How to find pH for acetate buffer?


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> Calculate the pH of the solution that results when $\pu{40.0 mL}$ of $\pu{0.100 M}$ $\ce{CH3COOH}$ is mixed with $\pu{30.0 mL}$ of $\pu{0.200 M}$ $\ce{NaCH3COO}.$ $K\_\mathrm{a}(\ce{CH3COOH}) = \pu{1.8E-5}.$
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Can I use Henderson–Hasselbalch equation to solve this?
It would be


$$\mathrm{pH} = \mathrm{p}K\_\mathrm{a} + \log\frac{[\ce{CH3COOH}]}{[\ce{NaCH3COO}]} \quad\implies\quad \mathrm{pH} = 4.44,$$


but if I set up the ICE table


\begin{array}{rrrrr}
\ce{&CH3COOH &-> &H+ &+ &CH3COO-}\\
\text{I} & 0.100 && 0 && 0.200 \\
\text{C} & -x && +x && -x \\
\text{E} & 0.100 - x && x && 0.200 + x
\end{array}


and solve for $x,$ it gives me a different $\mathrm{pH}.$


The volumes threw me off and make me think that I have to find the total volume and new molarity before solving for $\mathrm{pH}.$


