At 25°C, what are Kw and the relationship between pH and pOH in aqueous solutions?

At 25°C, what are Kw and the relationship between pH and pOH in aqueous solutions?

• A. Kw equals 1.0×10^-15; pH + pOH = 14.
• B. pH and pOH are always equal.
• C. Kw remains constant at 1.0×10^-14; pH + pOH = 14 at 25°C.
• D. Kw = 1.0×10^-14 and pH + pOH = 14 at 25°C.

Answer: (D)

The main idea is the autoionization of water and how pH and pOH relate to it. Water self-ionizes to give H+ and OH−, and their product is a constant at a given temperature: Kw = [H+][OH−]. At 25°C this constant has the value 1.0×10^−14. Relate pH and pOH through their definitions: pH = −log[H+] and pOH = −log[OH−]. Adding these gives pH + pOH = −log([H+][OH−]) = −log(Kw) = pKw. Since Kw = 1.0×10^−14 at 25°C, pKw = 14, so pH + pOH = 14 at 25°C. Kw does depend on temperature, so it isn’t a universal constant for all conditions; specifying 25°C is essential. That’s why the most complete and accurate statement is Kw = 1.0×10^−14 and pH + pOH = 14 at 25°C.

The main idea is the autoionization of water and how pH and pOH relate to it. Water self-ionizes to give H+ and OH−, and their product is a constant at a given temperature: Kw = [H+][OH−]. At 25°C this constant has the value 1.0×10^−14.

Relate pH and pOH through their definitions: pH = −log[H+] and pOH = −log[OH−]. Adding these gives pH + pOH = −log([H+][OH−]) = −log(Kw) = pKw. Since Kw = 1.0×10^−14 at 25°C, pKw = 14, so pH + pOH = 14 at 25°C.

Kw does depend on temperature, so it isn’t a universal constant for all conditions; specifying 25°C is essential. That’s why the most complete and accurate statement is Kw = 1.0×10^−14 and pH + pOH = 14 at 25°C.