pH for Strong Acids

[rita_debbaneh2G]

Why is the pH for strong acids just the negative log of the molar concentration of the acids?

[Sabina House 2A]

The pH for strong acids is just the negative log of the molar concentration because for strong acids, all of the acid dissociates into H3O+ ions and A- ions (A- refers to the conjugate base of an acid). If the strong acid completes dissociates into H3O+ and A-, then the initial concentration of the acid is the same as the equilibrium concentration of the H3O+ ions (because the ratio of acid to H3O+ ions is 1:1). Because pH is -log[H3O+] and [H3O+] = [HA] (HA is the strong acid), then pH of a strong acid is equal to the negative log of the initial concentration of that strong acid. Hope this helps!

[Brittney Nguyen 2L]

Hi,

The pH for strong acids is not necessarily the -log of the molar concentration of the acids; rather, it is the -log of the molar concentration of protons the acid gives off. This is because strong acids dissociate completely, so all of the acid will give off a proton, thereby completely contributing to the pH.

In the case of a monoprotic strong acid, such as HCl, HCl dissociates completely into H+ and Cl-, so the concentration of H+ is the same as the concentration of HCl. Therefore, pH = -log[H+] = -log[HCl]. However, in the case of a polyprotic strong acid, such as H2SO4, we cannot assume that pH = -log[acid] because H2SO4 can dissociate completely into 2H+ and SO4^2-. The concentration of H+ produced is twice that of the concentration of H2SO4. Thus, pH in this scenario = -log[H+] = -log(2[H2SO4]).

I hope this helps!

[Vince Li 2A]

You would assume that the strong acids completely dissociate, because these acids are more readily able to give off a proton, and that the anion is more stable afterwards. As a result, you would just use the concentration of the strong acid/base to calculate the pH or pOH. Also, make sure to look at the strong acid/base and make sure there are not 2HCl or Ba(OH)2. In both of those cases, you would multiply the molarity by two, and then proceed from there.