= 4-Nitrophenolate
{title2=$\mathrm{C_6H_4NO_3^-}$}
<4-Nitrophenolate> is the <conjugate base> formed when <4-Nitrophenol> loses its phenolic <proton>. Its <chemical formula> is $\mathrm{C_6H_4NO_3^-}$ and its net <electric charge> is $-1$. The acid and base share the same substituted <benzene> framework; removing a proton changes the charge and electronic structure without removing the <nitro group>.<4-Nitrophenolate in ChEBI>[CHEBI:57917, definition, formula, net charge and conjugate-base relation]{ref}
The distinction is experimentally useful because alkaline <4-Nitrophenol> assays exploit the strongly absorbing anionic form. A 1980 reference-material study measured the <molar absorption coefficient> under specified alkaline conditions at $401\,\mathrm{nm}$ and $25\,{}^{\circ}\mathrm C$. Matching the final reaction conditions mattered to the <spectrophotometric analysis>.<High-purity 4-nitrophenol: purification, characterization, and specifications for use as a spectrophotometric reference material by G. N. Bowers Jr and others>[Abstract, alkaline reference measurement and need for reaction-condition-specific coefficients]{ref}
An ideal dilute-solution calculation explains why changing <pH> can change colour without forming more product. Write $c_{\rm T}=[\mathrm{HA}]+[\mathrm{A^-}]$ for total acid-plus-base concentration. The <Henderson-Hasselbalch equation> gives the anionic fraction
$$
\alpha=\frac{[\mathrm{A^-}]}{c_{\rm T}}=\frac{1}{1+10^{\mathrm{p}K_a-\mathrm{pH}}}.
$$
At $\mathrm{pH}=\mathrm{p}K_a$, half is anionic; two pH units higher, $\alpha=100/101\approx0.9901$. Thus a hypothetical fixed total $20\,\mathrm{\mu M}$ contains $10\,\mathrm{\mu M}$ anion at the first condition and about $19.80\,\mathrm{\mu M}$ at the second. These are relative-to-<pKa> examples, not an asserted universal numerical pKa. Where the protonated form absorbs weakly, <absorbance> tracks the anionic concentration more directly than the unchanged total.
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