The equilibrium constants for the reactions are-H3PO4 loverset K- 1 rightleftharpoons H - H - 2 PO - 4 - K - 1HPO -2- 4 loverset K - 3 rightleftharpoons H - PO -3- 4 - K - 3The equilibrium constant for- H 3 P O 4- 3 H - P O 43- will be-A- K 1 - K 2 K 3B- K 1- K 2- K 3C- K 2 - K 1 K 3D- K 1- K 2- K 3

The correct option is B K_1 × K_2 × K_3 K_1 = [H^+][H_2PO^ _4]/[H_3PO_4]; K_2 = [H^+] [HPO_4^2 ]/[H_2PO_4^ ] K_3 = [H^+][PO_4^3 ]/HPO_4^2 Multiplying these th ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Effective Equilibrium Constant

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:

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

$K_{1} \times K_{2} \times K_{3} = \frac{[H^{+}]^{3} [P O_{4}^{3 -}]}{[H_{3} P O_{4}]}$

Suggest Corrections

Similar questions

Q. 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 :

Q. For the hypothetical reactions, the equilibrium constant (K) values are given:

A ⇌ B; K_{1} = 2; B ⇌ C; K_{2} = 4; C ⇌ D; K_{3} = 3

The equilibrium constant (K) for the reaction

A ⇌ D

is:

Q. For the hypothetical reactions, the equilibrium constant (K) values are given:

A ⇌ B; K_{1} = 2; B ⇌ C; K_{2} = 4; C ⇌ D; K_{3} = 3

The equilibrium constant (K) for the reaction

A ⇌ D

is:

Q. Consider the reactions

(i) C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g); K_{1} (i i) C H_{4} (g) + H_{2} O (g) ⇌ C O (g) + 3 H_{2} (g); K_{2} (i i i) C H_{4} (g) + 2 H_{2} O (g) ⇌ C O_{2} (g) + 4 H_{2} (g); K_{3}

which of the following is correct?

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

Join BYJU'S Learning Program

The equilibrium constants for the reactions are: H3PO4K1⇌H++H2PO−4;K1H2PO−4K2⇌H++HPO2−4;K2HPO2−4K3⇌H++PO3−4;K3 The equilibrium constant for, H3PO4⇌3H++PO3−4 will be:

The correct option is B K1×K2×K3 K1=[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: