Question:

Consider the equilibrium reactions:

\(H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-\) \((K_1)\)
\(H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}\) \((K_2)\)
\(HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}\) \((K_3)\)

The equilibrium constant, \(K_c\), for the following dissociation:

\(H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-}\)

is:

Show Hint

Remember the three fundamental rules for manipulating equilibrium constants: 1. If you add reactions, multiply their $K$ values. 2. If you reverse a reaction, take the reciprocal ($1/K$). 3. If you multiply a reaction by a factor $n$, raise $K$ to the power $n$ ($K^n$).
Updated On: Jun 3, 2026
  • \(K_1 + K_2 + K_3\)
  • \(K_1 K_2 K_3\)
  • \(\dfrac{K_1}{K_2 K_3}\)
  • \(\dfrac{K_1 K_2}{K_3}\)
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The Correct Option is B

Solution and Explanation

Concept: When a chemical reaction can be expressed as the sum of two or more individual reactions, the equilibrium constant for the net reaction is the product of the equilibrium constants of the individual steps.

Step 1: Identify the relationship between the reactions.
The three stepwise dissociation reactions of phosphoric acid are:
• \(H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-\) (Equilibrium constant = \(K_1\))
• \(H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}\) (Equilibrium constant = \(K_2\))
• \(HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}\) (Equilibrium constant = \(K_3\))

Step 2: Add the reactions together.
Summing the three equations:

\[ H_3PO_4 + H_2PO_4^- + HPO_4^{2-} \rightleftharpoons (H^+ + H_2PO_4^-) + (H^+ + HPO_4^{2-}) + (H^+ + PO_4^{3-}) \] \[ H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \] Intermediate species \(H_2PO_4^-\) and \(HPO_4^{2-}\) cancel out from both sides.

Step 3: Calculate the net equilibrium constant \((K_c)\).
\[ K_c = K_1 \times K_2 \times K_3 \]
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