Question

The ionic product of water at ordinary temperature (25°C) is a constant value of :

• A. $10^{-4}$
• B. $10^{-7}$
• C. $10^{-8}$
• D. $10^{-14}$

Hint:

Since $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$ at $25^\circ\text{C}$, taking the negative logarithm of both sides yields the familiar relationship: $\text{pH} + \text{pOH} = 14$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Water undergoes self-ionization (auto-protolysis) to a very small extent, where a water molecule transfers a proton to another water molecule, producing hydronium ($\text{H}_3\text{O}^+$) and hydroxide ($\text{OH}^-$) ions:
\[ 2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]
The equilibrium constant for this reaction is known as the ionic product of water ($K_w$).

Step 3: Detailed Explanation:
The ionic product of water ($K_w$) is mathematically defined as:
\[ K_w = [\text{H}^+][\text{OH}^-] \]
At ordinary room temperature ($25^\circ\text{C}$ or $298\text{ K}$), pure water is neutral, meaning the concentration of hydrogen ions equals the concentration of hydroxide ions:
\[ [\text{H}^+] = [\text{OH}^-] = 1.0 \times 10^{-7} \text{ mol/L} \]
Substituting these concentrations into the expression for $K_w$:
\[ K_w = (1.0 \times 10^{-7}) \times (1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \]
This value, $10^{-14}$, is a constant at $25^\circ\text{C}$ for all aqueous solutions.
If the concentration of $\text{H}^+$ increases (acidic solution), the concentration of $\text{OH}^-$ must decrease proportionally to maintain this constant product, and vice versa.

Step 4: Final Answer:
The ionic product of water at $25^\circ\text{C}$ is $10^{-14}$, corresponding to option (D).