Question

what is the equilibrium constant of pure water at 25°c?
○ 10¹⁴
○ 10⁻¹⁴
○ 10⁻⁷
○ 10⁷

Explanation:

For pure water at \(25^{\circ}\text{C}\), the auto - ionization reaction is \(\ce{H2O(l)
ightleftharpoons H+(aq) + OH-(aq)}\). The equilibrium constant for this reaction, \(K_w\) (ion - product constant of water), is defined as \(K_w = [\ce{H+}][\ce{OH-}]\). At \(25^{\circ}\text{C}\), the concentration of \(\ce{H+}\) and \(\ce{OH-}\) in pure water is each \(1.0\times 10^{-7}\space mol/L\). So \(K_w=(1.0\times 10^{-7})\times(1.0\times 10^{-7}) = 1.0\times 10^{-14}\).

Answer:

B. \(10^{-14}\)