vii. Derive the relation pH + pOH = 14.

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, \([H_3O^+][OH^-] = 1.0 \times 10^{-14}\)

Taking the logarithm of both 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, \(pH + pOH = 14\)