Explain the concentration of pure water and why the equilibrium of pure water lies on the left side.

(N/A) Calculation of the concentration of pure water:
Density of pure water $= 1.0 \ g \ mL^{-1} = 1000 \ g \ L^{-1}$.
Concentration of water $= \frac{\text{Mass of } 1 \ L \text{ water}}{\text{Molar mass of } H_2O} = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol \ L^{-1}$.
The dissociation equilibrium of water is $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
The ionic product of water at $298 \ K$ is $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$.
In pure water,$[H^+] = [OH^-] = 1.0 \times 10^{-7} \ M$.
The ratio of dissociated water to undissociated water is $\frac{1.0 \times 10^{-7}}{55.55} \approx 1.8 \times 10^{-9}$.
Since this value is extremely small,the equilibrium lies heavily towards the left side (undissociated water).