Question
Derive the relationship between $pH$ and $pОН.$

Answer

Relationship between $pH$ and $pOH:$
The ionic product of water is given as:
$K_w = [H_3O^+][OH^–]$
Now, $K_w=1 \times 10^{-14}$ at $298 K$
Thus, $\left[ H _3 O ^{+}\right]\left[ OH ^{-}\right]=1.0 \times 10^{-14}$
Taking the logarithm of both the sides, we write
$log_{10}[H_3O^+] + log_{10}[OH^-] = -14$
$-log_{10}[H_3O^+] + {-log_{10}[OH^-]} = 14$
$Now, pH = -log_{10}[H_3O^+]$ and $pOH = -log_{10}[OH^-]$
$\therefore pH + pOH = 14$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free