Question

Calculate the pHof pure water at (a)$\mathbf{25}\mathbf{°}\mathbf{C}$ and (b)$\mathbf{100}\mathbf{°}\mathbf{C}$.

Short answer

The pH value at $25°\mathrm{C}$is 6.998

The pH value at $100°\mathrm{C}$is 6.132

Step-by-step solution

Formula used

The pH of the solution can be calculated as:

$\mathbf{pH}\mathbf{=}\mathbf{-}\mathbf{log}\left[H^{+}\right]$

The formula of the ionic product of water is:

${\mathbf{K}}_{\mathbf{w}}\mathbf{=}\left[H^{+}\right]\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}$

Where,

${\mathbf{K}}_{\mathbf{w}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}$

Calculate the pH value at 25°C 

(a)

Firstly calculate the pH value at $25°\mathrm{C}$

${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-14}$

Let the concentration of $\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{‐}\right]=\mathrm{x}$

$$\begin{gathered} {\mathrm{x}}^2=1.0\times {10}^{-14} \\ \mathrm{x}=\left[{\mathrm{H}}^{+}\right]=1.005\times {10}^7 \\ \mathrm{pH}=-\log \left(1.005\times {10}^{-7}\right) \\ =6.998 \\ \approx 7 \end{gathered}$$

(b)

Calculate the pH value at $100°\mathrm{C}$

At ${\mathrm{K}}_{\mathrm{w}}=5.43\times {10}^{-14}$

${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=5.43\times {10}^{-13}$

Let the concentration of$\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{‐}\right]=\mathrm{x}$

$$\begin{gathered} {\mathrm{x}}^2=5.43\times {10}^{-13} \\ \mathrm{x}=\left[{\mathrm{H}}^{+}\right]=7.368\times {10}^7 \\ \mathrm{pH}=-\log \left(7.368\times {10}^{-7}\right) \\ \mathrm{pH}=6.132 \end{gathered}$$