 Solution

The all-carbonyl hydride, HCo(CO)${}_4$, is more acidic. Loss of ${\mathrm{H}}^{+}$ gives $[\mathrm{Co}(\mathrm{CO}{)}_4{]}^{-}$, whose negative charge is stabilized by pi backbonding into four carbonyl ligands. Replacing one CO by triphenyl phosphite reduces the net electron-withdrawing character of the ligand set relative to four carbonyls. The substituted hydride is more electron rich and its conjugate-base anion is less stabilized against protonation.Both conjugate bases have a conventional 18-electron count, so the difference is not a missing-electron argument. It is the relative stabilization of acid and base by the ligands in coordination chemistry. This concerns heterolytic deprotonation, not the homolytic Co–H bond-dissociation energy.