Sciencemadness Discussion Board

Electronegativity

CHRIS25 - 4-5-2014 at 11:43

Have not seen this written anywhere as I was reading about electronegativity; but was pondering this as a theory and really want to make sure that I have understood what the pauline scale is for, am I right to assume that: I am making KCl via the NaCl + KNO3 route and for the first time asked myself a question that I never asked before. When split up in boiling water What on earth makes the K+ ion migrate to the Cl- ion? why does the Na+ not return to the Cl-- ion when cooling? So I considered this: The Na has an electron charge that is more positive than the K and so the chlorine hunts down the weakest charge because Cl- has a powerful electronic pull and given the choice of K or Na chooses K because it is weaker than Na? Is this a good understanding? Because if it is, it really changes the way now I see chemistry and reactions.

[Edited on 4-5-2014 by CHRIS25]

Zyklon-A - 4-5-2014 at 12:28

It doesn't work that way at all. The ions don't migrate because one ion has a bigger affinity for another, they migrate only to precipitate the most insoluble ionic compound that can be made from the ions in solution.
At high temps, KNO3 is quite soluble, KCl is less soluble.
At low temps KNO3 is less soluble than KCl.
So at high temps the reaction might shift to the right, but at low temps, it will mostly stay on the left. NaCl + KNO3 ↔ KCl + NaNO3
However, that is a simple explanation, I doubt this reaction will ever work, because sodium chloride is not very soluble at any temp, and sodium nitrate is rater soluble at all temps. These property's probably keep the reaction on the left always.
The answer to your question is, Electronegativity has very little to do with double replacement reactions, aqueous or other wise. Single replacement? Absolutely.

CHRIS25 - 4-5-2014 at 12:48

Thankyou, glad that I posted this then. Should have added that I am just doing a trial with small amounts. It was the solubility table that I reasoned out that I should get the KCl precipitating if I kept the temp close to 90 and although KNO3 has a similar solubility as KCl I then reasoned that once the Ion bonds had been broken with the boiling water that the electronegativity aspect would come into play and the Cl would attract the K. Anyway, we'll see what happens in the meantime.

So a very nice theory - pity then that it does not work that way.

[Edited on 4-5-2014 by CHRIS25]

Zyklon-A - 4-5-2014 at 13:16

Quote: Originally posted by CHRIS25  
...I then reasoned that once the Ion bonds had been broken with the boiling water...

Ionic compounds don't have any bonds between the anion and the cation.
A crystal of NaCl is just am equal amount (by moles) of Na+ ions and Cl- ions, they are not connected at all.
For this reason it is better to write Na+Cl- as the formula, but this is rarely done, because it's not really necessary. NaCl is simply a pure ionic compound, it complete ionizes in water, so there are not any bonds there either.
This goes for all completely ionic compounds (salts).
The neutralization reaction: HCl + NaOH → NaCl + H2O, could be written like this: H+Cl- + Na=OH- → Na+Cl- + H2O.
So in reality, when you mix a strong acid ( an acid which completely ionizes) with a strong base (a base which completely ionizes), you are left with the same ions floating in solution.
So a salt water solution is really just a neutral mixture of hydrochloric acid and sodium hydroxide.

[EDIT]
Quote: Originally posted by CHRIS25  
So a very nice theory - pity then that it does not work that way.

Yeah, it is a pity, but chemistry is very (in fact completely) logical, so it only makes sense that ions floating in water will precipitate any insoluble compounds if possible. On the other hand, the fact that it works the way it does, is helpful, it allows the opposite reactions to happen.

[Edited on 5-5-2014 by Zyklonb]

Artemus Gordon - 4-5-2014 at 13:33

zyklonb is absolutely right, but just to make a slightly more direct reply to your situation: When you dissolve two salts in water, you have four ions floating freely. This means you really have four salts - in your case you have NaCl, KCl, NaNO3, and/or KNO3. So you need to look up the solubility for all four salts to see which one crashes out of solution first.

