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Topic: Solubility  (Read 3928 times)

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Offline Pirate

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Solubility
« on: February 19, 2009, 10:36:56 PM »
1.((Ca(OH)2 dissolves in water 0,908 g/dm^3 ( Temperature =25C ) What is the Ksp.

2.CO2 dissolves in water ( Temperature= 25C )at 0,1 atm pressure into 0,0037 mol liters. Lets assume that all CO2 that has dissolved in  water is in form of H2CO3. It is formed in:

CO2(aq) + H2O(l) < - > H2CO3(aq)
What is the ph of 0,0037M H2CO3.



Any help here? I don't even know where to start  ???

Offline Astrokel

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Re: Solubility
« Reply #1 on: February 20, 2009, 02:23:05 AM »
Quote
Any help here? I don't even know where to start
 
We will definitely help only if you show some attempts. Are you given assignments that you have never learn before?
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline Pirate

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Re: Solubility
« Reply #2 on: February 20, 2009, 04:39:03 AM »
I have missed a couple of classes, so I'm confused  :o
For the first question umm, if you could tell me something to get started, I would greatly appreciate it. Since I only have 0,908 g/dm^3 I don't know how to get

ksp= [Ca^2+] 2[OH^-1]
??




Second question seems harder, so I'll look into that later.

Offline Borek

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Re: Solubility
« Reply #3 on: February 20, 2009, 04:57:12 AM »
Since I only have 0,908 g/dm^3

Convert to molarity.

Quote
ksp= [Ca^2+] 2[OH^-1]

That's not correct. Stoichiometric coefficients are used as powers.
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Offline Pirate

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Re: Solubility
« Reply #4 on: February 20, 2009, 05:11:21 PM »
Ca(OH)2
O=16g/mol
H=1,008g/mol
Ca=40,08g/mol

40,08+2(16+1,008)
=75,096g/mol

so 0,908g/dm^3 * (1mol/74.096g)
= 1,22 *19^-2 mol/L

so according to: http://www.coolschool.ca/lor/CH12/unit3/U03L01.htm

Ksp = (solubility)^2
Ksp = 0,11M

that can't be right?

Offline Borek

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Re: Solubility
« Reply #5 on: February 20, 2009, 06:57:01 PM »
And it isn't right.

Ksp = (solubility)^2

That's wrong. It was OK on the page you linked to, but that was solution for a very particualr situation, not for a general case.

Solubility product for AmBn substance is

Ksp = [A]m[B ]n
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Offline Pirate

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Re: Solubility
« Reply #6 on: February 22, 2009, 05:24:37 PM »
Is this right?

O= 15.9994
Ca=40,078
H=1,00749

0.908 g / 74.092 g/mol=0.0123
Concentration=0,0123M

Ca2+= 0.0123 M
OH-= 2 x 0.0123 = 0.0246 M

Ksp = 0.0123 ( 0.0246)^2 =7.44 x 10^-6

Also, would anyone be kind enough to get me started on problem 2? It just seems like  :o to me

Offline Borek

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Re: Solubility
« Reply #7 on: February 22, 2009, 06:05:06 PM »
Looks OK to me. The only thing that is wrong is that you shouldn't round down intermediate results, only the final. All calculations should be done with full available accuracy.

Second question asks about pH of a weak acid. You will need pKa1 and pKa2 for that.

http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base
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Offline Pirate

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Re: Solubility
« Reply #8 on: March 01, 2009, 08:52:32 AM »
ph = -log[H+]

Assuming the H2CO3 fully dissolves, it will be 0.0037 * 2 = -log(0.0074)

Ugh... is this right?

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