Question:
Consider the equilibrium reactions:
The equilibrium constant, $K_c$ for the following dissociation $\text{H}_3\text{PO}_4 \rightleftharpoons 3\text{H}^+ + \text{PO}_4^{3-}$ is:}
Consider the equilibrium reactions:
The equilibrium constant, $K_c$ for the following dissociation $\text{H}_3\text{PO}_4 \rightleftharpoons 3\text{H}^+ + \text{PO}_4^{3-}$ is:}
Show Hint
Remember the golden rule of chemical equilibrium arithmetic:
- Adding chemical equations $\implies$ Multiply their equilibrium constants ($K$).
- Subtracting chemical equations $\implies$ Divide their equilibrium constants.
- Multiplying an equation by a factor $n \implies$ Raise the constant to power $n$ ($K^n$).
- Adding chemical equations $\implies$ Multiply their equilibrium constants ($K$).
- Subtracting chemical equations $\implies$ Divide their equilibrium constants.
- Multiplying an equation by a factor $n \implies$ Raise the constant to power $n$ ($K^n$).
Updated On: Jun 3, 2026
- $\frac{K_1}{K_2 K_3}$
- $K_1 K_2 K_3$
- $\frac{K_2}{K_1 K_3}$
- $K_1 + K_2 + K_3$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The question asks to find the relationship between the overall equilibrium constant ($K_c$) for the complete dissociation of phosphoric acid ($\text{H}_3\text{PO}_4$) and the individual stepwise dissociation constants ($K_1, K_2, K_3$).
Step 2: Key Formula or Approach:
For a multi-step chemical reaction, if the overall reaction is obtained by adding several stepwise reactions, the overall equilibrium constant is the product of the equilibrium constants of the individual steps:
\[ K_{overall} = K_1 \times K_2 \times K_3 \times \dots \times K_n \]
Step 3: Detailed Explanation:
• Let us write the mathematical expression for each of the given step-wise dissociation processes:
• For the first step:
\[ \text{H}_3\text{PO}_4 \rightleftharpoons \text{H}^+ + \text{H}_2\text{PO}_4^- \implies K_1 = \frac{[\text{H}^+][\text{H}_2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} \]
• For the second step:
\[ \text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} \implies K_2 = \frac{[\text{H}^+][\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} \]
• For the third step:
\[ \text{HPO}_4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-} \implies K_3 = \frac{[\text{H}^+][\text{PO}_4^{3-}]}{[\text{HPO}_4^{2-}]} \]
• The final overall dissociation reaction represents the complete deprotonation of phosphoric acid:
\[ \text{H}_3\text{PO}_4 \rightleftharpoons 3\text{H}^+ + \text{PO}_4^{3-} \]
• The equilibrium constant expression for this target reaction is:
\[ K_c = \frac{[\text{H}^+]^3 [\text{PO}_4^{3-}]}{[\text{H}_3\text{PO}_4]} \]
• Let us evaluate the mathematical product of the three step-wise constants, $K_1 \times K_2 \times K_3$:
\[ K_1 \cdot K_2 \cdot K_3 = \left( \frac{[\text{H}^+][\text{H}_2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} \right) \times \left( \frac{[\text{H}^+][\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} \right) \times \left( \frac{[\text{H}^+][\text{PO}_4^{3-}]}{[\text{HPO}_4^{2-}]} \right) \]
• Looking at the terms in the numerator and denominator, we can cancel out the intermediate species $[\text{H}_2\text{PO}_4^-]$ and $[\text{HPO}_4^{2-}]$ as they appear in both places.
• After cancellation, the remaining terms in the product are:
\[ K_1 \cdot K_2 \cdot K_3 = \frac{[\text{H}^+] \cdot [\text{H}^+] \cdot [\text{H}^+] \cdot [\text{PO}_4^{3-}]}{[\text{H}_3\text{PO}_4]} = \frac{[\text{H}^+]^3 [\text{PO}_4^{3-}]}{[\text{H}_3\text{PO}_4]} \]
• This simplified expression is exactly equal to the equilibrium constant of the overall reaction, $K_c$. Therefore, we have established that $K_c = K_1 K_2 K_3$.
Step 4: Final Answer:
The overall equilibrium constant $K_c$ is equal to $K_1 K_2 K_3$.
The question asks to find the relationship between the overall equilibrium constant ($K_c$) for the complete dissociation of phosphoric acid ($\text{H}_3\text{PO}_4$) and the individual stepwise dissociation constants ($K_1, K_2, K_3$).
Step 2: Key Formula or Approach:
For a multi-step chemical reaction, if the overall reaction is obtained by adding several stepwise reactions, the overall equilibrium constant is the product of the equilibrium constants of the individual steps:
\[ K_{overall} = K_1 \times K_2 \times K_3 \times \dots \times K_n \]
Step 3: Detailed Explanation:
• Let us write the mathematical expression for each of the given step-wise dissociation processes:
• For the first step:
\[ \text{H}_3\text{PO}_4 \rightleftharpoons \text{H}^+ + \text{H}_2\text{PO}_4^- \implies K_1 = \frac{[\text{H}^+][\text{H}_2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} \]
• For the second step:
\[ \text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} \implies K_2 = \frac{[\text{H}^+][\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} \]
• For the third step:
\[ \text{HPO}_4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-} \implies K_3 = \frac{[\text{H}^+][\text{PO}_4^{3-}]}{[\text{HPO}_4^{2-}]} \]
• The final overall dissociation reaction represents the complete deprotonation of phosphoric acid:
\[ \text{H}_3\text{PO}_4 \rightleftharpoons 3\text{H}^+ + \text{PO}_4^{3-} \]
• The equilibrium constant expression for this target reaction is:
\[ K_c = \frac{[\text{H}^+]^3 [\text{PO}_4^{3-}]}{[\text{H}_3\text{PO}_4]} \]
• Let us evaluate the mathematical product of the three step-wise constants, $K_1 \times K_2 \times K_3$:
\[ K_1 \cdot K_2 \cdot K_3 = \left( \frac{[\text{H}^+][\text{H}_2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} \right) \times \left( \frac{[\text{H}^+][\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} \right) \times \left( \frac{[\text{H}^+][\text{PO}_4^{3-}]}{[\text{HPO}_4^{2-}]} \right) \]
• Looking at the terms in the numerator and denominator, we can cancel out the intermediate species $[\text{H}_2\text{PO}_4^-]$ and $[\text{HPO}_4^{2-}]$ as they appear in both places.
• After cancellation, the remaining terms in the product are:
\[ K_1 \cdot K_2 \cdot K_3 = \frac{[\text{H}^+] \cdot [\text{H}^+] \cdot [\text{H}^+] \cdot [\text{PO}_4^{3-}]}{[\text{H}_3\text{PO}_4]} = \frac{[\text{H}^+]^3 [\text{PO}_4^{3-}]}{[\text{H}_3\text{PO}_4]} \]
• This simplified expression is exactly equal to the equilibrium constant of the overall reaction, $K_c$. Therefore, we have established that $K_c = K_1 K_2 K_3$.
Step 4: Final Answer:
The overall equilibrium constant $K_c$ is equal to $K_1 K_2 K_3$.
Was this answer helpful?
0
0