The equilibrium constant for the reaction given is 3.6 × 1 0 - 7 OCl − aq + H 2 O l ⇌ HOCl aq + OH − aq . What is Ka for HOCl?

K e q = K w K a 3.6 × 1 0 - 7 = K w K a K a = 1 0 - 14 3.6 × 1 0 - 7 = 2.77 × 1 0 - 8

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumHydrolysis of SaltsEasy2 minPYQ_2020

ChemistryEasysingle choice

The equilibrium constant for the reaction given is $3.6 \times 1 0^{- 7}$
${OCl}^{-} (aq) + H_{2} O (l) \overset{}{⇌} HOCl (aq) + {OH}^{-} (aq) .$
What is Ka for HOCl?

Options:

Answer:

Solution:

$K_{e q} = \frac{K_{w}}{K_{a}}$
$3.6 \times 1 0^{- 7} = \frac{K_{w}}{K_{a}}$
$K_{a} = \frac{1 0^{- 14}}{3.6 \times 1 0^{- 7}} = 2.77 \times 1 0^{- 8}$

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

⏱ 2mℹ️ Source: PYQ_2020

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