Can Henderson-Hasselbalch equation be used to determine the pH of a solution when a strong base and a strong salt of that base stay together?
The pH of the solution after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ is $13.12$. Your book or solution manual is wrong. The $\mathrm{pH} = 8.87$ at the equivalence point. 

The Henderson Hasselbalch equation only applies in the buffer region - the part of the titration curve before the equivalence point. I have worked out the pH for a number of volumes of titrant and graphed the curve. The buffer region is labeled. You can see that past the equivalence point, the pH jumps to above 12 and stays there. You determined the pH after 25 mL correctly.



There is a point where the $\mathrm{pH=8.87}$ - the equivalence point (labeled with its pH). After $\mathrm{5\ mL}$ of $\ce{NaoH}$ solution is added, all acetic acid has been converted into the acetate ion. We can solve an equilibrium problem to determine the $\mathrm{pOH}$ and $\mathrm{pH}$ at that point. The following equation applies:

$$\ce{CH3CO2- + H2O &lt;=&gt; CH3CO2H + OH-}\ \ \mathrm{K_b=10^{-9.26}}$$

Since all acetic acid was converted into acetate, and the volume has doubled, the initial concentration of acetate is $\mathrm{0.1\ M}$. The initial volume of acetic acid is $\mathrm{0\ M}$ and the initial concentration of hydroxide is $\mathrm{10^{-7}\ M}\approx 0$.

$$\begin{array}{|c|c|c|c|}\hline
\ &amp; [\ce{CH3CO2-}] &amp; [\ce{CH3CO2H}] &amp; [\ce{OH-}] \\ \hline
I &amp; 0.1 &amp; 0 &amp; 0\\
C &amp; -x &amp; +x &amp; +x \\
E &amp; 0.1 -x &amp; +x &amp; +x\\ \hline
\end{array}$$

If we plug in the values into the law of mass action, we can solve for $[\ce{OH}]$ and determine $\mathrm{pOH}$.

$$\begin{align}
K_b &amp;= \dfrac{[\ce{CH3CO2H}][\ce{OH-}]}{[\ce{CH3CO2-}]}\\
10^{-9.26} &amp;= \dfrac{x^2}{0.1-x}\end{align}$$

Solving for x gives 

$$\begin{align}
[\ce{OH-}] &amp;=x=7.41\times 10^{-6}\\
\mathrm{pOH} &amp;=5.13 \\
\mathrm{pH} &amp;=8.87
\end{align}$$

Even if I had not made the assumption that the initial hydroxide concentration was zero, I would get the same $\mathrm{pH}$.


  Your book has a typographical error. Either you were meant to determine the pH after $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ or the pH after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$. Both would yield a $\mathrm{pH} =8.87$ at the equivalence point.