Question:
Consider the equilibrium reactions:
Consider the equilibrium reactions:
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Important rules of equilibrium constants:
• When reactions are added $\rightarrow$ multiply equilibrium constants.
• When a reaction is reversed $\rightarrow$ take reciprocal of $K$.
• When coefficients are multiplied by $n$ $\rightarrow$ raise $K$ to the power $n$.
• When reactions are added $\rightarrow$ multiply equilibrium constants.
• When a reaction is reversed $\rightarrow$ take reciprocal of $K$.
• When coefficients are multiplied by $n$ $\rightarrow$ raise $K$ to the power $n$.
Updated On: May 13, 2026
- $ \dfrac{K_1}{K_2K_3} $
- $ K_1K_2K_3 $
- $ \dfrac{K_2}{K_1K_3} $
- $ K_1 + K_2 + K_3 $
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The Correct Option is B
Solution and Explanation
Concept:
Phosphoric acid ($H_3PO_4$) is a polybasic acid and dissociates stepwise. Each dissociation step possesses its own equilibrium constant.
Whenever individual chemical reactions are added to obtain an overall reaction, the equilibrium constants are multiplied.
Thus:
\[
K_{\text{overall}} = K_1 \times K_2 \times K_3
\]
Step 1: Writing the equilibrium constant expressions.
For the first dissociation: \[ H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \] \[ K_1 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \] For the second dissociation: \[ H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \] \[ K_2 = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} \] For the third dissociation: \[ HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \] \[ K_3 = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]} \]
Step 2: Multiplying the three equilibrium constants.
\[ K_1K_2K_3 = \left( \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \right) \left( \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} \right) \left( \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]} \right) \] Intermediate species cancel out: \[ [H_2PO_4^-] \text{ and } [HPO_4^{2-}] \] Hence, \[ K_1K_2K_3 = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]} \]
Step 3: Comparing with the overall equilibrium expression.
For the overall reaction: \[ H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \] the equilibrium constant is: \[ K_c = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]} \] Therefore, \[ \boxed{K_c = K_1K_2K_3} \] Hence, the correct answer is: \[ \boxed{(B)\ K_1K_2K_3} \]
Step 1: Writing the equilibrium constant expressions.
For the first dissociation: \[ H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \] \[ K_1 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \] For the second dissociation: \[ H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \] \[ K_2 = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} \] For the third dissociation: \[ HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \] \[ K_3 = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]} \]
Step 2: Multiplying the three equilibrium constants.
\[ K_1K_2K_3 = \left( \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \right) \left( \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} \right) \left( \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]} \right) \] Intermediate species cancel out: \[ [H_2PO_4^-] \text{ and } [HPO_4^{2-}] \] Hence, \[ K_1K_2K_3 = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]} \]
Step 3: Comparing with the overall equilibrium expression.
For the overall reaction: \[ H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \] the equilibrium constant is: \[ K_c = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]} \] Therefore, \[ \boxed{K_c = K_1K_2K_3} \] Hence, the correct answer is: \[ \boxed{(B)\ K_1K_2K_3} \]
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