In water, the following equilibrium exists: H⁺(aq) + OH⁻(aq) ⇄ H₂O(l)
In pure water at 25°C, the concentration of H⁺ ions is 1.00 × 10⁻⁷ mol/L. Calculate the value of the equilibrium constant for the reaction as written above.

Question

Quizplus · Accepted Answer

The equilibrium constant for the reaction is calculated using the concentrations of the products and reactants at equilibrium. Given that the concentration of $H^+$ ions is $1.00 \times 10^{-7}$ mol/L in pure water at 25°C, and knowing that water self-ionizes to produce equal concentrations of $H^+$ and $OH^-$ ions, the concentration of $OH^-$ ions is also $1.00 \times 10^{-7}$ mol/L. The equilibrium constant ($K$) for the reaction is the product of the concentrations of the reactants raised to their stoichiometric coefficients divided by the concentration of the products raised to their stoichiometric coefficients. Since water ($H_2O$) is a pure liquid, its concentration does not appear in the expression for $K$. Therefore, $K = [H^+][OH^-] = (1.00 \times 10^{-7})(1.00 \times 10^{-7}) = 1.00 \times 10^{-14}$.

In Water, the Following Equilibrium Exists: H+(aq) + OH−(aq) ⇄

In Water, the Following Equilibrium Exists: H⁺(aq) + OH⁻(aq) ⇄