Why is CrCl3 acidic?
For starters, note that you will probably be using the hexahydrate $\ce{CrCl3 * 6 H2O}$. In fact, there is not the hexahydrate, but three distinct hexahydrate crystal structures known. These can be described as $\ce{[CrCl2(H2O)4]Cl . 2 H2O}$, $\ce{[CrCl(H2O)5]Cl2 . H2O}$ and $\ce{[Cr(H2O)6]Cl3}$. In solution, these ions will dissociate, giving you the corresponding free cation, one to three free chloride anions and potentially additional water molecules. It is safe to say that these three anions will take part in an equilibrium reaction $(1)$.

$$\ce{[CrCl2(H2O)4]+ + 2 H2O &lt;=&gt; [CrCl(H2O)5]^2+ + H2O + Cl- &lt;=&gt; [Cr(H2O)6]^3+ + 2 Cl-}\tag{1}$$

It doesn’t really matter which one is the most prevalent species, the argument is generally the same for all three. In all three cases, we have a positively charged cation to which water forms a dative bond. In neutral solution, water is known to self-ionise according to equation $(2)$

$$\ce{H2O &lt;&lt;=&gt; H+ + OH-}\tag{2}$$

The displacement of a proton is facilitated by neighbouring electron withdrawing groups: They reduce the charge density making an anion more stable. And for all intents and purposes, cations are electron withdrawing, displaying a $-I$ effect (negative inductive effect). Therefore, forming a coordinate bond to a chromium(III) ion — even if the overall complex only has a $1+$ charge — will increase the tendency of a water molecule to deprotonate. Or, in equation terms, the equilibrium of equation $(3)$ is right-shifted when compared to equation $(2)$’s equilibrium. (I’ll only discuss this for the monochloridocomplex for simplicity.)
$$\ce{[CrCl(H2O)5]^2+ &lt;=&gt; [CrCl(OH)(H2O)4]+ + H+}\tag{3}$$

Since displacing a proton is easier, the solution is more acidic, the $\mathrm{pH}$ is lower.