September 17, 2024, 02:00:47 PM
Forum Rules: Read This Before Posting

### Topic: Solubility product constant  (Read 25025 times)

0 Members and 1 Guest are viewing this topic.

#### 777888

• Guest
##### Solubility product constant
« on: November 29, 2004, 10:13:36 AM »
1.Calculate the molar solubility of Zn(OH)2 at  25oC, where Ksp is 7.7x10-17.

[Zn2+]      [OH-]
Initial          0               0
Change      +x            +2x
Equilibrium    x              2x
Ksp=[Zn2+][OH-]=(x)(2x)
7.7x10-17=4x3
x=2.7x10-6

Therefore,
Concentrations at equilibrium:
[Zn2+]=2.7x10-6
[OH-]2(2.7x10-6)=5.4x10-6

[Zn(OH)2]=2.7x10-6mol/L and the solubility of Zn(OH)2 is 2.7x10-6mol/L.

[I have a question...Why does the concentration of Zn(OH)2 at equilibrium mean the molar solubility of Zn(OH)2? Should the concentration of Zn(OH)2 at equilbrium mean the Zn(OH)2 that doesn't dissolve? How could this be the value of the solubility?]
« Last Edit: November 29, 2004, 10:15:27 AM by 777888 »

#### jdurg

• Banninator
• Retired Staff
• Sr. Member
• Posts: 1366
• Mole Snacks: +106/-23
• Gender:
• I am NOT a freak.
##### Re:Solubility product constant
« Reply #1 on: November 29, 2004, 02:02:24 PM »
Well, if the Zn(OH)2 doesn't dissolve, then how can there be a concentration of it in solution?    If an item does not dissolve, then it is not in solution and therefore it's concentration is zero since 'concentration' is defined as the amount of a substance per specified amount of solution.  No dissolution = no solution.
"A real fart is beefy, has a density greater than or equal to the air surrounding it, consists

#### 777888

• Guest
##### Re:Solubility product constant
« Reply #2 on: November 29, 2004, 03:34:31 PM »
HI!

"[Zn(OH)2]=2.7x10-6mol/L..." ---(1)
I am OK up to here...

"...and the solubility of Zn(OH)2 is 2.7x10-6mol/L" ---(2)
How can I know that? Are (1) and (2) related in same way?

Say genereally for this one:
A3B4(s) <-> 3A(4+)(aq) + 4B(3-)(aq)
And calculated that x(change in concentration)=1mol/L
Therefore:
[A4+]=3mol/L
[B3-]=4mol/L
[A3B4]=1mol/L

Then which one is the solubility of A3B4? 1mol/L, 3mol/L, or 4mol/L? WHY?

« Last Edit: November 29, 2004, 03:35:38 PM by 777888 »

#### jdurg

• Banninator
• Retired Staff
• Sr. Member
• Posts: 1366
• Mole Snacks: +106/-23
• Gender:
• I am NOT a freak.
##### Re:Solubility product constant
« Reply #3 on: November 29, 2004, 05:27:52 PM »
Okay.  The first thing to remember is that the molar solubility = Ksp.  The only difference between molar solubility and normal solubility is that molar solubility REQUIRES the value to be given in moles per liter.  Normal solubility can be given in any mass/volume unit.  (So if they say that something has a solubility of .04 g/L, it's molar solubility would be (.04/molar mass)/L).

You can tell that (1) and (2) are the same based upon the units.  Number 1 states that the concentration of your zinc hydroxide in solution is 2.7x10-6 mol/L, and number 2 is giving the solubility in terms of moles/L.  Therefore in this instance the molar solubility and solubility given are the same.

For your hypothetical equation, every mole of A3B4 that dissolves gives 3 moles of A and 4 moles of B.  So your Ksp equation would be:

Ksp = [A+4 ]3 [B-3 ]4.  Your 'I.C.E.' chart would look like the following:

.[A+4 ].[B-3 ]
Initial:    0       0
Change:  +x    +1.3333x
Eq:  x   1.3333x

If x is calculated to be 1 mole/liter, your concentrations would be 1 mole/liter for the A ion, and 1.3333 moles for the B ion.  The solubility of the initial compound would then be 1/3rd the concentration of A, or 1/4 the concentration of B.  So in the end, the result would be 0.333 mole/L.
« Last Edit: November 29, 2004, 05:29:48 PM by jdurg »
"A real fart is beefy, has a density greater than or equal to the air surrounding it, consists

#### 777888

• Guest
##### Re:Solubility product constant
« Reply #4 on: November 30, 2004, 10:15:25 PM »
Thanks!
So is the solubility the same as its concentration at equilibrium?

But I thought that solubility is the amount of a salt that HAS BEEN dissolved in a solvent, that is, the amount of salt that is NOT present in equilibrium.

First, consider solubility in g/L, it is actually refering to the grams of salt that is dissolved in a solvent of a particular volume. But not the grams of salt that is LEFT, right?
« Last Edit: November 30, 2004, 10:43:59 PM by 777888 »

#### 777888

• Guest
##### Re:Solubility product constant
« Reply #5 on: November 30, 2004, 10:20:51 PM »
2.Name 2 compounds that will decrease the solubility of BaSO4(s)

I say:
-H2SO4(aq)
-Na2SO4(aq)
Are they right? Is it true that the 2 required compounds must be (aq) state and can dissociate into ions? (If no, what are the requirements?) What are some other examples that will decrease the solubility of BaSO4(s)?

