CH3COOH - H2O - CH3COO- - H3O- K1 - 1-8 - 10-52H2O - H3O- - OH- K2 - 1- 1014Equilibrium constant for reaction CH3COOH - OH- - CH3COO- - H2O is x - 109- Find -x-

Equilbrium constant for the reaction CH_3COOH + H_2O ⇌ CH_3COO^ + H_3O^+ K_1 =[CH_3COO^ ][H_3O^+][CH_3COOH][H_2O] Equilibrium constant for the reaction nbsp ...

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Standard XIII

Chemistry

Characteristics of Equilibrium Constant

CH3COOH + H2O...

Question

$C H_{3} C O O H + H_{2} O ⇌ C H_{3} C O O^{-} + H_{3} O^{+}; K_{1} = 1.8 \times 10^{- 5}$
$2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}, K_{2} = 1 \times 10^{14}$
Equilibrium constant for reaction
$C H_{3} C O O H + O H^{-} ⇌ C H_{3} C O O^{-} + H_{2} O$ is $x \times 10^{9}$ . Find 'x'.

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Solution

Equilbrium constant for the reaction
$C H_{3} C O O H + H_{2} O ⇌ C H_{3} C O O^{-} + H_{3} O^{+}$
$K_{1} = \frac{[C H_{3} C O O^{-}] [H_{3} O^{+}]}{[C H_{3} C O O H] [H_{2} O]}$

Equilibrium constant for the reaction
$2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$
$K_{2} = \frac{[H_{3} O^{+}] [O H^{-}]}{[H_{2} O]^{2}}$

Taking the ratio of $K_{1}$ and $K_{2}$
$\frac{K_{1}}{K_{2}} = \frac{[C H_{3} C O O^{-}] [H_{3} O^{+}]}{[C H_{3} C O O H] [H_{2} O]} \times \frac{[H_{2} O]^{2}}{[H_{3} O^{+}] [O H^{-}]}$
$\frac{K_{1}}{K_{2}} = \frac{[C H_{3} C O O^{-}] [H_{2} O]}{[C H_{3} C O O H] [O H^{-}]}$
Above ratio is the equilibrium constant of the reaction
$C H_{3} C O O H + O H^{-} ⇌ C H_{3} C O O^{-} + H_{2} O$
$∴ K = \frac{K_{1}}{K_{2}} = \frac{1.8 \times 10^{- 5}}{1 \times 10^{- 14}} = 1.8 \times 10^{9}$

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Similar questions

C H_{3} C O O H + O H^{-} ⇋ C H_{3} C O O^{-} + H_{2} O + q_{1}

and

H^{+} + O H^{-} ⇋ H_{2} O + q_{2}

Then what is the enthalpy for reaction ?

C H_{3} C O O H ⇋ C H_{3} C O O^{-} + H^{+}

C H_{3} C O O H + O H^{-} \to C H_{3} C O O^{-} + H_{2} O + q_{1}

H^{+} + O H^{-} \to H_{2} O + q_{2}

then the enthalpy change for the reaction

C H_{3} C O O H \to C H_{3} C O O^{-} + H^{+}

is equal to:

Q. For the reaction:

N H_{4}^{+} + 2 H_{2} O ⇌ N H_{4} O H + H_{3} O^{+}

K_{1} = 5.56 \times 10^{- 10}

H_{2} O + H_{2} O ⇌ H_{3} O^{+} + O H^{-}

K_{2} = 1 \times 10^{- 14}

Hence

K_{3}

of the reaction:

N H_{4} O H ⇌ N H_{4}^{+} + O H^{-}

is:

Q. The correct order of rate of

{S_{N}}^{2}

reactivity of

H_{2} O

^{-} O C H_{3},^{-} O H

and

C H_{3} C O O^{-}

Q. If equilibrium constant of

{C H}_{3} C O O H + H_{2} O ⇌ {C H}_{3} {C O O}^{-} + H_{3} O^{+}

1.8 \times 10^{- 5}

, equilibrium constant for

{C H}_{3} C O O H + {O H}^{-} ⇌ {C H}_{3} {C O O}^{-} + H_{2} O

is?

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Characteristics of Equilibrium Constant

Standard XIII Chemistry

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CH3COOH+H2O⇌CH3COO−+H3O+;K1=1.8×10−5 2H2O⇌H3O++OH−,K2=1×1014 Equilibrium constant for reaction CH3COOH+OH−⇌CH3COO−+H2O is x×109. Find 'x'.

$C H_{3} C O O H + H_{2} O ⇌ C H_{3} C O O^{-} + H_{3} O^{+}; K_{1} = 1.8 \times 10^{- 5}$
$2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}, K_{2} = 1 \times 10^{14}$
Equilibrium constant for reaction
$C H_{3} C O O H + O H^{-} ⇌ C H_{3} C O O^{-} + H_{2} O$ is $x \times 10^{9}$ . Find 'x'.