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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Jinn21 on December 03, 2006, 05:52:56 PM

Title: Question about Solubility Products
Post by: Jinn21 on December 03, 2006, 05:52:56 PM
Hey, I've just got a quesiton about soulbility products:

"One of the substances responsible for "hardness" of water is CaSO4. If a particular sample has 131 ppm of CaSO4, approximitly what fraction of water must be evaporated before CaSO4 (s) begins to deposit? Assume 1.0 L of water."

Thanks guys
Title: Re: Question about Solubility Products
Post by: Dan on December 03, 2006, 05:56:08 PM
Show your attempt. How do you think you could work it out?
Title: Re: Question about Solubility Products
Post by: Donaldson Tan on December 03, 2006, 06:38:04 PM
You will need the solubility product constant of calcium sulphate.
Title: Re: Question about Solubility Products
Post by: Jinn21 on December 04, 2006, 07:28:41 PM
ok well so far, I've recognized that CaSO4 is highly soluble in water. I would have to evaporate almost all of it before I would get any precipitation.

If I ignore supersaturation, a solute will begin to precipitate out of solution when its concentration exceeds its solubility.

But how do I include the math part to this quesiton?

*Also, thanks to Geodome, I need the solubility product constant as well, which is: 4.93 x 10^-5
Title: Re: Question about Solubility Products
Post by: Borek on December 04, 2006, 07:36:09 PM
Quote from: Jinn21 on December 04, 2006, 07:28:41 PM
ok well so far, I've recognized that CaSO4 is highly soluble in water.

It is not.

Quote
I need the solubility product constant as well, which is: 4.93 x 10^-5

So you know it, not need it :)

It is enough to take a look at the solubility product to see that saturated solution concentration must be below 0.01M - hardly much.

What is definition of solubility product?
Title: Re: Question about Solubility Products
Post by: Donaldson Tan on December 05, 2006, 05:44:13 AM
Quote from: Jinn21 on December 04, 2006, 07:28:41 PM
But how do I include the math part to this quesiton?

Don't you know the equilibrium expression relating to the solubility product constant?

You need to convert 131ppm to mol/dm3, ie. 131ppm = Y mol/dm3 (note ppm is mass ratio)

Ksp = [Ca2+][SO42-]

Find the saturated concentration of calcium sulphate from the above expression, which should be larger than Y.

From here, you should be able to derive the required fraction of water to be evaporated using the mass balance of calcium sulphate.

Title: Re: Question about Solubility Products
Post by: Jinn21 on December 05, 2006, 09:13:51 PM
wait. would the chemical equation just be the decomposition of CaSO4? so CaSO4 ----> Ca + SO4 ?
Or do i have to add water into this equation?
Title: Re: Question about Solubility Products
Post by: Borek on December 06, 2006, 04:46:53 AM
Quote from: Jinn21 on December 05, 2006, 09:13:51 PM
Ca + SO4

These are ions (Ca2+ & SO42-). And no such compound as SO4.
Title: Re: Question about Solubility Products
Post by: Jinn21 on December 06, 2006, 04:12:29 PM
ok. well Geodome says to convert 131 ppm into Y mol/dm3

but I have a problem here. My teacher did not even teach us about dm3 or dm3 conversion, so i am completely unfamiliar with the term dm3. Is that density or something? I was hoping that there is another way to convert 131 ppm into just mols or something. And is "Y" just a variable?
Title: Re: Question about Solubility Products
Post by: Borek on December 06, 2006, 05:10:16 PM
Y is just a variable (or an unknown if you prefer), dm3 is the same as L. dm is a length measure - 1/10th of a meter. dm3 is dm cubic.

http://en.wikipedia.org/wiki/Deci
Title: Re: Question about Solubility Products
Post by: Jinn21 on December 06, 2006, 05:22:20 PM
ok. well before i do all this i still need to make the chemical equation.
My equation goes like this:

CaSO4 + H2O ------> CaO + H2(SO4)

is this right or am i doing something wrong?
Title: Re: Question about Solubility Products
Post by: Borek on December 06, 2006, 06:06:22 PM
I am afraid your reaction is completely wrong. First: this is ionic compound that gets dissolved to ions. Two: you need net ionic reaction, as it will help determine how the solubility product for CaSO4 looks alike.
Title: Re: Question about Solubility Products
Post by: Jinn21 on December 06, 2006, 06:21:41 PM
Okay, so its an ionic compound. THe net-ionic equation would be:

CaSO4(s) <---> Ca2+ + SO42-

and so:

Qsp= [Ca2+] [SO42-]

so i somehow find thhe concentrations of both Ca and SO4 and sub them into that Qsp equation to find Qsp?
Title: Re: Question about Solubility Products
Post by: Borek on December 06, 2006, 06:24:06 PM
Look at the reaction equation - how is concentration of Ca2+ related to concentration of SO42- if solid CaSO4 is the only source of ions?
Title: Re: Question about Solubility Products
Post by: Jinn21 on December 06, 2006, 06:29:39 PM
ahh im so confused. i don't even know where to begin ???
Title: Re: Question about Solubility Products
Post by: Borek on December 06, 2006, 07:06:57 PM
It is all in solution stoichiometry - imagine 1 mol of CaSO4 was dissolved in some volume of water. How many moles of Ca2+ and how many moles of SO42- are present in the solution?
Title: Re: Question about Solubility Products
Post by: Jinn21 on December 06, 2006, 07:35:06 PM
Well, the Molar mass of  CaSO4 is 136g
Ca has a mass of 40g
and SO4 has 96g

So: Ca- 40g/136g = 0.29 mols -->0.30 mols
SO4- 96g/136g = 0.70 mols
Title: Re: Question about Solubility Products
Post by: Borek on December 06, 2006, 07:58:44 PM
No. Look at the reaction equation. And read second paragraph on this page:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations