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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumHydrolysis of SaltsMedium2 minQB

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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.

$K_{a 1}$ (H₂A) = 10⁻³, $K_{a 2}$ (H₂A) = 10⁻⁶.

H₂A + NaOH ⇌ NaHA + H₂O (methyl orange)

NaHA + NaOH ⇌ Na₂A + H₂O (phenolphthalein)

When 400 mL of NaOH is added in the presence of phenolphthalein indicator. The pH of solution is

Options:

Answer:

Solution:

Moles NaOH:

$0.4 \times 0.1 = 0.04$

Second equivalence.

Now the solution contains A²⁻ (basic salt).

$p K_{b} = 14 - p K_{a 2}$

$p K_{b} = 8$

Half neutralization of the second step:

$p H = 14 - p K_{b}$

$p H = 9.26$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Hydrolysis of Salts

⏱ 2mℹ️ Source: QB

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