pH of dilute HCl solution
Like you guessed, the pH should be around 7, since the concentration of HCl is very low. Obviously, it can't be 11.65, since it's still an acid. The reason for this discrepancy is that, for low concentrations, you must consider the self-ionization of water in order to get the correct concentration of $\ce{H3O+}$ ions. 

Consider the equilibrium:

$$\ce{2H2O &lt;=&gt; H3O+ + OH-}$$
$$K_w = \ce{[H3O+][OH^{-}]} = 1 \times 10^{-14} \tag 1$$

and also the dissociation of the acid:

$$\ce{H2O + HCl -&gt; H3O+ + Cl-}$$

Then
$$c_0 = \ce{[Cl^{-}]} = 2.26 \times 10^{-12} \tag 2 $$ 

and 

$$\ce{[H3O+] = [Cl^{-}] + [OH^{-}]} \Rightarrow \ce{[OH^{-}] = [H3O+] - [Cl^{-}]} \tag 3$$

If we replace $(2)$ in $(3)$ and that in $(1)$, we end up with

$$ K_w = \ce{[H3O+]} (\ce{[H3O+]} -c_0) $$

$$ \ce{[H3O+]} (\ce{[H3O+]} -c_0) - K_w = 0$$

using $ x = \ce{[H3O+]}$

$$x^2 - c_0x - Kw = 0$$

This can be easily solved. Taking only the positive value of $x$:

$$ x = \frac{c_0 + \sqrt{c_o^2 + 4K_w}}{2}$$

Inserting the values of $x$ and $K_w$, the result is:

$$ x = 1.000023 \times 10^{-07} = \ce{[H3O+]}$$

So

$$pH = -\log\ce{[H3O+]} = -\log{1.000011 \times 10^{-07}} = 6.999995 $$

Which is not surprising at all, but is the calculated answer.

For a less diluted solution of $\ce{HCl}$ with concentration of $1 \times 10^{-8} \frac{mol}{L}$, which still can't be calulated directly, the result is $pH = 6.98$.