Do you know how to find pH?

When dilutions are made on aqueous solutions of acids or bases, at some point the effect of the self-dissociation of water comes into play.

$\ce{2 H2O\space (l) -> H3O+ (aq) + OH- (aq)}$

The amount of $\ce{[H3O+]}$ from the dissociation of water will vary depending on the concentration of the solution, so the equilibrium expression for this dissociation must be used.

$\ce{[H3O+][OH- ]} = 1.00 \times 10^{-14}$

If $x$ is the amount of $\ce{H3O+}$ (and of $\ce{OH-}$ ) from the dissociation of water, the total concentration of the $\ce{H3O+}$ in the solution will be this amount plus the amount from the added acid which is $1.00 \times 10^{-8} \text{ M}$ :

$\ce{[H3O+]} = x + 1.00 \times 10^{-8} \text{ M}$

Therefore, $\begin{aligned} \ce{[H3O+][OH- ]} & = 1.00 \times 10^{-14} \\ (x+10^{-8})x & = 10^{-14} \\ x^2 + 10^{-8}x - 10^{-14} & = 0 \\ \Rightarrow x & = 9.512 \times 10^{-8} \end{aligned}$

And we have:

$\begin{aligned} \ce{[H3O+]} & = x + 1.00 \times 10^{-8} \\ & = 9.512 \times 10^{-8} + 1.00 \times 10^{-8} \\ & = 1.0512 \times 10^{-7} \\ & = \text{pH } 6.98 \end{aligned}$

Therefore, the $\ce{pH}$ of $10^{-8}$ M solution of $\ce{HCl}$ in water is at $25^\circ C$ is $\boxed{\text{between 6 and 7}}$ .

For more information, read Very Dilute Strong Acids and Bases .

No matter, what is the concentration of $\text{HCl}$ , its $\text{pH}$ will always be less than $7$ at $25℃$ .
In the present case, the solution is very dilute, $\text{pH}$ will be between $6$ and $7$