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:

• A. \(K_1 + K_2 + K_3\)
• B. \(K_1 K_2 K_3\)
• C. \(\dfrac{K_1}{K_2 K_3}\)
• D. \(\dfrac{K_1 K_2}{K_3}\)

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 $n$ ($K^n$).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When reactions are added, their equilibrium constants are multiplied.
Step 3: Detailed Explanation:
The net reaction is the sum of the three stepwise dissociations.
Summing reactions \( \implies \) Product of constants.
\( K_{net} = K_{1} \times K_{2} \times K_{3} \).
Step 4: Final Answer:
The result is \(K_{1}K_{2}K_{3}\).