The visual pigment rhodopsin is a seven-transmembrane (7-TM) G protein-coupled receptor (GPCR). Activation of rhodopsin involves two pH-dependent steps: proton uptake at a conserved cytoplasmic motif between TM helices 3 and 6, and disruption of a salt bridge between a protonated Schiff base (PSB) and its carboxylate counterion in the transmembrane core of the receptor. Formation of an artificial pigment with a retinal chromophore fluorinated at C14 decreases the intrinsic pK(a) of the PSB and thereby destabilizes this salt bridge. Using Fourier transform infrared difference and UV-visible spectroscopy, we characterized the pH-dependent equilibrium between the active photoproduct Meta II and its inactive precursor, Meta I, in the 14-fluoro (14-F) analogue pigment. The 14-F chromophore decreases the enthalpy change of the Meta I-to-Meta II transition and shifts the Meta I/Meta II equilibrium toward Meta II. Combining C14 fluorination with deletion of the retinal beta-ionone ring to form a 14-F acyclic artificial pigment uncouples disruption of the Schiff base salt bridge from transition to Meta II and in particular from the cytoplasmic proton uptake reaction, as confirmed by combining the 14-F acyclic chromophore with the E134Q mutant. The 14-F acyclic analogue formed a stable Meta I state with a deprotonated Schiff base and an at least partially protonated protein counterion. The combination of retinal modification and site-directed mutagenesis reveals that disruption of the protonated Schiff base salt bridge is the most important step thermodynamically in the transition from Meta I to Meta II. This finding is particularly important since deprotonation of the retinal PSB is known to precede the transition to the active state in rhodopsin activation and is consistent with models of agonist-dependent activation of other GPCRs.