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vii. Derive the relation pH + pOH = 14.

# vii. Derive the relation pH + pOH = 14.

### 3. Answer the following in brief :

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- vii. Derive the relation pH + pOH = 14.

**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, \(pH + pOH = 14\)