Q.i. The pH of $10^{-8}$ M of HCl is a. 8
b. 7
c. less than 7
d. greater than 7

Step 1. A naive calculation would say $[H^+]=10^{-8}$ M gives $pH=-\log_{10}(10^{-8})=8$ -- but this ignores water's own autoionization, which is not negligible when the acid concentration is this close to (or below) water's own $[H_3O^+]=1.0\times10^{-7}$ M at 298 K.

Step 2. At $10^{-8}$ M, the HCl contributes only a small amount of extra H+ on top of what pure water already supplies; the TOTAL [H+] is therefore not simply the acid's nominal concentration, but the sum of the acid's contribution and water's own autoionization contribution (properly, found by solving the combined equilibrium).

Step 3. Because water still supplies roughly $1\times10^{-7}$ M H+ on its own, and the acid adds a comparatively small extra $10^{-8}$ M, the total [H+] ends up only slightly ABOVE $1\times10^{-7}$ M -- definitely not as high as $10^{-8}$ M would suggest on its own, but also definitely more than pure water's neutral value.

Step 4. A [H+] slightly above $1\times10^{-7}$ M gives a pH slightly BELOW 7 -- so the correct qualitative answer is 'less than 7', never exactly 8 (which would wrongly ignore water's autoionization) and never above 7 (since adding any acid, however dilute, can only push [H+] up, never down).

Recognize that at concentrations near or below water's own [H+] (~10^-7 M), water's autoionization cannot be ignored, so a naive -log(c) calculation is invalid.

• Computing pH = -log10(10^-8) = 8 directly -- this silently assumes the acid is the only source of H+, which fails once the acid concentration approaches water's own ionization level.
• Concluding the solution is basic (pH > 7) simply because the acid is extremely dilute -- adding any amount of acid can only make a solution more acidic (or leave it effectively neutral), never basic.