Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Glorzifen on December 04, 2009, 11:01:21 PM
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I've been making my way through some acid/base chemistry and ran into some of these questions...everything was going so well! I'm not sure if I'm supposed to be integrating previous solubility expressions or old equations into these...or if I'm over complicating things?
1. A saturated aqueous solution of Ca(OH)2 has a pH of 12.35. What is the solubility of Ca(OH)2, expressed in milligrams per 100mL of solution?
Am I supposed to be using some sort of Ksp = (x)(2x)2 thing where x = [OH-]? I found that by:
pH = 12.35 = 14 - pOH
pOH = 1.65 = -log[OH-]
[OH-] = 44.67M (10^1.65)
This seems way to high which tells me I'm not doing this correctly. Solubility resurfacing in acids/bases kind of threw me off...a hint in the right direction would be a big help. I will proceed with the following 2 questions after I get the first one sorted out (I'm not going to ask for help until I've put some more thought into them).
2. What is the [H3O]+ in a solution obtained by dissolving 205mL HCl(g), measured at 23C and 751mmHg, in 4.25L of aqueous solution?
3. What is the pH of the solution obtained when 125mL of 0.606M NaOH is diluted to 15.0L with water?
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Am I supposed to be using some sort of Ksp = (x)(2x)2 thing where x = [OH-]? I found that by:
pH = 12.35 = 14 - pOH
pOH = 1.65 = -log[OH-]
[OH-] = 44.67M (10^1.65)
This step here, -log [OH-] = 1.65
[OH-]= 10-1.65 since it is a negative log
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Ah, of course. Provided that gives me the correct [OH-]...do I just calculate solubility like I suggested in my first post (Ksp = (x)(2x)^2)?
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Yes, [OH-] = 2x
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No actually, forget what I said, that would be calculating the solubility product (they are two different things you know?). The solubility of calcium hydroxide would be the same as [Ca2+]
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Okay...so much simpler than I thought. Can I just ask, in this example its the [Ca2+], but what's the general idea here...is it, "The solubility is equal to the concentration of the cation"? or, "The solubility is equal to the concentration of the the atom with the highest charge?" Just want to know why it is what you say it is.
Thanks very much :D
EDIT: Also...to get this thing into milligrams per 100mL of sol'n...do I just multiply by 1000 and divide by 0.1?
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I've got 2. figured out...except for one small problem:
2. What is the [H3O]+ in a solution obtained by dissolving 205mL HCl(g), measured at 23C and 751mmHg, in 4.25L of aqueous solution?
Use pv=nRT to find n = 8.33E-3. Next, C = n/V can be used to determine the concentration of HCl(aq) and therefore, [H3O]+ in my initial volume. In order to find [H3O]+ in the "4.25L of aqueous solution", I used C1V1=C2V2, solved for C2 and found my answer.
However, for C2, I added my volumes to get a C2 of (4.455L). The solution manual didn't do this, which resulted in a slightly different answer (though very similar). Why am I not supposed to add the volumes? Is it because the HCl is gaseous?
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Okay...so much simpler than I thought. Can I just ask, in this example its the [Ca2+], but what's the general idea here...is it, "The solubility is equal to the concentration of the cation"? or, "The solubility is equal to the concentration of the the atom with the highest charge?" Just want to know why it is what you say it is.
No, there is no 'rule' to look at which tells you this, you've just got to look at the species present.
Lemme give you an example, say Ks (Mg(OH)2) = 7.1 x 10-12 this is the solubility product.
The dissolving of the solid in water goes: Mg(OH)2 ::equil:: Mg2+ + 2OH-
Ks = [Mg2+][OH-]2
In a saturated solution, the concentration of magnesium ions or [Mg2+]= cube root (Ks/4), whereas the concentration of OH- ions would be double this number. If you look at the dissolving equation, one molecule of magnesium hydroxide gives one Mg2+ and 2 OH-, this means that the solubility of magnesium hydroxide solid is equal to [Mg2+] since it is a 1:1 ratio. The solubility of magnesium hydroxide is therefore cube root (7.1 x 10-12/4) = 1.21 x 10-4 mol L-1
The same applies to this here, you have found [OH-] and since Ca(OH)2:OH- = 1:2 you must halve this concentration to find the solubility of Ca(OH)2, this number is the same as [Ca2+] because Ca(OH)2:Ca2+ is a 1:1 ratio. Once you have got this concentration, which will be in mol L-1, divide by 10 to get answer in mol/100mL and then to find the mass of calcium hydroxide, you multiply the number of moles by the molar mass. See how you go....
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Why am I not supposed to add the volumes? Is it because the HCl is gaseous?
Yes, just like you usually ignore volume hange after dissolving solids you ignore volume change when dissolving gases. As long as the amount of dissolved substance is small (compared with amount of solvent) you are not making large errors.
what's the general idea here...is it, "The solubility is equal to the concentration of the cation"? or, "The solubility is equal to the concentration of the the atom with the highest charge?"
General idea is - take a look at the stoichiometry of dissolution reaction. How many moles of ions are produced per mole of dissolved substance?