Autoprotolysis of water — What is the ionic product (dissociation constant) of water (Kw) at 25°C expressed in molar units?

Introduction / Context:
The self-ionization of water is fundamental to acid–base chemistry. At 25°C, water undergoes a slight autoprotolysis forming hydronium and hydroxide, characterized by the ionic product Kw. This value anchors pH and pOH scales.

Given Data / Assumptions:

• Temperature is 25°C (standard laboratory conditions).
• Pure water has equal [H+] and [OH−].
• The question expects the numerical value of Kw.

Concept / Approach:
Kw = [H+][OH−]. At 25°C, Kw ≈ 1.0 × 10^−14 (M^2). In pure water, [H+] = [OH−] = 1.0 × 10^−7 M, hence pH = 7. The offered choice using scientific notation shorthand “10-14 M” indicates the magnitude expected for the product under these conditions.

Step-by-Step Solution:

Verification / Alternative check:
Temperature dependence shows Kw increases with temperature, but at 25°C it is conventionally 10^−14, aligning with pH 7 neutrality.

Why Other Options Are Wrong:

• 10^14 M or 10^7 M: orders of magnitude and units are incorrect.
• 10^−7 M: corresponds to [H+] or [OH−] each in pure water, not the product.

Common Pitfalls:
Confusing Kw with individual ion concentrations; forgetting temperature dependence.

Final Answer:
10-14 M