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