91Ó°ÊÓ

The objective is to prove why does the neutral solution has the pH value of$7.0$

The water's ion product constant, $K_{\mathrm{w}}$, is the sum of the molar concentrations of hydronium and hydroxide ions in the aqueous solution.

$K_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

The concentrations in moles per litre are indicated by square brackets $\left(M\right)$. Any aqueous solution, regardless of how acidic, basic, or neutral it is, $K_w$is $1.0×{10}^{-14}$at $25°\mathrm{C}$.

$$\begin{gathered} K_w=[H_3{O}^{+}]\left[O{H}^{-}\right] \\ =1.0×{10}^{-14} \end{gathered}$$

When the quantity of hydronium and hydroxide ions is equal (that is, when $\left[H_3{O}^{+}\right]=\left[O{H}^{-}\right]$, the solution is neutral.

Substitute $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]$in equation (1), we get$\left[{\mathrm{H}}_3^{+}{\mathrm{O}}^{+}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0×{10}^{-14}$

${\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^2=1.0×{10}^{-14}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0×{10}^{-7}\mathrm{M}$

As a result, we have a neutral answer.

The hydronium ion concentration's negative logarithm (base 10) is the pH.

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

By substituting the value of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$of a neutral solution as $1.0×{10}^{-7}\mathrm{M}$in the above equation, we get

$$\begin{gathered} \mathrm{pH}=-\log \left(1.0×{10}^{-7}\right) \\ =-(-7.0) \\ =7.0 \end{gathered}$$

The pH of a neutral solution is thus $7.0$.