[Edited on 4-5-2014 by Artemus Gordon]

CHRIS25 - 4-5-2014 at 15:22

That is precisely what I did before I started this, analysed the solubility. Now that the reaction is finished and filtrated I am almost certain that I do have Potassium Chloride, well very possibly. What I did was wait until there were signs of a precipitate, quite a noticeable precipitate, then I removed from boil, and added just enough water to re-dissolve all, just enough to ensure that any salt that was there did also re-dissolve. I then quickly cooled the product and put it into the freezer next to my garden peas! After 10 minutes there was a massive precipitate as expected. Why, because on the solubility graph one clearly sees that approaching zero degrees KCl insolubility is greater than NaCl by about 7 grams per 100mLs. Anyway I have to find out how to test this for either K or Na. I have learned quite a lot by what has been said actually, and I see the logic.
Actually it is probably just a big mess of everything, but at least I have gained more awareness of solubility and ions in solution and how things actually work against the perception about how I thought they worked. A good exercise.

[Edited on 4-5-2014 by CHRIS25]
This has opened up a whole new understanding and filled in some blanks in my thinking: Thankyou.

Ionic compounds don't have any bonds between the anion and the cation.
A crystal of NaCl is just am equal amount (by moles) of Na+ ions and Cl- ions, they are not connected at all.
For this reason it is better to write Na+Cl- as the formula, but this is rarely done, because it's not really necessary. NaCl is simply a pure ionic compound, it complete ionizes in water, so there are not any bonds there either.
This goes for all completely ionic compounds (salts).
The neutralization reaction: HCl + NaOH → NaCl + H2O, could be written like this: H+Cl- + Na=OH- → Na+Cl- + H2O.
So in reality, when you mix a strong acid ( an acid which completely ionizes) with a strong base (a base which completely ionizes), you are left with the same ions floating in solution.
So a salt water solution is really just a neutral mixture of hydrochloric acid and sodium hydroxide.


[Edited on 4-5-2014 by CHRIS25]

blogfast25 - 5-5-2014 at 04:57

Quote: Originally posted by CHRIS25  

Ionic compounds don't have any bonds between the anion and the cation.
A crystal of NaCl is just am equal amount (by moles) of Na+ ions and Cl- ions, they are not connected at all.


That's really only half true. It's true that there are no covalent bonds between the ions in an ionic lattice. But the strong electrostatic attraction is what we call the 'ionic bond' and it's what keeps the lattice so tightly 'bonded' together.

This is important to note because the electrostatic attraction is what gives salts a general property: high melting and boiling points.

It also explains why ionic compounds often have very high Enthalpies of Formation: when these ions first join up into the lattice a lot of energy is dissipated, known as the 'lattice energy'.
Quote: Originally posted by CHRIS25  
The neutralization reaction: HCl + NaOH → NaCl + H2O, could be written like this: H+Cl- + Na=OH- → Na+Cl- + H2O.
So in reality, when you mix a strong acid ( an acid which completely ionizes) with a strong base (a base which completely ionizes), you are left with the same ions floating in solution.
So a salt water solution is really just a neutral mixture of hydrochloric acid and sodium hydroxide.


You're getting it! Except for that last sentence: there's no hydrochloric acid and sodium hydroxide left after the neutralisation: only sodium ions, chloride ions and water

The Enthalpy that is dissipated when neutralising a strong acid with a strong base (both in solution, of course) is much, much smaller than the Enthalpy of Formation of the solid salt... because the solid salt doesn't form and no lattice energy is dissipated.

In fact the only enery that is dissipated is from OH<sup>-</sup> + H<sub>3</sub>O<sup>+</sup> === > 2 H2O. This is the so-called neutralisation enthalpy (about - 58 kJ/mol of OH<sup>-</sup>;).



[Edited on 5-5-2014 by blogfast25]

MrHomeScientist - 5-5-2014 at 07:32

CHRIS25: What did your crystals look like? I have a strong suspicion that cooling your solution as you did would favor KNO3 precipitation. KCl and NaCl are cubic, while KNO3 look like needles.
This is similar to how I make my potassium nitrate from instant cold packs via:
NH4NO3 + KCl == KNO3 + NH4Cl
Cooling the solution precipitates the least soluble of the four possible salts - KNO3.

