hw 12.39

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Parsia Vazirnia 2L
Posts: 44
Joined: Fri Jul 22, 2016 3:00 am

hw 12.39

Postby Parsia Vazirnia 2L » Thu Nov 24, 2016 2:43 pm

Hi for problem number 39 on chapter 12, why are we subtracting 14 from the pkA value? why is (Ch3)2NH2+ and +NH3OH bases? They have protons to donate

Same goes for problem number 41 except vice versa the question involves pKb

Pauline Tze 3B
Posts: 57
Joined: Sat Jul 09, 2016 3:00 am
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Re: hw 12.39

Postby Pauline Tze 3B » Fri Nov 25, 2016 5:29 pm

Hi Parsia,

If you look carefully at (CH3)2NH2 and NH2OH on Table 12.2, they are bases (we are given the conjugate acids in the problem (Ch3)2NH2+ and +NH3OH respectively) so therefore, in the table we are given pKb values. (Intuitively, you can also predict that these will be the weakest acids because we are give pKb rather than pKa values).

However, to compare the relative strengths of the acids, we need to convert them all to pKa.
To do that, the equation is pKw = 14 = pKa + pKb

This process also applies to 12.41, in which you need to convert them all to pKb instead.

I hope that helps!

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