Determining Solubility Product of \text{BaCO}_3_{(s)}

[broegger]

Hi,

I am told that $$\text{BaCO}_3_{(s)}$$ is sparingly soluble in water with a solubility product $$K_{sp} = 8.1\cdot10^{-9}~M^2$$. I have to write the expression for the solubility product, but how do I know which ions Ba and CO3 forms (e.g. $$Ba^{+},~Ba^{2+},~Ba^{3+},\ldots$$). I happen to know that the expression is

$$K_{sp} = [Ba^{2+}][{CO_3}^{2-}],$$​

but how can you find out?

 

[Borek]

These things - like $$Ba^{2+}$$ and $$CO_3^{2-}$$ ions charge - you have to memorise.

 

[Bladibla]

These things - like $$Ba^{2+}$$ and $$CO_3^{2-}$$ ions charge - you have to memorise.

Is that answering the question? It still doesn't tell us the reason for such formation of ions..

 

[The Bob]

Hi,...
...but how can you find out?

Well some are simple to find out. Anything in Group 1 of the periodic table will have a charge of +1. Anything in Group 2 will have a charge of +2. Anything in Group 7 has a charge of -1 and it is expected that anything in Group 6 will have a charge of -2 (although this is not always true).

So here are some examples of Group 1, 2, 6 and 7 elements in compounds:

$$LiCl = Li ^{+1} + Cl^{-1}$$

$$CaCl_2 = Ca ^{+2} + 2Cl^{-1}$$

$$Na_2O = 2Na ^{+1} + O^{-2}$$

$$MgO = Mg ^{+2} + O^{-2}$$

As you might have noticed the charges add together to make 0. This is how oxidation states work, you add the charges to find if there are changes or not but I don't know if you have done them yet. So lithium (+1) and chlorine (-1) add to make 0 (1 - 1 = 0). Calcium (+2) and chlorine (-1) add to make 0 (+2 - (2 x -1) = 2 - 2 = 0).

As I said, these are the simple ones. Now let's say you have $$CaCO_3$$.

We have already said that calcium has a charge of +2 so the remaining CO₃ must have a charge of -2. You can then say you know that oxygen has a charge of -2 (overall of -6 in the carbonate ion) and so carbon has a charge of +4 (as you need to make -2 for the overall charge of the ion).

Some examples of these are:

$$Na_2 CO_3 = 2Na ^{+1} + CO_3 ^{-2}$$

$$CaSO_4 = Ca ^{+2} + SO_4 ^{-2}$$

As you can see, yuo know one part of the compounds charge and so you can work out the rest by balancing out the charges.

Hope this helps a bit.

The Bob (2004 ©)

 

[broegger]

Thanks, Bob. It's really helpful, since most of the ions we encounter belong to Group 1 or 2. But what about Fe (group 8), Cu and Ag (group 11) and Zn (group 12), which are ions I frequently encounter. Any scheme here?

 

[Borek]

Is that answering the question? It still doesn't tell us the reason for such formation of ions..

Well, The Bob gave much more detailed explanation starting with

Well some are simple to find out. Anything in Group 1 of the periodic table will have a charge of +1. Anything in Group 2 will have a charge of +2. Anything in Group 7 has a charge of -1 and it is expected that anything in Group 6 will have a charge of -2 (although this is not always true).

which is absolutly true - but a step further and you have to memorise that sodium is in the first group, calcium in the second and so on - so no matter how you try, there are things that you just have to remember.

 

[The Bob]

...a step further and you have to memorise that sodium is in the first group, calcium in the second and so on - so no matter how you try, there are things that you just have to remember.

Good point. Forgot to say this.

Thanks, Bob. It's really helpful, since most of the ions we encounter belong to Group 1 or 2. But what about Fe (group 8), Cu and Ag (group 11) and Zn (group 12), which are ions I frequently encounter. Any scheme here?

I see. Well let's start with Zn (a transitition metal in Group 2b) normally has the charge of +2 e.g.

$$ZnO = Zn ^{+2} + O ^{-2}$$

$$ZnSO_4 = Zn ^{+2} + SO_4 ^{-2}$$

$$Zn_2 NO_3 = Zn ^{+2} + 2NO_3 ^{-1}$$

I am sure someone will come and say it can change but so far as I know it doesn't.

Iron is a little harder but it can be worked out using the other part of the compound, e.g.

$$FeF_2 = 2F ^{-1} + Fe ^{+2}$$

$$FeF_3 = 3F ^{-1} + Fe ^{+3}$$

$$FeO = O ^{-2} + Fe ^{+2}$$

$$Fe_2 O_3 = 3O ^{-2} + 2Fe ^{+3}$$

$$Fe_3 O_4 = 4O ^{-2} + 2Fe ^{+3} + Fe ^{+2}$$

As you can see, the last one shows iron to have two different charges. If you look at it you can see that you need both charges to make the compound up. Iron comes in charges of +2 and +3 (unless someone can enlighten me on a compound were it is not).

Copper is similar to iron but it only has charges of +1 and +2 (again, if anyone knows any others...) e.g.

$$CuF = F ^{-1} + Cu ^{+1}$$

$$CuF_2 = 2F ^{-1} + Cu ^{+2}$$

$$CuO = O ^{-2} + Cu ^{+2}$$

$$Cu_2 O = O ^{-2} + 2Cu ^{+1}$$

So far you can see that all you need to do is look at the part of the compound you know for sure (in the cases above you know what fluorine and oxygen have a charge of) and then work out what the over part from the part you do know.

