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4-Nitrophenol is phenol with a nitro group opposite the hydroxy group on the benzene ring: positions 1 and 4. Its chemical formula is ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{O}}_3$, and the alternative name p-nitrophenol describes the same substitution pattern.^[ref]

Removing the phenolic proton produces its conjugate base, 4-Nitrophenolate. This change matters in optical enzyme assays: alkaline phosphatase can release 4-Nitrophenol from 4-Nitrophenyl phosphate, and the alkaline mixture favours the strongly absorbing anionic form. A spectrophotometer follows product-associated absorbance; an enzyme-free reference blank distinguishes the reaction signal from background.^[ref]

A 1980 reference-material study measured a molar absorption coefficient of $18380\pm 90\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$ at $401\mathrm{nm}$ and $25{}^{\circ }\mathrm{C}$ under its specified alkaline conditions.^[ref] For an illustrative blank-corrected absorbance $A=0.3676$ and optical path $l=1\mathrm{cm}$ under matching conditions, the Beer-Lambert law gives $c=A/(\epsilon l)=20\mu \mathrm{mol}{\mathrm{L}}^{-1}$. Applying that coefficient at a different wavelength, pH or reaction composition needs validation: colour alone does not fix a concentration.

4-Nitrophenolate is the conjugate base formed when 4-Nitrophenol loses its phenolic proton. Its chemical formula is ${\mathrm{C}}_6{\mathrm{H}}_4\mathrm{N}{\mathrm{O}}_3^{-}$ 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.^[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.^[ref]

An ideal dilute-solution calculation explains why changing pH can change colour without forming more product. Write $c_{\mathrm{T}}=[\mathrm{HA}]+[{\mathrm{A}}^{-}]$ for total acid-plus-base concentration. The Henderson-Hasselbalch equation gives the anionic fractionAt $\mathrm{pH}=\mathrm{p}{K}_a$, half is anionic; two pH units higher, $\alpha =100/101\approx 0.9901$. Thus a hypothetical fixed total $20\mu \mathrm{M}$ contains $10\mu \mathrm{M}$ anion at the first condition and about $19.80\mu \mathrm{M}$ 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.