Given that Kb for CH3CH2NH2 is 6.3×10−4 at 25 °C, what is the value of Ka for CH3CH2NH3+ at 25 °C?

Answer 1

We can use the auto-ionization constant of water and is expressed as:

Kw = Ka* Kb

where Kw = 1.0 x 10^-14 and ka and kb are the acid and base constants, respectively. Another thing to note is that CH3CH2NH3⁺ and CH3CH2NH2 are a acid/conjugate base pair which means that the above equation can be used.

Given the value for Kb, we have to substitute in this value and solve for Ka.

Ka = Kw/Kb = 1.0 x 10^-14 /6.3 x 10^-4 = 1.58 x 10^-11

Answer 2

So, firstly, it's important to realise that CH₃CH₂NH₂ and CH₃CH₂NH₃⁺ are a conjugate acid/base pair... Meaning, that if we write the reaction:

CH₃CH₂NH₃⁺ + H₂O --> ?

The CH₃CH₂NH₃⁺ will act as an acid (we know it is an acid, given the fact that it has a Ka), giving off a hydrogen, and forming its conjugate base:

CH₃CH₂NH₃⁺ + H₂O --> CH₃CH₂NH₂ + H₃O⁺

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

CH₃CH₂NH₂ + H₂O --> CH₃CH₂NH₃⁺ + 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:

Ka * 6.3E-4 = 1.0E-14

then solve for Ka...:

Ka = (1.0E-14) / (6.3E-4)

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