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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumHydrolysis of SaltsEasy2 minPYQ_2020

ChemistryEasysingle choice

$H_{3} P O_{4} + H_{2} O ⇌ H_{3} O^{+} + H_{2} P O_{4}^{-}; p K_{1} = 2.15$
$H_{2} P O_{4}^{-} + H_{2} O ⇌ H_{3} O^{+} + H P O_{4}^{2 -}; p K_{2} = 7.20$
Hence pH of 0.01 M $N a H_{2} P O_{4}$ is

Options:

Answer:

Solution:

$H_{2} P O_{4}^{-}$ is amphiprotic
$H_{2} P O_{4}^{-} + H_{2} O ⇌ H_{3} P O_{4} + O H^{-}$
$H_{2} P O_{4}^{-} + H_{2} O ⇌ H P O_{4}^{2 -} + H_{3} O^{+}$
So $p H = \frac{p K_{1} + p K_{2}}{2} = \frac{2.15 + 7.2}{2} = 4.675$

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

⏱ 2mℹ️ Source: PYQ_2020

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