SATHEE: Equilibrium - Result Question 27

Equilibrium - Result Question 27

####28. The dissociation constants for acetic acid and $H C N$ at $25^{\circ} C$ are $1.5 \times 10^{- 5}$ and $4.5 \times 10^{- 10}$ respectively. The equilibrium constant for the equilibrium

[2009]

$C N^{-} + C H_{3} C O O H \to H C N + C H_{3} C O O^{-}$ would be:

(a) $3.0 \times 10^{- 5}$

(b) $3.0 \times 10^{- 4}$

(d) $3.0 \times 10^{5}$

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Solution:

$K_{a_{1}} = 1.5 \times 10^{- 5}$ $H C N \to H^{+} + C N^{-}; K_{a_{2}} = 4.5 \times 10^{- 10}$

or $H^{+} + C N^{-} \to H C N$ ;

$Double subscripts: use braces to clarify$

$∴$ From (i) and (ii), we find that the equilibrium constant $(K_{a})$ for the reaction,

$Double subscripts: use braces to clarify$

Equilibrium - Result Question 27

Solution:

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Equilibrium - Result Question 27

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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