Silver normally (well at A-Level) has charge of +1 but I did a little research and it seems it is more complicated then that ( like copper it can be +2 as well) e.g.

Simple:
$$Ag NO_3 = NO_3 ^{-1} + Ag ^{+1}$$

Others:
$$Ag_2 O = O ^{-2} + 2Ag ^{+1}$$

$$Ag F = F ^{-1} + Ag ^{+1}$$

$$Ag F_2 = 2F ^{-1} + Ag ^{+2}$$

And some that I have never seen before:
$$Ag_2 F = F ^{-1} + Ag ^{+1} + Ag ^{0}$$

$$Ag O = O ^{-2} + Ag ^{+3} + Ag ^{-1}$$

(If anyone could shed some light on these last two I would be grateful).

I hope this has shown you that when you get your chemical compound you should be able to find the charge of the other parts of the compound.

I hope this helps more. If you want I will give you some to do. I could do with the practise myself.

The Bob (2004 ©)

 

[broegger]

Again; thank you very much :) Yeah, throw in some exercises, that would be great!

 

[The Bob]

Here you go then:

1. What is the charge of the:
(a) Potassium in Potassium Chloride?
(b) Carbonate in Sodium Carbonate?
(c) Beryllium in Beryllium Oxide?
(d) Magnesium in Magnesium Nitrate?

(For each of the above write the chemical compound in symbol form and split the compound out into its different charges)

2.What compound is made when:
(a) Magnesium and Carbonate are combined?
(b) Lithium and Fluorine are combined?
(c) Zinc and Nitrate are combined?
(d) Copper (II) and Oxygen are combined?
(e) Iron (II) and Nitrate are combined?
(f) Ammonia and Sulphate are combined?
(g) Barium and Hydrogensulphate are combined?

(For this question you will simply need to write the chemical formula of the compound)

Try these. Most are considered easy for A-level students (in the UK) but believe me, some people in my group would struggle with them so don't worry if you don't get them all or any at all.

The point Borek made in this thread is good; you need to memorise as many of the charges and compounds as you can so it becomes easier.

The Bob (2004 ©)

 

[broegger]

Thanks for taking you time. Here goes:

1.
(a) $$KCl \rightarrow K^+ + Cl^-$$
(b) $$Na_2CO_3 \rightarrow 2Na^+ + {CO_3}^{2-}$$
(c) $$BeO \rightarrow Be^{2+} + O^{2-}$$
(d) $$Mg(NO_3)_2 \rightarrow Mg^{2+} + 2{NO_3}^-$$

2.
(a) $$MgCO_3$$
(b) $$LiF$$
(c) $$Zn(NO_3)_2$$
(d) $$Cu\text{(II)}O$$
(e) $$Fe\text{(II)}(NO_3)_2$$
(f) $$(NH_4)_2SO_4$$
(g) $$Ba(HSO_4)_2$$

 

[Bladibla]

Well, The Bob gave much more detailed explanation starting with



which is absolutly true - but a step further and you have to memorise that sodium is in the first group, calcium in the second and so on - so no matter how you try, there are things that you just have to remember.


Borek
--
http://www.chembuddy.com
Chemical calculators for labs and education
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I wasn't disagreeing about it at all. I'm just saying given that we (at least i think) arn't in a time-desperate situation to know about this, a detailed expalnation would be better.

But i do apologize if i made you misunderstand Its just that i hate people begging the question, as I've seen from my chemistry teachers from school.

 

[The Bob]

Thanks for taking you time. Here goes:

1.
(a) $$KCl \rightarrow K^+ + Cl^-$$
(b) $$Na_2CO_3 \rightarrow 2Na^+ + {CO_3}^{2-}$$
(c) $$BeO \rightarrow Be^{2+} + O^{2-}$$
(d) $$Mg(NO_3)_2 \rightarrow Mg^{2+} + 2{NO_3}^-$$

2.
(a) $$MgCO_3$$
(b) $$LiF$$
(c) $$Zn(NO_3)_2$$
(d) $$Cu\text{(II)}O$$
(e) $$Fe\text{(II)}(NO_3)_2$$
(f) $$(NH_4)_2SO_4$$
(g) $$Ba(HSO_4)_2$$

Nice. Even the ammonia one was right. Most people go 'AHHHHHHHHH' when they see it. I used to (maybe 8 months ago for 2 hours) but it is all about understanding.

Do you know what oxidation states are?

The Bob (2004 ©)

 

[broegger]

Hmm, oxidation states are the charge numbers you ascribe to atoms when you're doing redox-reactions, right?

 

[The Bob]

Hmm, oxidation states are the charge numbers you ascribe to atoms when you're doing redox-reactions, right?

They are used in redox-reactions but they can be used to work out charges and names of compounds. I only asked because it might be worth while you seeing how the charges of each bit are set up and this should help with an even better understanding...well either that or a confused understanding like I had for a month.

The Bob (2004 ©)

 

[broegger]

Well, I seem to manage this, thanks to you :)