Question

(a) At any instant, how many water molecules are ionized in 1 L of pure water at pH 7.0? (b) Express this number as a percentage of the total water molecules.

Short answer

(a) The number of water molecules ionized is $\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{16}}\mathbf{}\mathbf{molecules}$.

(b) The percentage of water molecules ionized is $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{7}}\mathbf{\%}$.

Step-by-step solution

(a) Calculation of the number of water molecules ionized in 1L.

One out of every 10⁷ water molecules is ionized by breaking up into hydrogen and hydroxide ions.

At a pH value of 7, the number of moles of water molecules that are ionized is 10^-7.

Calculation of the number of molecules that are ionized:

$\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{molecules}\mathbf{=}\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{\times }\mathbf{Avogadro}\mathbf{\text{'}}\mathbf{s}\mathbf{}\mathbf{Number}$

Substitute 10^-7 for the number of moles and 6.022x10²³ for Avogadro’s number.

$$\begin{gathered} \mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{molecules}\mathbf{=}{\mathbf{10}}^{\mathbf{‐}\mathbf{7}}\mathbf{moles}\mathbf{\times }\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{16}\mathbf{}}\mathbf{molecules}\mathbf{.} \end{gathered}$$

Hence, the number of ionized molecules of water is$\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{16}\mathbf{}}\mathbf{molecules}\mathbf{.}$

(b) Calculation of the percentage of water molecules ionized to total molecules.

The mass of 1 L of water is equal to 1000 g. The molecular mass of water is 18 g.

Calculation of the number of moles of water:

$\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{water}}{\mathbf{Molecular}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{water}}\mathbf{}$

Substitute 1000 g for the mass of water and 18 g for the molecular mass of water.

$$\begin{gathered} \mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{=}\frac{\mathbf{1000}\mathbf{}\mathbf{g}}{\mathbf{18}\mathbf{}\mathbf{g}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{55}\mathbf{.}\mathbf{55}\mathbf{}\mathbf{mol} \end{gathered}$$

Calculation of the total number of molecules in 1L:

$\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{molecules}\mathbf{=}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{\times }\mathbf{Avogadro}\mathbf{\text{'}}\mathbf{s}\mathbf{}\mathbf{Number}}{{\mathbf{10}}^{\mathbf{7}}}$

Substitute 55.55 mol for the number of moles and 6.022x10²³.

$$\begin{gathered} \mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{molecules}\mathbf{=}\mathbf{55}\mathbf{.}\mathbf{55}\mathbf{}\mathbf{moles}\mathbf{\times }\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{34}\mathbf{\times }{\mathbf{10}}^{\mathbf{25}}\mathbf{molecules}\mathbf{.} \end{gathered}$$

Calculation of the percentage of molecules ionized to total molecules:

$$\begin{gathered} \mathbf{Percentage}\mathbf{}\mathbf{of}\mathbf{}\mathbf{molecule}\mathbf{}\mathbf{ionized}\mathbf{=}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{molecules}\mathbf{}\mathbf{ionized}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{molecules}\mathbf{}} \\ \mathbf{Percentage}\mathbf{}\mathbf{of}\mathbf{}\mathbf{molecule}\mathbf{}\mathbf{ionised}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{16}}}{\mathbf{3}\mathbf{.}\mathbf{34}\mathbf{\times }{\mathbf{10}}^{\mathbf{25}}}\mathbf{\times }\mathbf{100} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{7}} \end{gathered}$$

Hence, the percentage of water molecules ionized is $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{7}}\mathbf{\%}$.