By introducing a weak acid as $HNO_2$ in water, we form among others its corresponding base $NO_2^-$, but it also remains the unionized acid $HNO_2$. Because each acid molecule that dissociates gave birth to a base molecule, it is clear that: The number of acid molecules initially introduced equals the number acid molecules remaining in non-ionized state in addition to the number of molecules of the newly formed base. $c_{HNO_2}$ $=$ $[HNO_2]\;+\;[NO_2^-]$
The same also applies to weak bases.
For Broenstedt acids or bases : Original molarity = sum of molarities of species after ionization
A winemaker bought a stock of 1.000 empty bottles. After harvest, he filled and corked 850 bottles, so it remained 150 empty bottles. Obviously the initial stock = Number of empty bottles that remain + the number of full bottles.
1) Aqueous ammonia solution: Conservation of matter: $c_{NH_3}$ $=$ $[NH_3]+[NH_4^+]$ 2) Chlorhydric acid solution (fully dissociated): Conservation of matter: $c_{HCl}$ $=$ $[Cl^-]$ 3) Aqueous sulfuric acid solution (fully dissociated into $HSO_4^-$, which in turn partially dissociates in $SO_4^{2-}$) Conservation of matter: $c_{H_2SO_4}$ $=$ $[HSO_4^-]+[SO_4^{2-}]$