is ph at eq point dependent on acid concentration at all
The concentration of both acid and base solution matters - or rather the ratio of concentrations matters. Let's look at some simple examples titrating acetic acid, which has a $K_a = 1.74 \times 10^{-5}$ with sodium hydroxide.

$$\ce{HC2H3O2 + NaOH -&gt; H2O + NaC2H3O2}$$


  Example 1 - 20.0 mL of 0.100 M acetic acid titrated with 0.100 M NaOH


In this example, since the concentrations are the same and the stoichiometric ratio is 1:1, we will need 10.0 mL of NaOH solution to reach the end point. The new volume is 40.0 mL and the concentration of acetate anion is 0.0500 M (volume doubled, concentration halved). Some of that acetate ion reacts with water in the following equilibrium:

$$\ce{C2H3O2- +H2O &lt;=&gt; HC2H3O2 + OH-}\\ K_b=\dfrac{[\ce{HC2H3O2}][\ce{OH-}]}{[\ce{C2H3O2-}]}=5.76\times 10^{-10}$$

To figure out how much hydroxide is formed (and thus eventually the pH), we need to do an ICE problem:

$$\begin{array}{|c|c|c|c|}
 &amp; [\ce{C2H3O2-}] &amp; [\ce{HC2H3O2}] &amp; [\ce{OH-}] \\ \hline
I &amp; 0.0500\ \text{M} &amp; 0\ \text{M} &amp; 1\times 10^{-7}\ \text{M} \\
C &amp; -x\ \text{M} &amp; +x\ \text{M} &amp; +x\ \text{M} \\ 
E &amp; 0.0500-x\ \text{M} &amp; x\ \text{M} &amp; 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$

Then we plug in the concentrations into the law of mass action:

$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0500-x)}$$ 

Solving for $x$ (since I am not making any approximations, I am letting Wolphram Alpha do the grind:

$$x=5.32\times 10^{-6}\ \text{M} \\
[\ce{OH-}] = 5.42 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 1.85 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.734$$


  Example 2 - 20.0 mL of 0.100 M acetic acid titrated with 0.200 M NaOH


In this example, the NaOH solution is twice as concentrated, so it only takes 10.0 mL to titrate the acid. The concentration of acetate anion is 0.0667 M.

$$\begin{array}{|c|c|c|c|}
 &amp; [\ce{C2H3O2-}] &amp; [\ce{HC2H3O2}] &amp; [\ce{OH-}] \\ \hline
I &amp; 0.0667\ \text{M} &amp; 0\ \text{M} &amp; 1\times 10^{-7}\ \text{M} \\
C &amp; -x\ \text{M} &amp; +x\ \text{M} &amp; +x\ \text{M} \\ 
E &amp; 0.0667-x\ \text{M} &amp; x\ \text{M} &amp; 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$

Then we plug in the concentrations into the law of mass action:

$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0667-x)}$$ 

$$x=6.15\times 10^{-6}\ \text{M} \\
[\ce{OH-}] = 6.25 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 1.60 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.796$$


  Example 3 - 20.0 mL of 0.200 M acetic acid titrated with 0.100 M NaOH


In this example, the NaOH solution is half as concentrated as the acetic acid solution, so it takes 40.0 mL to titrate the acid. The concentration of acetate anion is 0.0333 M.

$$\begin{array}{|c|c|c|c|}
 &amp; [\ce{C2H3O2-}] &amp; [\ce{HC2H3O2}] &amp; [\ce{OH-}] \\ \hline
I &amp; 0.0333\ \text{M} &amp; 0\ \text{M} &amp; 1\times 10^{-7}\ \text{M} \\
C &amp; -x\ \text{M} &amp; +x\ \text{M} &amp; +x\ \text{M} \\ 
E &amp; 0.0333-x\ \text{M} &amp; x\ \text{M} &amp; 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$

Then we plug in the concentrations into the law of mass action:

$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0333-x)}$$ 

$$x=4.33\times 10^{-6})\ \text{M} \\
[\ce{OH-}] = 4.43 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 2.26 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.646$$