The calculation of pH for a diluted solution of strong acid like hydrochloric acid (HCl) requires consideration not only of the initial concentration of the acid but also the autoionization of water, which can significantly affect pH when dealing with very dilute solutions of strong acids. HCl is a strong acid, which means it completely dissociates in water: HCl⟶H++Cl−
Therefore, in a solution of 10−8M‌HCl, we would initially think the concentration of H+ions to also be 10−8M, potentially yielding a pH of 8 . However, we must also take into account the water dissociation: H2O⇌H++OH−
For pure water, the ion product of water (Kw) at 25∘C is: Kw=[H+][OH−]=10−14 In pure water, [H+]=[OH−]=10−7M, giving a pH of 7 . When we add a small amount of HCl , the initial thought would be that the hydrogen ion concentration simply increases to 10∧−8M. However, since water itself contributes hydrogen ions and hydroxide ions of equal concentration ( 10∧−7M ), the total concentration of hydrogen ions becomes slightly more complicated. Adding the hydrogen ion contributions from both the dissociation of HCl and the autoionization of water: [H+]‌total ‌=10−7+10−8=1.1×10−7M Now, calculating pH:
pH=−log(1.1×10−7) To solve this, we remember that log(1.1)≈0.0414, so: pH=−log(10−7)−log(1.1)≈7−0.0414=6.9586
Thus, the pH of a 10−8M‌HCl solution is about 6.96 , which means it is just a little lower than 7 due to the slight excess of hydrogen ions contributed by the very diluted HCl solution in combination with the hydrogen ions from autoionization of water. The correct answer would be: Option D: greater than 6, less than 7