Using paulings rules to predict the pKa of oxoacids?
Using paulings rules to predict the pKa of oxoacids? How do I do this?
For example, predict the pKa of H2SeO4
For example, predict the pKa of H2SeO4
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. 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).
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
Thanks
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.
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).
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?
Quick Reply
Related discussions
- Chemistry a level buffers questions
- acids and bases question
- Pre-registration Postgraduate Funding
- why are sp orbital bonds stronger than sp3 orbital bonds
- St Paul's school 16+ entry exam
- Chemistry - acids
- Help urgent chem
- Determine the pH of a 100mM aq solution of Tris at 298K
- English grammar
- CCEA Religious Studies. GCSE and A-Level thread
- Sixth Form Entrance Exam for Top UK Private Schools (Westminster, KCS etc.)
- i need help with my law module problem based question !!!
- Help! Can't manage class!
- Student Finance application & Parent's redundancy
- alevel chemistry question
- Oxidative agents
- Glasgow School of Art Interviews 2024
- Paul O Grady has Passed Away
- Thoughts on the new Beatles song "Now and Then"?
- Ask Us Anything!
Latest
Trending
Last reply 1 week ago
Im confused about this chemistry question, why does it form these productsTrending
Last reply 1 week ago
Im confused about this chemistry question, why does it form these products
