The increasing order of the #"p"K_"b"# value is: 1) #"HC≡C"^"-"#; 2) #"H"^"-"#; 3) #"NH"_2^"-"#; 4) #"CH"_3^"-"#?

1 Answer
Dec 14, 2016

#"pK"_b( :"CH"_3^(-)) < "pK"_b( :"NH"_2^(-)) < "pK"_b( "H":^(-)) < "pK"_b( "HC"-="C":^(-))#


Well, the #"pK"_b# of the base is the #"pK"_a# of the conjugate acid. So, first, let's compare the #"pK"_a#s.

1) #"pK"_a = 25#, for #"HC"-="CH"#
2) #"pK"_a = 35#, for #"H"_2#
3) #"pK"_a = 36#, for #:"NH"_3#
4) #"pK"_a ~~ 48#, for #"CH"_4#

Assuming that the #"pK"_a#s are reasonably accurate between hydrogen and ammonia, let's find the #"pK"_b#s and compare. But first, let's rationalize this.

A stronger acid is a weaker base. The higher the #"pK"_a#, the lower the #"pK"_b#.

Thus, we expect that the lowest #"pK"_b# goes to #"CH"_3^(-)#, i.e. it is the strongest base (just like a low #"pK"_a# indicates a strong acid).

That should make sense, because really, when do you ever see methane deprotonated? Acetylene, however, tends to be deprotonated by, say, #"NaNH"_2#, or #"NaH"#, to react with alkyl halides and form a lengthened alkyne. That does in fact demonstrate the stronger basicity of #"NH"_2^(-)# or #"H"^(-)# relative to acetylide.

1) #bb("pK"_b) = 14 - "pK"_a = bb(-11)#

2) #bb("pK"_b) = 14 - "pK"_a = bb(-21)#

3) #bb("pK"_b) = 14 - "pK"_a = bb(-22)#

4) #bb("pK"_b) = 14 - "pK"_a = bb(-34)#

And checking these with a second method, we should get the same thing.

#10^(-"pK"_a) = K_a#

1) #K_a = 10^(-25)#

2) #K_a = 10^(-35)#

3) #K_a = 10^(-36)#

4) #K_a = 10^(-48)#

#K_w = K_aK_b => K_b = K_w/K_a#

1) #K_b = 10^(-14)/(10^(-25)) = 10^11#

2) #K_b = 10^(-14)/(10^(-35)) = 10^21#

3) #K_b = 10^(-14)/(10^(-36)) = 10^22#

4) #K_b = 10^(-14)/(10^(-48)) = 10^34#

Thus:

1) #bb("pK"_b) = -log(10^11) = bb(-11)#

2) #bb("pK"_b) = -log(10^21) = bb(-21)#

3) #bb("pK"_b) = -log(10^22) = bb(-22)#

4) #bb("pK"_b) = -log(10^34) = bb(-34)#

Same result. So yes, our qualitative prediction made sense. We knew that the #"pK"_b# of #"CH"_3^(-)# would be lowest, i.e. that it would be the strongest base.

Therefore, in increasing order of #"pK"_b#, i.e. decreasing basicity, it would be:

#color(blue)("pK"_b( :"CH"_3^(-)) < "pK"_b( :"NH"_2^(-)) < "pK"_b( "H":^(-)) < "pK"_b( "HC"-="C":^(-)))#