Given that 𝐾a for HBrO is 2.8×10−9 at 25 °C, what is the value of 𝐾b for BrO− at 25 °C? Kb=

Question

Given that 𝐾b for (CH3)2NH is 5.4×10−4 at 25 °C, what is the value of 𝐾a for (CH3)2NH+2 at 25 °C?

Ka=

Answer 1

You should recall the relationships between Ka and Kb and pKa and pKb.

Ka x Kb = 1x10^-14

pKa + pKb = 14

HBrO(aq) <==> H⁺(aq) + BrO^-(aq)

Ka HBrO = 2.8x10^-9

Kb BrO^- = 1x10^-14 / 2.8x10^-9

Kb BrO- = 3.6x10^-6