How to find the pH of acetic acid after adding sodium hydroxide? $3\cdot 10^{-2}\ \mathrm{mol}$ 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, $\ce{NaOH + CH3COOH -&gt; H2O + CH3COONa}$, 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, $\ce{CH3COONa -&gt; Na+ + CH3COO-}$, and $\ce{CH3COO- + H2O &lt;=&gt; CH3COOH + OH-}$

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?
If you add $3 \cdot 10^{-2}~\mathrm{mol}$ $\ce{NaOH}$ to $1~\mathrm{L}$ of $2 \cdot 10^{-2}~\mathrm{M}$ $\ce{AcOH}$ you end up with a solution that contains $2 \cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{AcO-}$ and $1\cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{OH-}$. So the pH is 12.