The dissociation constant for acetic acid and HCN at 25∘C are 1.5×10−5 and 4.5×10−10 respectively. The equilibrium constant for the equilibrium,CN−+CH3COOH⇌HCN+CH3COO−would be

CH3COOH⇌CH3COO−+H+Kacetic acid =1.5×10−5=[CH3COOH−][CH3COO−][H+]HCN⇌H++CN−KHCN=4.5×10−10=[HCN−][H+][CN−]For equilibrium :CN−+CH3COOH⇌HCN+CH3COO−K=[CN−][CH3COOH][CH3COO−][HCN]=KHCNKacetic acid =4.5×10−101.5×10−5=3.3×104≃3.0×104

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Ionisation constants of weak acids

The dissociation constant for acetic acid and {HCN} at 25^{} {

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The dissociation constant for acetic acid and $HCN$ at $2 5^{\circ} C$ are $1.5 \times 1 0^{- 5}$ and $4.5 \times 1 0^{- 10}$ respectively. The equilibrium constant for the equilibrium,
$CN^{-} + CH_{3} COOH ⇌ HCN + CH_{3} COO^{-}$ would be

3.0 \times 1 0^{5}

3.0 \times 1 0^{- 5}

3.0 \times 1 0^{- 4}

3.0 \times 1 0^{4}

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Updated on: Nov 4, 2023

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New Pattern JEE Problems Chemistry for JEE Main and Advance( Master Problem Package) (Arihant)

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Eleanor Discussed

The dissociation constant for acetic acid and

HCN

2 5^{\circ} C

are

1.5 \times 1 0^{- 5}

and

4.5 \times 1 0^{- 10}

respectively. The equilibrium constant for the equilibrium,

CN^{-} + CH_{3} COOH ⇌ HCN + CH_{3} COO^{-}

would be

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

CH_{3} COOH ⇌ CH_{3} COO^{-} + H^{+}

K_{acetic acid} = 1.5 \times 1 0^{- 5} = \frac{[ CH _{3} COO ^{-} ] [ H ^{+} ]}{[ CH _{3} COOH ^{-} ]}

HCN ⇌ H^{+} + CN^{-}

K_{HCN} = 4.5 \times 1 0^{- 10} = \frac{[ H ^{+} ] [ CN ^{-} ]}{[ HCN ^{-} ]}

For equilibrium :

CN^{-} + CH_{3} COOH ⇌ HCN + CH_{3} COO^{-}

K = \frac{[ CH _{3} COO ^{-} ] [ HCN ]}{[ CN ^{-} ] [ CH _{3} COOH ]} = \frac{K _{acetic acid}}{K _{HCN}}

= \frac{1.5 \times 1 0 ^{- 5}}{4.5 \times 1 0 ^{- 10}} = 3.3 \times 1 0^{4} ≃ 3.0 \times 1 0^{4}

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Question Text	The dissociation constant for acetic acid and $HCN$ at $2 5^{\circ} C$ are $1.5 \times 1 0^{- 5}$ and $4.5 \times 1 0^{- 10}$ respectively. The equilibrium constant for the equilibrium, $CN^{-} + CH_{3} COOH ⇌ HCN + CH_{3} COO^{-}$ would be
Updated On	Nov 4, 2023
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 24
Upvotes	2648
Avg. Video Duration	9 min

The dissociation constant for acetic acid and $HCN$ at $2 5^{\circ} C$ are $1.5 \times 1 0^{- 5}$ and $4.5 \times 1 0^{- 10}$ respectively. The equilibrium constant for the equilibrium,
$CN^{-} + CH_{3} COOH ⇌ HCN + CH_{3} COO^{-}$ would be

Text solutionVerified

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The dissociation constant for acetic acid and $HCN$ at $2 5^{\circ} C$ are $1.5 \times 1 0^{- 5}$ and $4.5 \times 1 0^{- 10}$ respectively. The equilibrium constant for the equilibrium,
$CN^{-} + CH_{3} COOH ⇌ HCN + CH_{3} COO^{-}$ would be