SATHEE: Equilibrium Question 836

Equilibrium Question 836

Question: The degree of hydrolysis in hydrolytic equilibrium $A^{-} + H_{2} O \to H A + O H^{-}$ at salt concentration of 0.001 M is : $(K_{a} = 1 \times 10^{- 5})$

Options:

A) $1 \times 10^{- 3}$

B) $1 \times 10^{- 4}$

C) $5 \times 10^{- 4}$

D) $1 \times 10^{- 6}$

Show Answer

Answer:

Correct Answer: A

Solution:

$\begin{matrix} \underset{(A t e q .)}{\underset{(I n i t i a l)}{},}, & \underset{\begin{matrix} 0.001 M \\ (0.001 - R) \end{matrix}}{A^{-}}, + H_{2} O \to \underset{(0.001, \times, h)}{\underset{0}{H A},}, + \underset{(0.001, \times, h)}{\underset{0}{H A},}, \end{matrix}$

$K_{h} = \frac{(0.001 \times h) (0.001 \times h)}{(0.001 - h)}$ or, $K_{h} = 0.001 \times h^{2} (a s, 0.001 - h \approx 0.001)$

$10^{- 9} = 0.001 \times h^{2}$

$(K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{10^{- 5}} = 10^{- 9})$

$h^{2} = 10^{- 6};$

$∴ h = 10^{- 3}$

Equilibrium Question 836

Question: The degree of hydrolysis in hydrolytic equilibrium $A^{-} + H_{2} O \to H A + O H^{-}$ at salt concentration of 0.001 M is : $(K_{a} = 1 \times 10^{- 5})$

Options:

Answer:

Solution:

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Equilibrium Question 836

Question: The degree of hydrolysis in hydrolytic equilibrium A− +H2O→HA+OH− at salt concentration of 0.001 M is : (Ka=1×10−5)

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The degree of hydrolysis in hydrolytic equilibrium $A^{-} + H_{2} O \to H A + O H^{-}$ at salt concentration of 0.001 M is : $(K_{a} = 1 \times 10^{- 5})$