What is the ionic product of water (Kw) and its units?

What is the ionic product of water (Kw) and its units?

• A. Kw = [H+][OH-] = 1.0×10^-14 (mol^2/L^2) at 25°C.
• B. Kw = [H+]+[OH-] = 1.0×10^-14 (mol/L).
• C. Kw = [H+][OH-] = 1.0×10^-7 (mol^2/L^2).
• D. Kw is not fixed.

Answer: (A)

The key idea is that Kw is the equilibrium constant for the autoionization of water, defined as the product of the hydrogen ion and hydroxide ion concentrations: Kw = [H+][OH−]. Since both [H+] and [OH−] are in molarity (mol/L), their product has units of mol^2/L^2. At 25°C, this product is 1.0 × 10^−14, so Kw = 1.0 × 10^−14 (mol^2/L^2). In pure water, [H+] = [OH−] = 1.0 × 10^−7 M, and Kw = (1.0 × 10^−7)^2 = 1.0 × 10^−14. Kw depends on temperature, so its numeric value isn’t fixed for all temperatures, only at a given temperature like 25°C. The idea of Kw as a sum of concentrations or as 1.0 × 10^−7 misstates the definition.

The key idea is that Kw is the equilibrium constant for the autoionization of water, defined as the product of the hydrogen ion and hydroxide ion concentrations: Kw = [H+][OH−]. Since both [H+] and [OH−] are in molarity (mol/L), their product has units of mol^2/L^2. At 25°C, this product is 1.0 × 10^−14, so Kw = 1.0 × 10^−14 (mol^2/L^2). In pure water, [H+] = [OH−] = 1.0 × 10^−7 M, and Kw = (1.0 × 10^−7)^2 = 1.0 × 10^−14. Kw depends on temperature, so its numeric value isn’t fixed for all temperatures, only at a given temperature like 25°C. The idea of Kw as a sum of concentrations or as 1.0 × 10^−7 misstates the definition.