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'Extreme Dissociation' printed from https://nrich.maths.org/
1 molecule of water dissociated into $H^+$ and $OH^-$ in 100ml of
water.
Thus in 1l of water, there will be 10 molecules of $H^+$.
$\therefore [H^+] = 10\ molecules/l$
By dividing by Avogadro's number, this gives:
$ [H^+] = 1.66 \times 10^{-23}\ mol/l$
$$\therefore pH = -log_{10}[H^+] = 22.78\ (2\ d.p.)$$
Under the condition that every molecule of water dissociates, we
need to make an assumption that 100ml of water weighs approximately
100g.
Therefore 100ml of water is $\frac{100}{18} = 5.56$ moles
Each mole of water gives 1 mole of $[H^+]$.
Therefore $$[H^+] = \frac{5.56}{0.1} = 55.6 mol/l$$
$$\therefore pH = -log_{10}[H^+] = -1.74$$
Under real world situations, the pH of water is 7.
Therefore $[H^+] = 10^{-7} mol/l = 6.02 \times 10^{16} molecules
/l$
Therefore there is a single $H^+$ in $\frac{1}{6.02 \times 10^{16}}
= 1.66 \times 10^{-17} l$