Question

What is the ion-product constant for water?

Short answer

The ion-product constant for water at 25 degrees Celsius is \(1.0 * 10^{-14}\).

Step-by-step solution

Identify the auto-ionization of water equation at 25 degree Celsius

The auto-ionization of water is represented by the following equation: \(2H_2O(l) \leftrightarrow H_3O^{+}(aq) + OH^{-}(aq)\). This equation describes the state of equilibrium that exists in water between water molecules and H+ and OH- ions. At 25 degree Celsius, the concentration of \(H_3O^{+}\) ions and \(OH^{-}\) ions in pure water are both equal to \(1.0 * 10^{-7}\) M.

Apply the ion-product constant for water

The ion-product constant for water (Kw) is the product of the concentration of the hydrogen ions and hydroxide ions. Kw = \([H_3O^{+}][OH^{-}]\). It's given that at 25°C, Kw = \(1.0 * 10^{-14}\).

Make conclusions

So, the ion-product constant for water (Kw) is \(1.0 * 10^{-14}\) at 25 degrees Celsius.