# Using paulings rules to predict the pKa of oxoacids?

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#1
Using paulings rules to predict the pKa of oxoacids? How do I do this?

For example, predict the pKa of H2SeO4
0
13 years ago
#2
Pauling's rules for pKa values are really simple, namely:

1. the pKa of an acid (O)pE(OH)q is pKa = 8-5p
2. the successive pKa's go up by 5 per ionisation (i.e. for the second, third, fourth proton etc.)

So, for H2SeO4:
1. first rewrite it in the form used above: O2Se(OH)2
2. calculate the pKa by Pauling's rules: pKa = 8 - 5*2 = -2
3. the second pKa (there are two protons that can be removed) is thus -2 + 5 = 3.

Easy.

The rationalisation is that the more oxo groups there are, the more canonical forms you can write of the conjugate base, and thus the more stable it is. But as you ionise further, you clearly have fewer canonicals, as they are 'used up' by negative charges, as it were.
There's no 'theoretical' basis for the rule; it's just an empirical rule that often works quite well, but it should be borne in mind that it's very approximate (it isn't dependent on the central atom at all, for example).
1
#3
Yeah, I understand the rationalisation, I just didn't see where the 8-5p part was coming in before. I realise not that p represtents oxo groups.

Thanks
0
3 years ago
#4
anyone know why hno3 has a p value of 2?????
0
2 years ago
#5
P stands here for the number of oxygen without having hydrogen attached to it,in case of HNo3 out of 3 oxygen one is attached to oxygen so.
0
1 year ago
#6
(Original post by Sinuhe)
Pauling's rules for pKa values are really simple, namely:

1. the pKa of an acid (O)pE(OH)q is pKa = 8-5p
2. the successive pKa's go up by 5 per ionisation (i.e. for the second, third, fourth proton etc.)

So, for H2SeO4:
1. first rewrite it in the form used above: O2Se(OH)2
2. calculate the pKa by Pauling's rules: pKa = 8 - 5*2 = -2
3. the second pKa (there are two protons that can be removed) is thus -2 + 5 = 3.

Easy.

The rationalisation is that the more oxo groups there are, the more canonical forms you can write of the conjugate base, and thus the more stable it is. But as you ionise further, you clearly have fewer canonicals, as they are 'used up' by negative charges, as it were.
There's no 'theoretical' basis for the rule; it's just an empirical rule that often works quite well, but it should be borne in mind that it's very approximate (it isn't dependent on the central atom at all, for example).
How about HSeO2CH3? How to convert it into (O)pE(OH)q?
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