Why is #"HCCl"_3# a stronger acid than #"HCF"_3#?

We can compare the acidities of acids by examining the stabilities of their conjugate bases.

#"HA" ⇌ "H"^"+" + A^"-"#

The more stable the anion, the greater the acidity will be.

We have the equilibria

#"Cl"_3"C-H" ⇌ "H"^"+" + "Cl"_3"C:"^"-"#

and

#"F"_3"C-H" ⇌ "H"^"+" + "F"_3"C:"^"-"#

#"F"# is highly electronegative, so we would expect it to stabilize the negative charge on the carbanion and make fluoroform the stronger acid.

This is NOT what we observe!

The #"p"K_"a"# of fluoroform is 30.5, while the #"p"K_"a"# of chloroform is 24.4.

Chloroform is a stronger acid them fluoroform by six orders of magnitude!

The explanation is that #"C"# and #"F"# atoms are about the same size.

Thus, the lone pairs on #"F"# are quite close to the lone pair on the carbanion.

The cumulative lone pair-lone pair electron repulsions destabilize the carbanion.

In chloroform, the #"Cl"# atoms can still stabilize by the inductive effect, but the #"Cl"# atoms are much bigger.

The lone pairs on the #"Cl"# are so far away that lone pair-lone pair electron repulsion is much less than in fluoroform.

Thus, the #"Cl"_3"C:"^"-"# ion is more stable and #"HCCl"_3# is the stronger acid.