Acid-Base buffer question Calculate the change in pH of a buffer made by combining $\pu{500 ml}$ of $\pu{0.1 M}$ $\ce{CH_3CH_2COOH}$ and $\pu{500 ml}$ of $\ce{CH_3CH_2COONa}$, when $0.04$ moles of $\ce{NaOH}$ is added to the solution. Assume the addition of sodium hydroxide has no effect on volume.
  ($\pu{\mathit{K_a} = 1.3 \times 10^{-4}}$)


My solution:  

$$\pu{0.10 M \times ~ 500 ml = 0.05 mol} \ \ce{CH3CH2COOH}$$
$$\pu{0.12 M \times ~ 500 ml = 0.06 mol} \ \ce{CH3CH2COO-}$$ 

$$\pu{pH = p\mathit{K_a} + \log \left(\frac{0.06}{0.05}\right) = 3.965}$$  

Using the equation again where the salt is $\pu{0.1M}$ and the acid is $\pu{0.01M}$ due to the addition of $\ce{NaOH}$ yields a wrong result equal to $1$.

I honestly find the wrong answer to be logical, however I can't think of a way to find the right one.
The reaction is $\ce{NaOH + CH3CH2COOH -&gt; Na+ + H2O + CH3CH2COO-},$ so adding $\pu{0.04}$ moles $\ce{NaOH}$ will result in the solution having $\pu{0.06 +0.04 = 0.1 mol CH_{3}CH_{2}COO^-}$ and $0.05 - 0.04 = 0.01\, \pu{mol} \, \ce{CH3CH2COOH}.$ Then, you can use the Henderson-Hesselbach equation to find the final pH is equal to $-\log(1.3*10^{-4})+\log\left(\frac{0.1}{0.01}\right) \approx 4.886.$