Given that Ka for HClO is 4.0×10−8 at 25 °C, what is the value of Kb for ClO− at 25 °C?

Answer 1

So, firstly, it's important to realise that HClO and ClO^- are a conjugate acid/base pair? Meaning, that if we write the reaction:

HClO + H₂O --> ?

The HClO will act as an acid, giving off a hydrogen, and forming its conjugate base

HClO + H₂O --> ClO^- + H₃O⁺

Likewise, we can write the base's equation as such:

ClO- + H₂O --> HClO + OH^-

Now here's the trick with Ka and Kb: the willingness of the acid to let go of the H+, and form the conjugate base, and the willingness of the conjugate base to pick up the H+, and form the conjugate acid are only being compared to the dissociation (read: willingness to separate) of water.

Ka * Kb = Kw

Where Ka is how much the acid is willing to let go of its H+ in water, Kb is how much the conjugate base is willing to pick up the H+ in water, and Kw is 1.0E-14.

We can substitute in the values you've been given:

4.0E-8 * Kb = 1.0E-14

then solve for Kb...:

Kb = (1.0E-14) / (4.0E-8)

Please let me know if that helps, and if not, send me a message--I'm happy to follow-up!