Equilibrium

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Equilibrium — solved practice questions

8 NEET Chemistry questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1medium

   For the reaction N2(g) + 3H2(g) <-> 2NH3(g), the relation between Kp and Kc is:

   • AKp = Kc (RT)^2
   • BKp = Kc (RT)
   • CKp = Kc (RT)^(-2)
   • DKp = Kc
   Show answer & solution

   Correct answer: (C) Kp = Kc (RT)^(-2)

   Kp = Kc (RT)^(delta n), where delta n = 2 - 4 = -2, so Kp = Kc (RT)^(-2).

2. Q2easy

   According to Le Chatelier's principle, increasing the pressure on the equilibrium N2(g) + 3H2(g) <-> 2NH3(g) will:

   • AShift equilibrium toward NH3 (forward)
   • BShift equilibrium toward reactants
   • CHave no effect
   • DDecrease the value of Kp
   Show answer & solution

   Correct answer: (A) Shift equilibrium toward NH3 (forward)

   Increasing pressure shifts the equilibrium toward the side with fewer gas moles. The product side has 2 moles versus 4 on the reactant side, so the equilibrium shifts forward (toward NH3).

3. Q3easy

   The pH of a 0.001 M HCl solution at 25 C is:

   • A2
   • B3
   • C1
   • D11
   Show answer & solution

   Correct answer: (B) 3

   HCl is a strong acid, fully dissociated, so [H+] = 0.001 = 10^-3 M and pH = -log(10^-3) = 3.

4. Q4medium

   The conjugate base of the bicarbonate ion (HCO3^-) is:

   • AH2CO3
   • BH2O
   • COH-
   • DCO3^2-
   Show answer & solution

   Correct answer: (D) CO3^2-

   A conjugate base is formed by removing one H+. Removing H+ from HCO3^- gives the carbonate ion CO3^2-.

5. Q5medium

   For a weak acid with dissociation constant Ka and concentration C, the degree of dissociation (alpha) is given by Ostwald's dilution law as (for small alpha):

   • Aalpha = sqrt(Ka/C)
   • Balpha = Ka/C
   • Calpha = sqrt(C/Ka)
   • Dalpha = Ka x C
   Show answer & solution

   Correct answer: (A) alpha = sqrt(Ka/C)

   For a weak acid, Ka = C alpha^2 / (1 - alpha). When alpha is small, 1 - alpha ≈ 1, so alpha = sqrt(Ka/C).

6. Q6hard

   The solubility product (Ksp) of AgCl is 1.8 x 10^-10 at 25 C. The molar solubility of AgCl in pure water is:

   • A1.8 x 10^-10 mol/L
   • B1.8 x 10^-5 mol/L
   • C1.34 x 10^-5 mol/L
   • D3.6 x 10^-5 mol/L
   Show answer & solution

   Correct answer: (C) 1.34 x 10^-5 mol/L

   For AgCl, Ksp = s^2, so s = sqrt(1.8 x 10^-10) ≈ 1.34 x 10^-5 mol/L.

7. Q7medium

   A buffer solution is made by mixing acetic acid (pKa = 4.74) with sodium acetate in equal concentrations. The pH of this buffer is:

   • A7.00
   • B4.74
   • C9.26
   • D5.74
   Show answer & solution

   Correct answer: (B) 4.74

   By the Henderson–Hasselbalch equation, pH = pKa + log([salt]/[acid]). With equal concentrations the log term is 0, so pH = pKa = 4.74.

8. Q8medium

   Which of the following salts, when dissolved in water, gives a basic solution?

   • ANH4Cl
   • BNaCl
   • CNH4CH3COO
   • DCH3COONa
   Show answer & solution

   Correct answer: (D) CH3COONa

   CH3COONa is a salt of a weak acid (acetic acid) and a strong base (NaOH); the acetate ion hydrolyzes to give a basic solution. NH4Cl is acidic, NaCl is neutral, and NH4CH3COO is roughly neutral.

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