Answer
6.02 x $10^{13}$ H+ ions.
Work Step by Step
1. Based on the pH of pure water, find the concentration of $H^{+}$.
pH of pure water: 7
[H+] = $10^{-pH}$
[H+] = $10^{-7}$ mol/l
2. Calculate the amount of mols in 1 ml of pure water.
Concentration = nº of moles $\div$ volume(L)
$10^{-7}$ = $\frac{x}{0.001}$
0.001 x $10^{-7}$ = x
x = $10^{-10}$ mol.
3. With the number of moles of H+, find the amount of ions.
nº of atoms (ions) = nº of moles $\times$ 6.02 $\times$ $10^{23}$
nº of atoms (ions) = $10^{-10}$ $\times$ 6.02 $\times$ $10^{23}$
nº of ions = 6.02 $\times$ $10^{13}$
