What is the ionic product for water (Kw) at 25°C?

Prepare for the NCEA Level 3 Chemistry – Aqueous Chemistry (AS91393) Exam. Utilize flashcards and multiple choice questions with hints and explanations. Excel on your exam!

The ionic product for water (Kw) at 25°C is defined as the equilibrium constant for the self-ionization of water. This process can be expressed by the following equilibrium equation:

[ H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq) ]

At equilibrium, the concentration of hydrogen ions ([H^+]) and hydroxide ions ([OH^-]) in pure water at 25°C is both 1 x 10^-7 mol/L. The ionic product (Kw) can be calculated using the equation:

[ K_w = [H^+][OH^-] ]

Substituting the concentrations at 25°C:

[ K_w = (1 \times 10^{-7})(1 \times 10^{-7}) = 1 \times 10^{-14} ]

This means that the actual value for Kw at this temperature is 1 x 10^-14. This value is critical in understanding the acid-base balance in aqueous solutions and is a reference point in many calculations involving pH and pOH.

Understanding that at 25°C the ionic product is a constant aids in predicting the behavior of acids and bases in water,

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