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

Dec 14, 2016

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

Well, the ${\text{pK}}_{b}$ of the base is the ${\text{pK}}_{a}$ of the conjugate acid. So, first, let's compare the ${\text{pK}}_{a}$s.

1) ${\text{pK}}_{a} = 25$, for $\text{HC"-="CH}$
2) ${\text{pK}}_{a} = 35$, for ${\text{H}}_{2}$
3) ${\text{pK}}_{a} = 36$, for $: {\text{NH}}_{3}$
4) ${\text{pK}}_{a} \approx 48$, for ${\text{CH}}_{4}$

Assuming that the ${\text{pK}}_{a}$s are reasonably accurate between hydrogen and ammonia, let's find the ${\text{pK}}_{b}$s and compare. But first, let's rationalize this.

A stronger acid is a weaker base. The higher the ${\text{pK}}_{a}$, the lower the ${\text{pK}}_{b}$.

Thus, we expect that the lowest ${\text{pK}}_{b}$ goes to ${\text{CH}}_{3}^{-}$, i.e. it is the strongest base (just like a low ${\text{pK}}_{a}$ 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, ${\text{NaNH}}_{2}$, or $\text{NaH}$, to react with alkyl halides and form a lengthened alkyne. That does in fact demonstrate the stronger basicity of ${\text{NH}}_{2}^{-}$ or ${\text{H}}^{-}$ 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.

${10}^{- {\text{pK}}_{a}} = {K}_{a}$

1) ${K}_{a} = {10}^{- 25}$

2) ${K}_{a} = {10}^{- 35}$

3) ${K}_{a} = {10}^{- 36}$

4) ${K}_{a} = {10}^{- 48}$

${K}_{w} = {K}_{a} {K}_{b} \implies {K}_{b} = {K}_{w} / {K}_{a}$

1) ${K}_{b} = {10}^{- 14} / \left({10}^{- 25}\right) = {10}^{11}$

2) ${K}_{b} = {10}^{- 14} / \left({10}^{- 35}\right) = {10}^{21}$

3) ${K}_{b} = {10}^{- 14} / \left({10}^{- 36}\right) = {10}^{22}$

4) ${K}_{b} = {10}^{- 14} / \left({10}^{- 48}\right) = {10}^{34}$

Thus:

1) $\boldsymbol{{\text{pK}}_{b}} = - \log \left({10}^{11}\right) = \boldsymbol{- 11}$

2) $\boldsymbol{{\text{pK}}_{b}} = - \log \left({10}^{21}\right) = \boldsymbol{- 21}$

3) $\boldsymbol{{\text{pK}}_{b}} = - \log \left({10}^{22}\right) = \boldsymbol{- 22}$

4) $\boldsymbol{{\text{pK}}_{b}} = - \log \left({10}^{34}\right) = \boldsymbol{- 34}$

Same result. So yes, our qualitative prediction made sense. We knew that the ${\text{pK}}_{b}$ of ${\text{CH}}_{3}^{-}$ would be lowest, i.e. that it would be the strongest base.

Therefore, in increasing order of ${\text{pK}}_{b}$, i.e. decreasing basicity, it would be:

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