The equilibrium constants for the reactions are H3PO4 K 1 -H- - H2PO4- K1 H2PO4- K 2 -H- - HPO42- K2 HPO42- K 3 -H- - PO43- K3 The equilibrium constant for- H3PO4- 3H- - PO43- will be -

The correct option is B K_1×K_2×K_3 K_1 = [H^+][H_2PO_4^ ]/[H_3PO_4]; nbsp; nbsp; nbsp;K_2 = [H^+][HPO_4^2 ]/[H_2PO_4^ ]; nbsp; nbsp; ...

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Byju's Answer

Standard XII

Chemistry

Arrhenius Equation

The equilibri...

Question

The equilibrium constants for the reactions are
$H_{3} P O_{4}_{1} H^{+} + H_{2} P O_{4}^{-}; K_{1}$
$H_{2} P O_{4}^{-}_{2} H^{+} + H P O_{4}^{2 -}; K_{2}$
$H P O_{4}^{2 -}_{3} H^{+} + P O_{4}^{3 -}; K_{3}$
The equilibrium constant for,
$H_{3} P O_{4} ⇌ 3 H^{+} + P O_{4}^{3 -}$ will be :

K_{1} / K_{2} K_{3}

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K_{1} \times K_{2} \times K_{3}

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K_{2} / K_{1} K_{3}

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K_{1} + K_{2} + K_{3}

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Solution

The correct option is B $K_{1} \times K_{2} \times K_{3}$
$K_{1} = \frac{[H^{+}] [H_{2} P O_{4}^{-}]}{[H_{3} P O_{4}]}; K_{2} = \frac{[H^{+}] [H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]}; K_{3} = \frac{[H^{+}] [P O_{4}^{3 -}]}{[H P O_{4}^{2 -}]}$
Multiplying these three
$K_{1} \times K_{2} \times K_{3} = \frac{{[H^{+}]}^{3} [P O_{4}^{3 -}]}{[H_{3} P O_{4}]}$

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Similar questions

Q. For

H_{3} P O_{4}, H_{3} P O_{4} ⇋ H_{2} P O_{4}^{-} + H^{+} (K_{1})

H_{2} P O_{4}^{-} ⇋ H P O_{4}^{2 -} + H^{+} (K_{2})

H P O_{4}^{2 -} ⇋ P O_{4}^{3 -} + H^{+} (K_{3})

, then

The equilibrium constants for the reactions are:

$H_{3} P O_{4}_{1} H^{+} + H_{2} P O_{4}^{-}; K_{1} H_{2} P O_{4}^{-}_{2} H^{+} + H P O_{4}^{2 -}; K_{2} H P O_{4}^{2 -}_{3} H^{+} + P O_{4}^{3 -}; K_{3}$

The equilibrium constant for, $H_{3} P O_{4} ⇌ 3 H^{+} + P O_{4}^{3 -}$ will be:

For $C u^{2 +}$ ion in $[C u (H_{2} O)_{4} (e n)]^{2 +}, [C u (H_{2} O)_{2} (e n)_{2}]^{2 +}$ and $[C u (e n)_{3}]^{2 +}$ with the stability constant $k_{1}, k_{2}$ and $k_{3}$ respectively, which of the following holds true?

Q. The following equilibrium are given

N_{2} + 3 H_{2} \to 2 N H_{3} - K_{1}

N_{2} + O_{2} \to 2 N O - K_{2}

H_{2} + 1 / 2 O_{2} \to H_{2} O - K_{3}

The equilibrium constant of reaction,

3 H_{2} + 5 / 2 O_{2} \to 2 N O + 3 H_{2} O

, in terms of

K_{1}, K_{2}, K_{3}

If $k_{1}, k_{2}, k_{3}$ are real numbers and the equation $k_{1} x^{2} + k_{2} x + k_{3} = 0$ have three roots, then value of $k_{1} + k_{2} + k_{3}$ is

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The equilibrium constants for the reactions are H3PO4K1⇌H++H2PO−4;K1 H2PO−4K2⇌H++HPO2−4;K2 HPO2−4K3⇌H++PO3−4;K3 The equilibrium constant for, H3PO4⇌3H++PO3−4 will be :

The correct option is B K1×K2×K3K1=[H+][H2PO−4][H3PO4];K2=[H+][HPO2−4][H2PO−4];K3=[H+][PO3−4][HPO2−4]Multiplying these three K1×K2×K3=[H+]3[PO3−4][H3PO4]

The equilibrium constants for the reactions are
$H_{3} P O_{4}_{1} H^{+} + H_{2} P O_{4}^{-}; K_{1}$
$H_{2} P O_{4}^{-}_{2} H^{+} + H P O_{4}^{2 -}; K_{2}$
$H P O_{4}^{2 -}_{3} H^{+} + P O_{4}^{3 -}; K_{3}$
The equilibrium constant for,
$H_{3} P O_{4} ⇌ 3 H^{+} + P O_{4}^{3 -}$ will be :