Artemus Gordon is right in that often when you mix two ionic salts together in solution, there isn't really a reaction, per se. You just have a soup of ions floating around. You only observe something happening when one of the possible combinations of ions makes something that is insoluble under the current conditions of the system. In my example above, mixing those two ionic salts results in a system where everything is dissolved and nothing really happens, but cooling in a fridge causes the K+ and NO3- ions to come together and drop out as solid crystals.

That all being said, there are times when a reaction does indeed happen upon mixing ionic salts - like neutralization reactions where water or a gas is a product.

neptunium - 5-5-2014 at 08:08

electronegativity (in case you wonder) define the ability of the nucleus to pull electron towards it and keep them there, much like electropositivity define how willing an atom is to give (or get rid of) 1 or more electron .

all of this is an effort to return to the rare gaz complete electronic comfiguration.
Fluorine has a strong tendency to get electron and keep them! because its a small atom where the electronic cloud is much closer to the positive nucleus than iodine for example.
Cesium is more than happy to give its lonely electron away because Cs is a big atom where that electron is far away and relatively unbonded to its nucleus.

CHRIS25 - 5-5-2014 at 08:24

Quote: Originally posted by MrHomeScientist  
CHRIS25: What did your crystals look like? I have a strong suspicion that cooling your solution as you did would favor KNO3 precipitation. KCl and NaCl are cubic, while KNO3 look like needles.
This is similar to how I make my potassium nitrate from instant cold packs via:
NH4NO3 + KCl == KNO3 + NH4Cl
Cooling the solution precipitates the least soluble of the four possible salts - KNO3.

Artemus Gordon is right in that often when you mix two ionic salts together in solution, there isn't really a reaction, per se. You just have a soup of ions floating around. You only observe something happening when one of the possible combinations of ions makes something that is insoluble under the current conditions of the system. In my example above, mixing those two ionic salts results in a system where everything is dissolved and nothing really happens, but cooling in a fridge causes the K+ and NO3- ions to come together and drop out as solid crystals.

Interestingly I will tell you what happened: Appearance, yes indeed needle crystals, but mixed in with what were not needle crystals, 50:50 subjectively speaking. So of course being two edible salts there I did the QIOT test (that's the Quick In and Out Taste) I tasted saltiness accompanied very quickly by a sour ting on the tongue (lack of chemical terminology here) - ok so this is not destined for my chips tonight:) I then did a flame test, it stayed yellow confirming that this had to be sodium not potassium, but for curiosity I continued to apply direct heat for 1 minute at 1000c. The flame went from yellow to melting everything into a clear aqueous liquid with yellow flame disappearing and after about another 20 seconds the whole liquid turned into a clean white flake with the same consistency and thickness as a good quality Egg (they have thicker shells). It was easy to break with the fingers, it was easy dissolve in cold tap water if you broke it up.
Conclusion: Everyone is right, I am wrong, but not a waste of 45 minutes by any means.

Zyklon-A - 5-5-2014 at 09:39

Quote: Originally posted by blogfast25  
That's really only half true. It's true that there are no covalent bonds between the ions in an ionic lattice. But the strong electrostatic attraction is what we call the 'ionic bond' and it's what keeps the lattice so tightly 'bonded' together.

Sorry, that's my mistake, I was typing fast and miss-spoke.

Rogeryermaw - 7-5-2014 at 10:54

a yellow flame does not necessarily indicate and absence of potassium but the presence of sodium. it doesn't take much sodium to discolor any flame. even at 10:1 in favor of K you would still only notice the color produced by the Na.

CHRIS25 - 7-5-2014 at 13:13

Quote: Originally posted by Rogeryermaw  
a yellow flame does not necessarily indicate and absence of potassium but the presence of sodium. it doesn't take much sodium to discolor any flame. even at 10:1 in favor of K you would still only notice the color produced by the Na.
Good to know then. Thanks. The flame test was really to establish any absence of sodium which would mean Potassium in this case. But anyway, all things learned, it was more of an experiment while I was waiting for other things to happen.