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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Pirate on February 19, 2009, 10:36:56 PM

Title: Solubility
Post by: Pirate 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  ???

Title: Re: Solubility
Post by: Astrokel 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?

Title: Re: Solubility
Post by: Pirate 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.

Title: Re: Solubility
Post by: Borek on February 20, 2009, 04:57:12 AM

Quote from: Pirate on February 20, 2009, 04:39:03 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.

Title: Re: Solubility
Post by: Pirate 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?

Title: Re: Solubility
Post by: Borek on February 20, 2009, 06:57:01 PM

And it isn't right.

Quote from: Pirate on February 20, 2009, 05:11:21 PM

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 A_mB_n substance is

K_sp = [A]^m[B ]ⁿ

Title: Re: Solubility
Post by: Pirate 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

Title: Re: Solubility
Post by: Borek 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

Title: Re: Solubility
Post by: Pirate 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?