Question

What is Ka at 25 °C for the following equilibrium? CH3NH3+(aq) + H2O(l) CH3NH2(aq) + H3O+(aq) Kb (CH3NH2) = 4.4 × 10–4 at 25 °C. a. 4.4 × 10–4 b. 2.3 × 103 c. 4.4 × 10–10 d. 4.4 × 104 e. 2.3 × 10–11

I have no idea how to do this problem please help me understand how it is done

Answer 1

`ka=2.3*10^-11`

Explanation:

`Ka*Kb=Kw`
Here's why:
Look at the expression for Ka:
`Ka=([CH_3NH_2][H_3O^+])/([CH_3NH_3^+])`

Look at the expression for Kb
`Kb= ([OH^-][CH_3NH_3^+])/([CH_3NH_2])`

And the value for Kb is what we are given

Now when I put these expression together and multiply them to find Kw, stuff does cancel out and you see a familiar expression:
`(cancel([CH_3NH_2])[H_3O^+])/cancel([CH_3NH_3^+])* ([OH^-]cancel([CH_3NH_3^+]))/cancel([CH_3NH_2])= kw`

`kw= [H_3O^+][OH^-]` and we are left with autoionization constant of water....which is a constant at 25 degrees Celsius: `1.0*10^-14`

So now that I have shown why Ka*Kb=Kw works, let's apply it to this problem:
`Ka=(1.0*10^-14)/(4.4*10^-4)`
`ka=2.3*10^-11`

Answer 2

Well, for a general monoprotic acid...

`"HA"(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A"^(-)(aq)`

`K_a = (["H"_3"O"^(+)]["A"^(-)])/(["HA"])`

And for a general base that grabs one proton...

`"A"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HA"(aq) + "OH"^(-)(aq)`

`K_b = (["HA"]["OH"^(-)])/(["A"^(-)])`

Now what if we do this...

`K_aK_b = (["H"_3"O"^(+)]cancel(["A"^(-)]))/cancel(["HA"])(cancel(["HA"])["OH"^(-)])/(cancel(["A"^(-)])) = ["H"_3"O"^(+)]["OH"^(-)]`

That looks just like `K_w`.

`=> K_w = K_aK_b`

Therefore, since `K_w = 10^(-14)` at `25^@ "C"`...

`K_a = K_w/K_b = 10^(-14)/K_b = ???`