# Equilibrium, Solubility and Precipitates

1. Aug 5, 2007

### skander

Can someone please explain to me:

1) How to determine an eqilibirium constant

2) How to calculate Ksp from solubility values

3) How to determine if a precipitate will form in a reaction

I am doing a chemistry course without a teacher and the textbook doesn't properly explain it.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 6, 2007

### Gokul43201

Staff Emeritus
Note that we typically require you to show your own effort before we can add anything to the discussion.

You may find these useful:
http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium.html [Broken]
...
http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium-Constant.html [Broken]
http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Calc-K-from-equilib-conc.html [Broken]
...
http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Writing-Ksp-expression.html [Broken]
http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Calc-Ksp-FromMolSolub.html [Broken]

Last edited by a moderator: May 3, 2017
3. Aug 6, 2007

### skander

Okay, thanks for the links. So let me try to work out some answers over here.

1)A(g) --> 2B(g) + C(g)
<--

When 1 mol of A is placed in a 4 L container at temperature t the concentration of C at equilibibrium is 0.050 mol/L. What is the equiliibrium constant for the reaction at temperature t?

Well the links didn't tell me how to find the eqilibrium contant with unknowns.

2) Calculate the Ksp for each of these salts

A) CaSO4 = 3.3 * 10 -3 mol/L
B) MgF2 = 2.7 * 10 -3 mol/L

So for a) CaSO4 would dissociate to form [Ca] and [SO4] meaning
(3.3 * 10 -3) (3.3 * 10 -3) = 1.089 -5 Ksp = 1.089 -5

and for b) MgF2 would become [Mg] and [F]2 so we would have a 1:2 ratio so,
(2.7 * 10 -3) (5.4 * 10 -3) = 1.458 -5 Ksp = 1.458 -5

Am I right with these two?

Also for determing a precipitate they had an article titled that on their main page but no link so I still am not sure how to do that.

4. Aug 6, 2007

### Gokul43201

Staff Emeritus
You're making a mistake on (b). First write down the expression for Ksp before you substitute values in.

5. Aug 7, 2007

### skander

Ksp = [Mg] [F2]
Ksp = (2.7 * 10 -3) (2.7 * 10 -3) =7.9 *10 -6
Ksp = 7.9 *10 -6

Is this better?

6. Aug 7, 2007

### Gokul43201

Staff Emeritus
No, it should be

$$K_{sp}=[Mg^{2+}][F^-]^2$$

Go back and read the definition of the equilibrium constant again.
http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium-Constant.html [Broken]

Last edited by a moderator: May 3, 2017
7. Aug 7, 2007

### skander

Okay, I got the exponent 2 the first time but I'm not sure how you determine that MgF2 will become [Mg2+] and [F-]2. How did you determine the ions? And shouldn't Ca and SO4 also have them then?

Also, I think I found how to determine if a precipitate will form, can you check this work for me?

The solubility product of Ca(OH)2 is 7.9 * 10 -6 at 25C. Will a precipitate form when 100 ml of 0.10 mol/L of CaCl2 and 50.0 ml of 0.070 mol/L of NaOH are combined?

So [CaCl2] = 0.10 mol/L * 100 ml / 150 ml = 0.6666666667
[NaOH] = 0.070 mol/L * 50 ml / 150 ml = 0.0233333333

(0.6666666667) (0.0233333333) = 0.01555555563

Since Ktrial > Ksp a precipitate will form.
Is this right?