Benzoic Acid most soluble in 0.10 m KOH

[ToKingdomCome]

I don't get why it's the most soluble in this case. The carboxy side chain is polar, while the benzene ring tends to be non polar. If base is added, it deprotantes the carboy making it charged, and thus soluble in water.

Well I guess I kind of answered my own question.

The BR review question asks In which of the following solvents or solutions is 4-hydroxybenzoic acid most soluble?

1. 0.10 M HCl
2. 0.10 M KOH
3. Water
4. Diethyl ether

When the question asks most soluble, is that implying water? So is the question really asking what conditions will make it most soluble in H2O?

---

Members don't see this ad.

 

[ToKingdomCome]

Can a brotha get some help?

 

[ChemTutor]

1. is mostly water with some acid
2. is mostly water with some base
3. is pure water
4. is a non-polar organic solvent (no water is present)

 

[cheesier]

Polarity doesn't have to do with it. Any acid, weak or strong, is most soluble in the most basic solution, go through your options and cross out anything that's not a base, then choose the strongest of the bases available, and then choose the one that's most concentrated.

 

[discowisco]

polarity doesn't have to do with it. Any acid, weak or strong, is most soluble in the most basic solution, go through your options and cross out anything that's not a base, then choose the strongest of the bases available, and then choose the one that's most concentrated.

+1

 

[chem4ever]

I don't get why it's the most soluble in this case. The carboxy side chain is polar, while the benzene ring tends to be non polar. If base is added, it deprotantes the carboy making it charged, and thus soluble in water.

Well I guess I kind of answered my own question.

The BR review question asks In which of the following solvents or solutions is 4-hydroxybenzoic acid most soluble?

1. 0.10 M HCl
2. 0.10 M KOH
3. Water
4. Diethyl ether

When the question asks most soluble, is that implying water? So is the question really asking what conditions will make it most soluble in H2O?

If it is in 0.10 molar KOH solution, that tends to mean 0.10 moles of KOH per 1 liter of water.

So answer 3 isn't the best answer because there's a large non-polar group (the ring). And answer 4 isn't the best because there's a carb. acid and a hydroxyl group that are polar and diethyl ether is nonpolar. Answer 1 is pretty much the same as putting it in water because the HCl is an acid and 4-hydroxybenzoic acid is not a very good base so not much of a rxn will occur.

But answer 2 is like you said: the KOH will deprotonate the carb. acid (and the hydroxyl group to a certain extent) making the compound charged which is when it will be very soluble in the water part of the 0.10 M KOH solution.

 

[dsoz]

If it is in 0.10 molar KOH solution, that tends to mean 0.10 moles of KOH per 1 liter of water.

Not quite. It is in water, but it should read 0.10 moles of KOH in 1 liter of SOLUTION There is a slight difference because if you added 0.10 mol of KOH to 1 liter of water, the final volume would be slightly more than 1L.

Molar implies aqueous solution. If it were molality, then it could be any solvent.

As to the original question... Adding a weak acid to a strong base will make a salt that is highly soluble in water. This is the same idea why caffeine is extracted using a strong acid, then neutralized to allow it to dissolve into an organic solvent. (I think I remember that correctly, it was many years ago that I did this extraction in O-chem).

dsoz

 

[chem4ever]

Not quite. It is in water, but it should read 0.10 moles of KOH in 1 liter of SOLUTION There is a slight difference because if you added 0.10 mol of KOH to 1 liter of water, the final volume would be slightly more than 1L.

Molar implies aqueous solution. If it were molality, then it could be any solvent.

As to the original question... Adding a weak acid to a strong base will make a salt that is highly soluble in water. This is the same idea why caffeine is extracted using a strong acid, then neutralized to allow it to dissolve into an organic solvent. (I think I remember that correctly, it was many years ago that I did this extraction in O-chem).

dsoz

True. I was only trying to point out that it is in water but good point,