# Calculating pH of extremely dilute HCl solution [duplicate]

The $$pH$$ of a $$10^{-8} M$$ (M = moles per liter) $$HCl$$ solution is to be calculated. The plan is to calculate the ionization constant ($$K_a$$/acidity) first, and then derive the $$pH$$.

However, as the solution is extremely dilute, autoionization of water will have to be considered. I can't figure out how to do that.

Additionally, is it possible to derive $$K_a$$ or $$pH$$ from just the molarity of an acidic solution?

This problem is from a textbook, and the following solution is provided (which doesn't make sense, to me at least):

$$2H_2O$$ $$\rightarrow$$ $$H_3O^+ + OH^-$$

$$K_w = [OH^-][H_3O^+] = 10^{-14}$$

Let $$x = [OH^-] = [H_3O^+]$$ from the ionization of $$H_2O$$. $$H_3O^+$$ is generated from $$HCl$$ as well as the above autoionization of water:

$$HCl + H_2O$$ $$\rightarrow$$ $$H_3O^+ + Cl^-$$

Both must be considered as the solution is extremely dilute.

$$[H_3O^+] = 10^{-8} + x$$ ($$10^{-8}$$ comes from $$HCl$$ and $$x$$ comes from water.)

(Below is the part that I don't understand a bit)

$$K_w = (10^{-8} + x)(x) = 10^{-14}$$ $$x^2 + 10^{-8}x - 10^{-14} = 0$$

(I have no idea why the book solution even mentions the above equation).

$$[OH^-] = x = 9.5 * 10^{-8}$$ $$pOH =7.02$$ and $$pH = 6.98$$.

I have copied the book's solution as exactly as I could. This doesn't make any sense to you. Does it make any sense to you? Are there any alternative ways to solve this problem? Would the alternative require other pieces of information?