In a qualitative analysis, Bi³⁺ is detected by appearance of precipitate of BiO(OH)(s). Calculate pH when the following equilibrium exists at 298 K: BiO(OH)(s) leftharpoons BiO^+(aq) + OH^-(aq), K = 4 × 10⁻¹⁰ (Given: 2 = 0.3010)

Solution & Explanation

Related Formula
Kₛₚ = s² s = Kₛₚ [H^+] = (Kw)/([OH^-]) pH = - ₁₀ [H^+]
Core Logic
K = [BiO^+][OH^-] = s² = 4 × 10⁻¹⁰ s = [OH^-] = 2 × 10⁻⁵ M [H^+] = 10⁻¹⁴2 × 10⁻⁵ = 0.5 × 10⁻⁹ = 5 × 10⁻¹⁰ M pH = - (5 × 10⁻¹⁰) = 10 - 5 = 9.301
Step 1: Final Calculation
pH = 9 + 2 = 9 + 0.3010 = 9.301
Pattern Recognition

For 1:1 salt equilibrium, s = √(K). Then pOH = - s, pH = 14 - pOH.

Chapter Mix

Class 11 Chemistry: Ionic Equilibrium

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Ionic Equilibrium

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Q68 neet_2024_05_may_morning Acid-Base Titration Indicators
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:
  • A. (1) pinkish red to yellow
  • B. (2) yellow to pinkish red
  • C. (3) colourless to pink
  • D. (4) pink to colourless

Solution

Related Formula
Phenolphthalein: pH < 8.3 (Colourless), pH > 10.0 (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 arrow Pink.

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Class 11 Chemistry: Equilibrium

Q78 neet_2024_05_may_morning Buffer Solutions and pH
At 298 K, a certain buffer solution contains equal concentrations of X^- and HX. Kb for X^- is 10⁻¹⁰. What is the pH of this buffer solution?
  • A. (1) 2
  • B. (2) 10
  • C. (3) 4
  • D. (4) 6

Solution

Related Formula
Kₐ × Kb = Kw = 10⁻¹⁴ pH = pKₐ + ([X^-])/([HX])
Core Logic
Kₐ = 10⁻¹⁴10⁻¹⁰ = 10⁻⁴ pKₐ = 4

Given [X^-] = [HX]:

pH = 4 + (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 = pKₐ = 14 - pKb.

Chapter Mix

Class 11 Chemistry: Ionic Equilibrium

Q87 neet_2024_05_may_morning Relation between Kp and Kc
Given below are certain reactions. Identify the reaction for which KP ≠ KC.
  • A. (1) H₂(g) + I₂(g) leftharpoons 2HI(g)
  • B. (2) N₂(g) + O₂(g) leftharpoons 2NO(g)
  • C. (3) N₂(g) + 3H₂(g) leftharpoons 2NH₃(g)
  • D. (4) H₂O(g) + CO(g) leftharpoons H₂(g) + CO₂(g)

Solution

Related Formula
KP = KC(RT)Δ ng
Core Logic

KP ≠ KC whenever Δ ng ≠ 0. (1) Δ ng = 2 - 2 = 0 KP = KC (2) Δ ng = 2 - 2 = 0 KP = KC (3) Δ ng = 2 - (1 + 3) = -2 ≠ 0 KP ≠ KC (4) Δ ng = 2 - 2 = 0 KP = KC

Step 1: Conclusion

Reaction (3) has KP ≠ KC.

Pattern Recognition

If sum of gaseous moles on reactant side equals product side, KP = KC.

Chapter Mix

Class 11 Chemistry: Equilibrium

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