Exact sequence multiplies an ideal

Question

let $A,B,C$ be $R$-modules and $I\subset R$ be an ideal. Given that $A\xrightarrow{f} B\xrightarrow{g} C\to 0$ is exact, do we have that $IA\xrightarrow{f'} IB\xrightarrow{g'} IC\to 0$ is exact? I have shown that $g'$ is surjective since $Im(g')=I\cdot Im(g)$. However, I have trouble showing that $\ker{g'}=I\cdot\ker{g}$. By definition $\ker{g'}=\{\sum i_k b_k\in IB: g'(\sum i_k b_k)=\sum i_k g(b_k)=0\}$, but this doesn't imply that $g(b_k)=0$. Any hints?