Question:
In a solution of pure water, the dissociation of water can
Last updated: 8/4/2022
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In a solution of pure water, the dissociation of water can be expressed by the following: H₂O(l) + H₂O(l) H₂O'(aq) + OH(aq) The equilibrium constant for the ionization of water, K, is called the ion-product of water. In pure water at 25 °C, K has a value of 1.0 10¹. The dissociation of water gives one H₂O' ion and one OH ion and thus their concentrations are equal. The concentration of each is 1.0 x 10 M. K - [HO][OH^] K(1.0 10 )(1.0 × 10) - 1.0 × 10 ¹¹ X X [H₂O'][OH]- 1.0 × 10 ¹ X A solution has a [OH ]-3.4 10 M at 25 °C. What is the [HO] of the solution? O O 2.9 × 109 M X 3.4 × 10⁹ M 2.9 × 10-¹5 M 2.9 × 10 10 M I DON'T KNOW YET