$3\cdot 10^{-2}\ \mathrm{mol}$ of $\ce{NaOH}$ are added to a $1\,\mathrm{l}$ solution of $2\cdot10^{-2}~\mathrm{M}\ \ce{CH3COOH}$. Find the $\mathrm{pH}$ of the solution.
I don't understand why my solution is wrong:
I said this, $\ce{NaOH + CH3COOH -> H2O + CH3COONa}$, so I'm left with $2\cdot10^{-2}~\mathrm{mol}$ of $\ce{CH3COONa}$ and $1\cdot10^{-2}\ \mathrm{mol}$ of $\ce{NaOH}$.
Now, from those $1\cdot10^{-2}\ \mathrm{mol}$ of $\ce{NaOH}$ I get the same moles of $\ce{OH-}$ (from dissociation).
Then, $\ce{CH3COONa -> Na+ + CH3COO-}$, and $\ce{CH3COO- + H2O <=> CH3COOH + OH-}$
Setting up the equilibrium equation: $K_h = \frac {\text{Products}}{\text{Reactants}}$, I get (where $K_h= K_\mathrm{w}/K_\mathrm{a}, K_\mathrm{a} = 1.8\cdot 10^{-5}$):
$$K_h=\frac {(x)(1\cdot 10^{-2}+x)}{x-1\cdot10^{-2} }=5.55\cdot 10^{-10}$$
Which yields $[\ce{OH-}]= x= 1.11\cdot 10^{-9}$ and so $\mathrm{pH}=5.04$ which is wrong. I know the answer should be around 12 from the answer sheet.
What am I doing wrong?