# Finding a stable conjugate base when comparing nitrogen vs. a triple bond

I am trying to understand acid-base reactions and I am trying to figure out which conjugate base is more stable.

The molecule on the top with dashed lines around the $\ce{H}$ is the original molecule and then below it I remove $\ce{H}$ and see what the resultant conjugate base looks like. I don't understand why the bottom left conjugate base is less stable than the one on the right. Why doesn't nitrogen create a more stable conjugate base than the triple bond?

• Maybe it's due to geometry? The triple bonded C has a single lone pair to donate, while the N has two. So that would allow a wider range of acceptable orientations when considering N-/H+ collisions? – Dan Burden Apr 1 '17 at 23:51
• Hey, just so you know, the homework tag is being phased out, so please don't tag questions with it. Thanks! – heather Apr 1 '17 at 23:58
• @heather thanks! I won't do that from now on. – ahat Apr 2 '17 at 0:06
• @DanBurden could you elaborate some more? – ahat Apr 2 '17 at 0:08
• Well I think the crux of this lies in the fact that ammonia has pKa 33 while ethyne has 25, which you can apply here. Why is it so, because I guess a sp carbon is more electronegative than a sp3 nitrogen perhaps. – Sawarnik Apr 2 '17 at 13:21