Question
Pure water at 25^circ C ionizes in the presence of acid to form an equilibrium in which [H_(3)O^+]=[OH^-]=10^7M ionizes in the presence of acid to form an equilibrium in which [H_(3)O^+]=[OH^-]=10^-7M self-ionizes to form an equilibrium in which [H_(3)O^+]=[OH^-]=10^7M self-ionizes to form an equilibrium in which [H_(2)O^+]=[OH^-]=10^-7M
Answer
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Bradford
Master · Tutor for 5 years
Answer
Pure water at 25°C self-ionizes to form an equilibrium in which
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Explanation
Pure water has a property known as self-ionization or autoionization, where water molecules dissociate into hydroxide ions (OH−) and hydronium ions (H3O+). At a temperature of 25°C, the concentrations of these ions in pure water are equal, and the product of their concentrations is known as the ion-product constant for water (Kw). The value of Kw at 25°C is 1 x 10^-14 M^2. This means that the concentration of each ion in pure water at equilibrium is the square root of Kw, which is 1 x 10^-7 M.When an acid is added to water, it increases the concentration of H3O+ ions. This shift in equilibrium would cause the concentration of OH− ions to decrease, as the product of the concentrations must still equal 1 x 10^-14 M^2. Therefore, the statement that the concentration of both H3O+ and OH− is 10^7 M or 10^-7 M in the presence of an acid is incorrect because the presence of an acid would disrupt the equilibrium.The correct statement is that in pure water at 25°C, without the addition of acid or base, the self-ionization equilibrium results in equal concentrations of H3O+ and OH− ions, each at a concentration of 1 x 10^-7 M.