Finding pH of acid in water
Most of $\mathrm{pH}$ calculations make simplifying assumptions and one must determine if they are justified for the particular case.
Simplification for strong acids  $$\mathrm{pH}=-\log{[\mathrm{acid}]}$$
implies $\pu{[\mathrm{acid}] &gt;&gt; 10^{-7} mol/L}$.
Because water dissociation constant $K_\mathrm{w}=\ce{[H+][OH-]}=\pu{10^{-14} mol^2/L^2}$,
neutral water contains $\pu{10^{-7} mol/L}$ of $\ce{H+}$.
That means the very most $\ce{H+}$ ion must come from acid dissociation and not from water dissociation.

If OTOH very most $\ce{H+}$ are from water dissociation as in your case, another simplification says $\mathrm{pH}=7$, no matter how much of acid you added.

For general case:
$$c_{\mathrm{acid}}=\ce{[A-]}=\ce{[H+]} - \ce{[OH-]} = \ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}}$$
$${\ce{[H+]}}^2 - \ce{[H+]}\cdot c_{\mathrm{acid}} - K_\mathrm{w}=0$$
$$\ce{[H+]} = \frac{c_\mathrm{acid} + \sqrt{{ c_\mathrm{acid}}^2 + 4\cdot K_\mathrm{w}}}{2}$$
The limit case for $c_\mathrm{acid} &lt;&lt; \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = \sqrt{K_\mathrm{w}}$$
$$\mathrm{pH}=\frac 12 \mathrm{p}K_\mathrm{w}$$
The limit case for $c_\mathrm{acid} &gt;&gt; \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = c_\mathrm{acid}$$
$$\mathrm{pH} = -\log{(c_\mathrm{acid})}$$

Simplification for weak acids $$\mathrm{pH}=0.5 \left(\mathrm{p}K_\mathrm{a}-\log {\ce{[acid]}}\right)$$
implies negligible water dissociation and also negligible relative acid dissociation $\ce{[HA]} &gt;&gt; \ce{[A-]}$.
So one can assume:

The acid concentration $\ce{[HA]}$ is not affected by acid dissociation.
$\ce{[A-]} = \ce{[H+]}$