Final answer:
The relationship between Ka and Kb for a conjugate acid-base pair at 25°C is KaKb = Kw, which is 1.0 × 10^-14. Using this, we can calculate that the Ka for the dimethylammonium ion is 1.9 × 10^-11.
Step-by-step explanation:
Relationship Between Ka and Kb
The balance between the ionization constants of a conjugate acid-base pair is maintained through the equation KaKb = Kw, where Ka represents the acid dissociation constant of the conjugate acid, Kb represents the base dissociation constant of the conjugate base, and Kw is the ion-product constant of water at a given temperature.
At 25°C, Kw is 1.0 × 10-14. To find the Ka for dimethylammonium ion ((CH3)2NH2+), the conjugate acid of dimethylamine ((CH3)2NH), we would use the given Kb for dimethylamine and solve the equation Ka * 5.9 × 10-4 = 1.0 × 10-14. This yields a Ka value of 1.9 × 10-11 for the dimethylammonium ion.