3.Name 2 compounds that will decrease the solubility of CuCO3(s)
I say:
-Na2CO3(aq)
-K2CO3(aq)
Are they right? Is it true that the 2 required compounds must be (aq) state and can dissociate into ions? What are some other examples that will decrease the solubility of CuCO3(s)?

Thank you for helping

#### 777888

• Guest
##### Re:Solubility product constant
« Reply #6 on: November 30, 2004, 10:26:52 PM »
4.Calculate the solubility of calcium sulfate in 0.010mol/L calcium nitrate at SATP(Ksp=7.1x10-5)

My work:
ICE table:
[Ca2+]  [SO4(2-)]
I      0.010     0
C       +x        +x
E      0.010+x   x

Ksp=[Ca2+][SO4(2-)]
7.1x10-5=(0.010+x)(x)
Assume 0.010+x=0.010 snce the value of Ksp is very small.
Then, x=7.1x10-3

[SO4(2-]=7.1x10-3

Therefore [CaSO4]=7.1x10-3 mol/L(solubility)

But the answer is 3.6x10-3 mol/L! Did I do something wrong? I check my work a few times but I don't know where I did wrong!

Can someone correct me? Thank you!

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7978
• Mole Snacks: +555/-93
• Gender:
##### Re:Solubility product constant
« Reply #7 on: December 01, 2004, 09:56:36 AM »
Assume 0.010+x=0.010 snce the value of Ksp is very small.

This assumption is not true!
« Last Edit: December 01, 2004, 10:22:52 AM by AWK »
AWK

#### 777888

• Guest
##### Re:Solubility product constant
« Reply #8 on: December 02, 2004, 03:35:54 PM »
2.Name 2 compounds that will decrease the solubility of BaSO4(s)

I say:
-H2SO4(aq)
-Na2SO4(aq)
Are they right? Is it true that the 2 required compounds must be (aq) state and can dissociate into ions? (If no, what are the requirements?) What are some other examples that will decrease the solubility of BaSO4(s)?

3.Name 2 compounds that will decrease the solubility of CuCO3(s)
I say:
-Na2CO3(aq)
-K2CO3(aq)
Are they right? Is it true that the 2 required compounds must be (aq) state and can dissociate into ions? What are some other examples that will decrease the solubility of CuCO3(s)?

Thank you for helping

Can someone help me with this problem please?

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7978
• Mole Snacks: +555/-93
• Gender:
##### Re:Solubility product constant
« Reply #9 on: December 03, 2004, 01:27:40 AM »
1. All soluble compounds that contain Ba(2+) or SO4(2-) in solution

2. All soluble compounds that contain Cu(2+) or CO3(2-) in solution
AWK

#### 777888

• Guest
##### Re:Solubility product constant
« Reply #10 on: December 04, 2004, 04:47:40 PM »
Thank you~
I have another question:
Consider the equilibrium: AgCl(s) <-> Ag+(aq) + Cl-(aq)
During the processing of photographic film, UNREACTED SILVER COMPOUNDS are removed from the film by the addition of thiosulfate ions, S2O3(2-)(aq), to produce the ion Ag(S2O3)2(3-)(aq). Using Le Chatelier's principle, explain why the addition of a sodium thiosulfate solution makes the solid silver compounds dissolve.

[For the UNREACTED SILVER COMPOUNDS, does it refer to AgCl(s)? (compounds right?) How can the removal of AgCl(s) make the solid silver compounds dissolve? Should the removal of AgCl(s) make crystallization instead?]
« Last Edit: December 04, 2004, 04:48:02 PM by 777888 »

#### 777888

• Guest
##### Re:Solubility product constant
« Reply #11 on: December 05, 2004, 01:28:22 AM »
How do the concepts of a saturated solution and a supersaturated solution help explain the formation of the precipitate of a slightly soluble salt when certain soluble salt solutions are mixed?

Can someone teach me? I would appreciate

#### 777888

• Guest
##### Re:Solubility product constant
« Reply #12 on: December 05, 2004, 04:06:36 PM »
Thank you~
I have another question:
Consider the equilibrium: AgCl(s) <-> Ag+(aq) + Cl-(aq)
During the processing of photographic film, UNREACTED SILVER COMPOUNDS are removed from the film by the addition of thiosulfate ions, S2O3(2-)(aq), to produce the ion Ag(S2O3)2(3-)(aq). Using Le Chatelier's principle, explain why the addition of a sodium thiosulfate solution makes the solid silver compounds dissolve.

[For the UNREACTED SILVER COMPOUNDS, does it refer to AgCl(s)? (compounds right?) How can the removal of AgCl(s) make the solid silver compounds dissolve? Should the removal of AgCl(s) make crystallization instead?]
Can someone help me? I am confused...

#### Donaldson Tan

• Editor, New Asia Republic
• Retired Staff
• Sr. Member
• Posts: 3177
• Mole Snacks: +261/-13
• Gender:
##### Re:Solubility product constant
« Reply #13 on: December 05, 2004, 06:26:08 PM »
thiosulphate forms complex with silver ion, thus reducing the concentration of Ag+ in solution. Hence, more silver chloride can dissolve.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### 777888

• Guest
##### Re:Solubility product constant
« Reply #14 on: December 05, 2004, 10:08:33 PM »
thiosulphate forms complex with silver ion, thus reducing the concentration of Ag+ in solution. Hence, more silver chloride can dissolve.
Oh, I see!
I thought that "UNREACTED SILVER COMPOUNDS" refers to AgCl. I did not get the question right.

How do you know that it is the Ag+ being removed instead? (I thought that ions are not compound...)