What is the value of the ion-product of water, Kw, at 25°C?

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Multiple Choice

What is the value of the ion-product of water, Kw, at 25°C?

Explanation:
The ion-product of water (Kw) is the equilibrium constant for the autoionization of water, Kw = a(H+) a(OH−). At 25°C in pure water, the concentrations of hydronium and hydroxide are equal, each being 1.0 × 10−7 M, because every ionization event produces one of each ion. Multiplying these gives Kw = (1.0 × 10−7) × (1.0 × 10−7) = 1.0 × 10−14 (in units of M^2, though often treated with activities). This also matches pH = 7 and pOH = 7, since pKw = 14 at this temperature.

The ion-product of water (Kw) is the equilibrium constant for the autoionization of water, Kw = a(H+) a(OH−). At 25°C in pure water, the concentrations of hydronium and hydroxide are equal, each being 1.0 × 10−7 M, because every ionization event produces one of each ion. Multiplying these gives Kw = (1.0 × 10−7) × (1.0 × 10−7) = 1.0 × 10−14 (in units of M^2, though often treated with activities). This also matches pH = 7 and pOH = 7, since pKw = 14 at this temperature.

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