You are, I believe, referring to the auto-dissociation reaction of water:
$$2\mathrm{H_2O}\text(l) \leftrightarrow \mathrm{H_3O^+}\text(aq)+\mathrm{OH^-}\text(aq)$$
This equilibrium reaction obeys the following equation (the square bracketed quantities are concentrations):
$$K_W=[\mathrm{H_3O^+}][\mathrm{OH^-}]=10^{-14}\:\mathrm{mol^2/L^2}$$
At $pH=7.00$ and in pure water this means:
$$[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=10^{-7}\:\mathrm{mol/L}$$
$$pH = -\log[\mathrm{H_3O^+}]=7.00$$
The concentration of water in water is $55\:\mathrm{mol/L}$, so the ratio of $\mathrm{H_3O^+}$ to $\mathrm{H_2O}$ in pure and neutral water is about: $$1.8 \times 10^{-9}$$
Naked (non-solvated) protons $\mathrm{H^+}$ do basically not exist in water because $\mathrm{H_2O}$ is such a hard Lewis acid (nucleophile), which immediately bonds to any free protons.
The $pH$ of water is determined by means of a $\mathrm{H_3O^+}$-selective electrode, commonly known as a $pH$-meter.