Am I solving this pH problem correctly?
Nope.

While $\ce{NaOH}$ is a strong base, the mixture of $\ce{C6H5COOH}$ and $\ce{C6H5COONa}$ is a  buffer solution and tends to resist large changes in pH. 

After adding the $\ce{NaOH}$, the extra $\ce{OH-}$ ions would react with the undissociated $\ce{C6H5COOH}$, effectively being removed from the solution. The solution would now be $1.1 \text{L}$ and contain a bunch of ions, like $\ce{Na+}$, $\ce{C6H5COO-}$, $\ce{H+}$, and $\ce{OH-}$. You can calculate the value of $\ce{[Na+]}$ now. 

Use the Henderson's buffer equation:

$$\text{pH}=\text{p}K_a + \log{\left(\frac{\ce{[C6H5COO^-] }}{\ce{[C6H5COOH]}}\right)}$$

I think you can continue from here.