Consider the equilibrium reactions:
The equilibrium constant, $K_c$, for the overall dissociation
\[
H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-}
\]
is:
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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 $n$.
Step 1: Understanding the Concept:
When a chemical reaction is composed of multiple intermediate steps, its overall equilibrium constant is mathematically related to the constants of those steps.
Phosphoric acid \( (H_3PO_4) \) is a polyprotic acid, meaning it can donate more than one proton per molecule.
Each dissociation step has its own specific equilibrium constant (\( K_1, K_2, K_3 \)).
A fundamental rule in chemical kinetics and equilibrium states that if reactions are added, their respective equilibrium constants must be multiplied to find the constant for the resulting reaction. Step 2: Key Formula or Approach:
For Reaction 1: \( A \rightleftharpoons B \) with \( K_1 \).
For Reaction 2: \( B \rightleftharpoons C \) with \( K_2 \).
The sum: \( A \rightleftharpoons C \) will have \( K = K_1 \times K_2 \). Step 3: Detailed Explanation:
Let's write down the expressions for each individual equilibrium constant:
\[ K_1 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \]
\[ K_2 = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} \]
\[ K_3 = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]} \]
The target overall reaction is:
\[ H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \]
The equilibrium expression for this overall reaction is:
\[ K_c = \frac{[H^+]^3 [PO_4^{3-}]}{[H_3PO_4]} \]
Now, let's multiply the three stepwise constants together:
\[ K_1 \times K_2 \times K_3 = \left( \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \right) \times \left( \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} \right) \times \left( \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]} \right) \]
Notice the cancellation of terms that appear in both the numerator and the denominator (the intermediate species):
- \( [H_2PO_4^-] \) cancels out.
- \( [HPO_4^{2-}] \) cancels out.
Remaining terms:
\[ K_1 \times K_2 \times K_3 = \frac{[H^+] \times [H^+] \times [H^+] \times [PO_4^{3-}]}{[H_3PO_4]} \]
\[ K_1 \times K_2 \times K_3 = \frac{[H^+]^3 [PO_4^{3-}]}{[H_3PO_4]} \]
This is exactly the expression for \( K_c \).
Thus, we have proven that \( K_c = K_1 K_2 K_3 \). Step 4: Final Answer:
The overall equilibrium constant is the product of the stepwise constants \( K_1, K_2, \) and \( K_3 \).
This is Option (B).
