[Diamonds]

Which solid is much more soluble in 1M HCl than in water?

a) CaHPO₄
b) CaCl₂
c) BaBr₂
d) BaSO₄

I was thinking of the reactions that each would make with the HCl

a) CaHPO₄ + 2HCl  CaCl₂ + H₃PO₄
b) CaCl₂ + 2HCl CaCl₄^2- + H₂
c) BaBr₂ + 2HCl BaCl₂ + 2HBr
d) BaSO₄ + 2HCl BaCl ₂ + H₂SO₄

I am confused on how to use these equations to determine which is more soluble in HCl than in water.

[Borek]

a) CaHPO₄ + 2HCl  CaCl₂ + H₃PO₄

Makes sense.

b) CaCl₂ + 2HCl CaCl₄^2- + H₂

Not balanced (and impossible to balance - you have charge on the RHS only).

c) BaBr₂ + 2HCl BaCl₂ + 2HBr

HBr is a stronger acid than HCl, so the protonation is unlikely.

d) BaSO₄ + 2HCl BaCl ₂ + H₂SO₄

H₂SO₄ is a strong acid, so protonation won't go that far, but HSO₄^- is quite likely...

In general this is just about acid/base equilibrium - you are trying to dissolve salt of a weak (or relatively weak) acid in the strong acid, so protonating the anion can shift the dissolution equilibrium to the right.

[Diamonds]

So then:

a) CaHPO₄ + 2HCl  CaCl₂ + H₃PO₄
b) CaCl₂ + 2HCl CaCl₄^2- + 2H⁺
c) BaBr₂ + 2HCl BaCl₂ + 2H⁺ + 2Br^-
d) BaSO₄ + 2HCl BaCl₂ + H⁺ + HSO₄^-

So now that I have these (hopefully) correct chemical equations, I need to know which of the equations will proceed the most to the right in order to find which of the four compounds will dissolve the most in HCl (as opposed to their dissolutions into water).
I am still confused on how to figure out which equation will proceed the furthest to the right.

[Borek]

b) CaCl₂ + 2HCl CaCl₄^2- + 2H⁺

Problem with balancing were not the only problem with this reaction. There is no CaCl₄^2- complex.

c) BaBr₂ + 2HCl BaCl₂ + 2H⁺ + 2Br^-

This is wrong - but it is at least partially a problem with all your reactions. What is BaBr₂? Is it solid? Or a dissolved, undissociated molecule? And what is BaCl₂? And why there is HCl on the left, but H⁺+Br^- on the right? Make all these reaction net ionic, and mark what is solid, what is not.

d) BaSO₄ + 2HCl BaCl₂ + H⁺ + HSO₄^-

So now that I have these (hopefully) correct chemical equations, I need to know which of the equations will proceed the most to the right in order to find which of the four compounds will dissolve the most in HCl (as opposed to their dissolutions into water).
I am still confused on how to figure out which equation will proceed the furthest to the right.

Whenever you remove one of the dissolution products from the reaction mixture, you shift the equilibrium, don't you? For example, calcium carbonate dissolves producing two ions:

CaCO₃(s)    Ca²⁺ + CO₃^2-

In the presence of any acid strong enough CO₃^2- gets protonated:

CO₃^2- + H⁺  HCO₃^-

and the overall reaction of CaCO₃ dissolution in the presence of another acid is:

CaCO₃ + H⁺  Ca²⁺ + HCO₃^-

with the equilibrium shifted (when compared to the first CaCO₃ dissolution reaction) to the right, as protonation removes CO₃^2- from the solution.

[Ben Cohen]

So, for example choice a)

Equilibrium 1: CaHPO₄ + 2HCl  CaCl₂ + H₃PO₄

So does this mean that if the H₃PO₄ will protonize to give:

Equilibrium 2: H₃PO₄  H⁺ + H₂PO₄^2-

So the net equation will be

CaHPO₄ + 2HCl H⁺ + H₂PO₄^2-

Meaning the dissolving it in HCl acid will increase the [HCl] on the right, driving the reaction to the right and increasing the solubility of CaHPO₄ ?

[Borek]

So, for example choice a)

Equilibrium 1: CaHPO₄ + 2HCl  CaCl₂ + H₃PO₄

So does this mean that if the H₃PO₄ will protonize to give:

Equilibrium 2: H₃PO₄  H⁺ + H₂PO₄^2-

So the net equation will be

CaHPO₄ + 2HCl H⁺ + H₂PO₄^2-

Meaning the dissolving it in HCl acid will increase the [HCl] on the right, driving the reaction to the right and increasing the solubility of CaHPO₄ ?

Where did you got H₃PO₄ in your second reaction from - that's not what you start with. What are the products of CaHPO₄ dissolution?

Note: you are still ignoring fact some of the substances involved are always dissociated, it doesn't help you.

[Ben Cohen]

So, for example choice a)

Equilibrium 1: CaHPO₄ + 2HCl  CaCl₂ + H₃PO₄

So does this mean that if the H₃PO₄ will protonize to give:

Equilibrium 2: H₃PO₄  H⁺ + H₂PO₄^2-

So the net equation will be

CaHPO₄ + 2HCl H⁺ + H₂PO₄^2-

Meaning the dissolving it in HCl acid will increase the [HCl] on the right, driving the reaction to the right and increasing the solubility of CaHPO₄ ?

Where did you got H₃PO₄ in your second reaction from - that's not what you start with. What are the products of CaHPO₄ dissolution?

Note: you are still ignoring fact some of the substances involved are always dissociated, it doesn't help you.

The products of CaHPO₄ dissolution are

CaHPO₄  Ca²⁺ + HPO₄^2-

The HPO₄^2- is a very weak acid and practically does not protonize.

[Borek]

The HPO₄^2- is a very weak acid and practically does not protonize.

Never heard a word protonize, what you mean is HPO₄^2- doesn't dissociate (well... it does, to some extent, but we can assume it doesn't). But it doesn't matter much here, as you got it all reversed. In acidic solution HPO₄^- will get PROTONATED - that is, it will accept a proton.

[Ben Cohen]

The HPO₄^2- is a very weak acid and practically does not protonize.

Never heard a word protonize, what you mean is HPO₄^2- doesn't dissociate (well... it does, to some extent, but we can assume it doesn't). But it doesn't matter much here, as you got it all reversed. In acidic solution HPO₄^- will get PROTONATED - that is, it will accept a proton.

http://www.chemicalforums.com/Themes/Borek/images/bbc/sub.gif
OH!

So you mean:

CaHPO4 Ca2⁺ + HPO₄^2-
And then it is in an acidic solution, so the HPO₄^2- will

HPO₄^2- + H⁺ H₂PO₄^-
Which will take the HPO₄^2- out of solution.
And that means that, according to Le Chatelier's, more CaHPO₄ will dissolve to accomodate the loss of the HPO₄^2- ion?

Is this reasoning correct?

[Borek]

You got it now.