pure water at 25°c\no ionizes in the presence of acid to form an equilibrium in which h₃o⁺=oh⁻=10⁷m.\no…

pure water at 25°c\no ionizes in the presence of acid to form an equilibrium in which h₃o⁺=oh⁻=10⁷m.\no ionizes in the presence of acid to form an equilibrium in which h₃o⁺=oh⁻=10⁻⁷m.\no self - ionizes to form an equilibrium in which h₃o⁺=oh⁻=10⁷m.\no self - ionizes to form an equilibrium in which h₃o⁺=oh⁻=10⁻⁷m.

pure water at 25°c\no ionizes in the presence of acid to form an equilibrium in which h₃o⁺=oh⁻=10⁷m.\no ionizes in the presence of acid to form an equilibrium in which h₃o⁺=oh⁻=10⁻⁷m.\no self - ionizes to form an equilibrium in which h₃o⁺=oh⁻=10⁷m.\no self - ionizes to form an equilibrium in which h₃o⁺=oh⁻=10⁻⁷m.

Answer

Brief Explanations:

Pure water at 25°C self - ionizes according to the reaction $H_2O + H_2O\rightleftharpoons H_3O^+ + OH^-$. The equilibrium constant for this self - ionization, $K_w=[H_3O^+][OH^-]=1.0\times10^{- 14}$ at 25°C. In pure water, $[H_3O^+]=[OH^-]$, and solving $K_w=[H_3O^+][OH^-]$ with $[H_3O^+]=[OH^-]$ gives $[H_3O^+]=[OH^-]=10^{-7}M$. It is self - ionization, not ionization in the presence of acid to reach this equilibrium.

Answer:

self - ionizes to form an equilibrium in which $[H_3O^+]=[OH^-]=10^{-7}M$