Achieve Week 2 #7

[Hannah Kim 3C]

HClO is a weak acid ( Ka=4.0×10−8 ) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.032 M in NaClO at 25 °C?

Hi,
I honestly have no idea how to solve this problem.
What is a spectator ion? How do I find Kb of ClO-?
It would be helpful if you can walk me through this problem!

[Lexi Tempera 3B]

Hi!

I'm not sure either, but I can help with what a spectator ion is. It's an ion that does not participate in a chemical reactions (acid/base), in this case Cl-; I-, Cl-, and Br- typically do not participate. I'm not sure why, but this is what I can offer!

[Edwin Montalvo 1G]

hi! so the way to understand this problem is to understand 1) why NaClO acts as a weak base and 2) the processes that take NaClO to produce OH-.

1) Because NaClO is a salt, it will completely dissociate into Na+ and ClO-. We know/assume that Na+ is derived from the strong base NaOH in the reaction NaOH + H20 --> Na+ + H30+ so Na+ is a really weak base to the extent that it does not contribute to pH. Therefore, Na+ is a spectator ion because the NaOH reaction strongly favors the forward reaction, so Na+ will not react with anything (but is present). ClO-, however, is derived from the weak acid HClO in the reaction HClO + H20 --> ClO- + H3O+. Because in this reaction neither the forward/reverse reaction is favored, it is possible for ClO- to react and contribute to pH.

2) Therefore, we see ClO- + H20 --> HClO + OH-, showing us how NaClO act as a weak base. Since ClO- is a base, we need the Kb value. Because HClO is the conjugate acid of ClO- and we know that Ka x Kb = 10^-14, Kb = 10^14/Ka = 2.5 x 10^-7. Now we just need to set up our ICE table to find the [OH-] concentration from the reaction. We get x^2 / 0.032 = 2.5 x 10^-7, so x = 8.9 x 10^-5 = [OH-]. So, pOH = -log(8.9 x 10^-5) = 4.05, and since pH = 14 - pOH, pH = 9.95.