Chemistry - Ionic Equilibrium Question with Solution | TestHub
ChemistryIonic EquilibriumBuffer solutionMedium2 minQB
ChemistryMediumsingle choice
Passage / Comprehension
Aqueous 200 mL of 0.1 M H₂A solution, when titrated against 0.1 M NaOH, shows different results in the presence of different indicators. For phenolphthalein indicator, it is converted to Na₂A, and with methyl orange, it is converted to NaHA.
(H₂A) = 10⁻³, (H₂A) = 10⁻⁶.
H₂A + NaOH ⇌ NaHA + H₂O (methyl orange)
NaHA + NaOH ⇌ Na₂A + H₂O (phenolphthalein)
When 100 mL of NaOH is added in the presence of methyl orange indicator pH of solution is
Options:
Answer:
C
Solution:
Concepts: Buffer solution, Henderson-Hasselbalch equation, pKₐ.
When 0.01 mol NaOH reacts with 0.02 mol H₂A, 0.01 mol HA⁻ forms. This creates a buffer where , so .
Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Buffer solution
⏱ 2mℹ️ Source: QB
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