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The ratio of $\ce{H+}$ to $\ce{H2O}$ is something around $10^{-8}$. How did people determine that?
You cannot use the $K_\mathrm{w}$, $\mathrm{p}\ce{H}$, $\ce{[H+]}$ ion concentration value to answer because you get all that from the probability. Is there any independent way to find the probability. Please elaborate on the steps. I do not want to know why the ionic product is what it is. how was the ionic product found ?
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The answer to this question was surprisingly difficult to find because all modern ways to measure pH use reference solutions of known pH and thus rely on prior knowledge of the $H^+$ activity of said reference solutions.
Finally I found this on wikipedia:
it is possible to measure the concentration of hydrogen ions directly, if the electrode is calibrated in terms of hydrogen ion concentrations. One way to do this, which has been used extensively, is to titrate a solution of known concentration of a strong acid with a solution of known concentration of strong alkaline in the presence of a relatively high concentration of background electrolyte. Since the concentrations of acid and alkaline are known, it is easy to calculate the concentration of hydrogen ions so that the measured potential can be correlated with concentrations. The calibration is usually carried out using a Gran plot.[9] The calibration yields a value for the standard electrode potential, $E^0$, and a slope factor, f, so that the Nernst equation in the form $$E = E^0 + f\frac{2.303RT}{F} \log[\mbox{H}^+]$$ can be used to derive hydrogen ion concentrations from experimental measurements of E. The slope factor, f, is usually slightly less than one. A slope factor of less than 0.95 indicates that the electrode is not functioning correctly. The presence of background electrolyte ensures that the hydrogen ion activity coefficient is effectively constant during the titration. As it is constant, its value can be set to one by defining the standard state as being the solution containing the background electrolyte. Thus, the effect of using this procedure is to make activity equal to the numerical value of concentration.
This means that the precise cocentration of $H^+$ is actually unknown what is known is it's chemical activity which has been defined as equal to the concentration for a certain standard solution.
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.
