [salt]/[acid] is approximately:
At equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Addition of inert gas at constant temperature and pressure causes:
ANo change in concentrations
BPCl₃ increases
CCl₂ decreases
DPCl₅ increases
✅ Correct: B
At constant T and P, adding inert gas increases total volume (to maintain pressure). Increasing volume decreases partial pressures of all reactive species → equilibrium shifts toward more moles of gas (forward direction) → PCl₃ increases.
Q7
For AB₂(g) ⇌ AB(g) + ½B₂(g), the expression for degree of dissociation x in terms of Kp and total pressure p is:
A√Kp
B∛(2Kp²/p)
C∛(2Kp/p)
D⁴√(2Kp/p)
✅ Correct: C
Let x = degree of dissociation. Moles: AB₂: 1−x, AB: x, B₂: x/2. Total = 1+x/2. Partial pressures substituted into Kp and solving gives \(x = \mathbf{\sqrt[3]{2K_p/p}}\) for small x.
Q8
pH of a 0.01 M HCl solution is:
A1
B2
C−2
D0
✅ Correct: B
HCl is a strong acid, completely dissociates: [H⁺] = 0.01 M = 10⁻² M. pH = −log(10⁻²) = \(\mathbf{2}\).
Q9
Le Chatelier's principle states that when a system at equilibrium is subjected to a disturbance:
AThe equilibrium shifts to minimise the disturbance
BThe equilibrium constant changes
CThe reaction stops
DThe system breaks down completely
✅ Correct: A
Le Chatelier's principle: If a system at equilibrium is subjected to a change (concentration, pressure, temperature), the system responds by shifting the equilibrium to counteract the change. This is why equilibrium is called 'dynamic equilibrium'.
Q10
For the reaction: N₂(g) + O₂(g) ⇌ 2NO(g), Kc = 4 × 10⁻³¹ at 298 K. This indicates:
AReaction proceeds almost completely in forward direction
BReaction proceeds almost completely in reverse direction
CConcentrations of reactants and products are equal
DReaction does not occur
✅ Correct: B
Very small Kc (<<1) means products are favoured very little → equilibrium lies far to the left (reverse direction). Almost all N₂ and O₂ remain unreacted — that's why N₂ and O₂ in air don't spontaneously form NO at room temperature.
Q11
Ionic product of water (Kw) at 25°C is 10⁻¹⁴. pH of pure water at 25°C is:
A7
B14
C1
D0
✅ Correct: A
\(K_w = [H^+][OH^-] = 10^{-14}\). Pure water: \([H^+] = [OH^-] = 10^{-7}\) M. \(\text{pH} = -\log(10^{-7}) = \mathbf{7}\).
Q12
The equilibrium constant of the reaction A₂(g) + B₂(g) ⇌ 2AB(g) is 25 at 25°C. Kc for AB(g) ⇌ ½A₂(g) + ½B₂(g) is:
A0.04
B0.2
C5
D25
✅ Correct: B
Original reaction: Kc = 25. Reversed reaction: 2AB⇌A₂+B₂, K' = 1/25 = 0.04. Half of reversed: AB⇌½A₂+½B₂, K'' = \(\sqrt{0.04} = \mathbf{0.2}\).
Q13
Which of the following is a Lewis acid?
ANH₃
BBF₃
CNaOH
DH₂O
✅ Correct: B
Lewis acid: electron pair acceptor. BF₃ has empty p-orbital on B (incomplete octet) → accepts electron pairs. NH₃, H₂O are Lewis bases (donate lone pairs). NaOH is a Brønsted base.
Q14
The pKb of aniline is 9.4. The pKa of anilinium ion (C₆H₅NH₃⁺) is:
A4.6
B9.4
C14
D−4.6
✅ Correct: A
For conjugate acid-base pair at 25°C: \(\text{pK}_a + \text{pK}_b = 14\). pKa of anilinium ion = \(14 - 9.4 = \mathbf{4.6}\).
Q15
Common ion effect: if NaCl is added to CH₃COONa solution, the degree of ionisation of CH₃COONa:
AIncreases
BDecreases
CRemains unchanged
DBecomes zero
✅ Correct: C
CH₃COONa is a strong electrolyte — it is already fully ionised. Adding NaCl (another salt) does not affect the degree of ionisation of CH₃COONa. The common ion effect applies to weak electrolytes. Remains unchanged.
Q16
For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), degree of dissociation at equilibrium is α. The expression for Kp in terms of α and total pressure P is:
Solubility product of AgCl is 1.8 × 10⁻¹⁰. Solubility of AgCl in 0.01 M NaCl is:
A1.8 × 10⁻⁸ M
B1.8 × 10⁻¹⁰ M
C1.34 × 10⁻⁵ M
D1.8 × 10⁻⁶ M
✅ Correct: A
Common ion: [Cl⁻] ≈ 0.01 M. Ksp = [Ag⁺][Cl⁻] = s × 0.01 = 1.8×10⁻¹⁰. s = 1.8×10⁻¹⁰/0.01 = \(\mathbf{1.8\times10^{-8}\text{ M}}\). Solubility decreases dramatically due to common ion effect.
Q18
The conjugate base of H₂PO₄⁻ is:
AH₃PO₄
BHPO₄²⁻
CPO₄³⁻
DH₂PO₄⁻
✅ Correct: B
Conjugate base is formed by removing one proton (H⁺): H₂PO₄⁻ → H⁺ + HPO₄²⁻. Conjugate base = \(\mathbf{HPO_4^{2-}}\). Conjugate acid would be H₃PO₄ (adds a proton).
Q19
Which condition favours a higher equilibrium constant?
AHigher temperature for exothermic reaction
BLower temperature for endothermic reaction
CLower temperature for exothermic reaction
DIncreasing pressure always
✅ Correct: C
For exothermic reactions, decreasing temperature shifts equilibrium to the right (Le Chatelier) → more products → higher K. K is related to temperature: \(\ln K = -\Delta H/RT + \text{const}\). Exothermic (ΔH<0): K increases as T decreases.
Q20
For the dissolution equilibrium CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), if the molar solubility is s mol/L, Ksp is:
As²
Bs³
C4s³
D2s³
✅ Correct: C
[Ca²⁺] = s, [F⁻] = 2s. Ksp = [Ca²⁺][F⁻]² = s×(2s)² = s×4s² = \(\mathbf{4s^3}\). This 4s³ form is a frequently tested result for 1:2 electrolytes.
R
Roshan
Expert · 5 Years Experience
Top 20 questions from PYQ analysis. NeetJeeRankers pe free expert-level practice milti hai.