what is the equilibrium constant of pure water at 25°c?\n10^14\n10^-14\n10^-7\n10^7

what is the equilibrium constant of pure water at 25°c?\n10^14\n10^-14\n10^-7\n10^7

what is the equilibrium constant of pure water at 25°c?\n10^14\n10^-14\n10^-7\n10^7

Answer

Answer:

  1. $10^{-14}$

Explanation:

Step1: Recall water - autoionization

The auto - ionization of water is given by the equation $H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$. The equilibrium constant expression for this reaction, $K_w = [H^+][OH^-]$.

Step2: Know concentrations at 25°C

At $25^{\circ}C$, in pure water, $[H^+]=[OH^-]=1.0\times10^{-7}\ M$.

Step3: Calculate $K_w$

$K_w=(1.0\times 10^{-7})\times(1.0\times 10^{-7}) = 1.0\times10^{-14}$.