4. Let $I$ be the incenter of $\triangle A B C$, and $3 \overrightarrow{I A}+4 \overrightarrow{I B}+5 \overrightarrow{I C}=0$, then the size of angle $C$ is
$$
\begin{array}{l}
\text { As shown in the figure, } \frac{B D}{D C}=\frac{c}{b} \Rightarrow B D=\frac{a c}{b+c}, D C=\frac{a b}{b+c} . \\
\overrightarrow{B D}=\frac{c}{b+c} \overrightarrow{B C} \Rightarrow \overrightarrow{I D}=\frac{c}{b+c} \overrightarrow{I C}+\frac{b}{b+c} \overrightarrow{I B}, \\
\text { Also, } \frac{I D}{A I}=\frac{a}{b+c} \Rightarrow \overrightarrow{I D}=\frac{a}{b+c} \overrightarrow{A I},
\end{array}
$$

Thus, we have $a \overrightarrow{I A}+b \overrightarrow{I B}+c \overrightarrow{I C}=\mathbf{0}$. According to the problem, 
$$
a: b: c=3: 4: 5 \Rightarrow \cos C=0 \Rightarrow C=\frac{\pi}{2} .
$$