The equilibrium constants for the reactions are: H ₃ P O ₄ ⇌ K 1 H ⁺ + H ₂ P O ₄ ⁻ ; K 1 H 2 P O 4 - ⇌ K 2 H + + H P O 4 2 - ; K 2 H P O 4 2 - ⇌ K 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

Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumpH of Polyprotic Acids and BasesMedium2 minQB

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The equilibrium constants for the reactions are:

$H ₃ P O ₄ \overset{K_{1}}{⇌} H ⁺ + H ₂ P O ₄ ⁻$ ; $K_{1}$

$H_{2} P O_{4}^{-} \overset{K_{2}}{⇌} H^{+} + H P O_{4}^{2 -}$ ; $K_{2}$

$H P O_{4}^{2 -} \overset{K_{3}}{⇌} H^{+} + P O_{4}^{3 -}$ ; $K_{3}$

The equilibrium constant for

$H_{3} P O_{4} ⇌ 3 H^{+} + P O_{4}^{3 -}$ will be

Answer:

Solution:

When reactions are added, their equilibrium constants are multiplied. $k = k_{1} \times k_{2} \times k_{3}$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:pH of Polyprotic Acids and Bases

⏱ 2mℹ️ Source: QB

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