We compare acidity by stability of the conjugate base.

For each compound, the acidic proton is the hydrogen attached to carbon. On deprotonation:

$\text{R-CH} \rightarrow \text{R-C}^- + H^+$

The stronger acid is the one whose conjugate base is more stabilized.

Examine each option.

(A) $CHBr_3$

Its conjugate base is $CBr_3^-$. The three bromine atoms exert a $-I$ (electron-withdrawing inductive) effect, which stabilizes the negative charge.

(B) $CHI_3$

Its conjugate base is $CI_3^-$. Iodine also shows a $-I$ effect, but it is weaker than bromine and chlorine due to lower electronegativity. So $CHI_3$ is less acidic than $CHBr_3$ and $CHCl_3$.

(D) $CHCl_3$

Its conjugate base is $CCl_3^-$. Chlorine has a stronger $-I$ effect than bromine and iodine, so among haloforms:

$CHCl_3 > CHBr_3 > CHI_3$

in acidity.

(C) $CH(CN)_3$

Its conjugate base is $(CN)_3C^-$. Each $-CN$ group is a very strong electron-withdrawing group due to both:

• strong inductive effect ($-I$), and
• additional stabilization because the adjacent nitrile groups strongly delocalize/polarize electron density.

Thus the carbanion is much more stabilized than in haloforms.

Compare overall acid strengths.

Since $-CN$ is much more electron-withdrawing than halogens, three cyano groups stabilize the conjugate base far more effectively than three halogens.

Therefore:

$CH(CN)_3 \text{ is the strongest acid}$

Option check

• (A) $CHBr_3$ — not strongest
• (B) $CHI_3$ — not strongest
• (C) $CH(CN)_3$ — strongest
• (D) $CHCl_3$ — stronger than (A), (B) but weaker than (C)

Final answer

The strongest acid is:

$\boxed{CH(CN)_3}$