August 09, 2020, 01:33:55 PM
Forum Rules: Read This Before Posting

### Topic: Solubility  (Read 3938 times)

0 Members and 1 Guest are viewing this topic.

#### Pirate

• Regular Member
• Posts: 10
• Mole Snacks: +0/-0
##### 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

#### Astrokel

• Full Member
• Posts: 989
• Mole Snacks: +65/-10
• Gender:
##### 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!!!!!!!!!!!!

#### Pirate

• Regular Member
• Posts: 10
• Mole Snacks: +0/-0
##### Re: Solubility
« Reply #2 on: February 20, 2009, 04:39:03 AM »
I have missed a couple of classes, so I'm confused
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.

#### Borek

• Mr. pH
• Deity Member
• Posts: 25879
• Mole Snacks: +1692/-401
• Gender:
• I am known to be occasionally wrong.
##### 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Pirate

• Regular Member
• Posts: 10
• Mole Snacks: +0/-0
##### 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?

#### Borek

• Mr. pH
• Deity Member
• Posts: 25879
• Mole Snacks: +1692/-401
• Gender:
• I am known to be occasionally wrong.
##### 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
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Pirate

• Regular Member
• Posts: 10
• Mole Snacks: +0/-0
##### 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  to me

#### Borek

• Mr. pH
• Deity Member
• Posts: 25879
• Mole Snacks: +1692/-401
• Gender:
• I am known to be occasionally wrong.
##### 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
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Pirate

• Regular Member
• Posts: 10
• Mole Snacks: +0/-0
##### 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?