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

#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#

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 = ???#