Acid-Base buffer question
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 $-\log(1.3*10^{-4})+\log\left(\frac{0.1}{0.01}\right) \approx 4.886.$