Q.iv. Write a reaction in which water acts as a base.

$pOH=-\log(2\times10^{-4})=3.7$; $pH=14-3.7=10.3$.

Conjugate acid-base pair: by the Bronsted–Lowry theory, when an acid donates a proton it becomes its conjugate base, and this acid–base pair (differing only by $H^+$) is called a conjugate acid-base pair, e.g. $CH_3COOH$ (acid) / $CH_3COO^-$ (its conjugate base).

pH calculation: given $[OH^-] = 2\times10^{-4}\ mol\,dm^{-3}$

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Bronsted acid donates H$^+$, base accepts it. From $K_w=[H^+][OH^-]$: $pH+pOH=pK_w=14$ at 25°C.

Bronsted–Lowry theory: an acid is a species that donates a proton ($H^+$) to another species; a base is a species that accepts a proton. Every acid, on losing a proton, forms its conjugate base, and every base, on gaining a proton, forms its conjugate acid.

Relationship between pH and pOH: water undergoes self-ionisation:

$H_2O \rightleftharpoons H^+ + OH^-$, with $K_w = [H^+][OH^-] = 10^{-14}$ at 25°C.

Taking $-\log$ of both sides:

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