# Ksp question

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

#### KappaSig

##### Future Dentist!
10+ Year Member
15+ Year Member
Question: The solubility of Fe(OH)3 in an aqueous solution was determine to be 4.5*10^-10 mol/L. What is the vaule of the Ksp for Fe(OH)3?

so i got the part of :

Fe(OH)3(s) <---> Fe^3+ (aq) + 3OH^- (aq)

Ksp= [Fe^3+][OH^-]^3

i dont get this part here how they got [OH^-]= 3[Fe^3+] where is this coming from

then it comes out to
Can someone please show me how to work out this calculation without a freaking calculator?
Ksp=[Fe^3+](3[Fe^3+])^3=27[Fe^3+]^4
Ksp= (4.5*10^-10)[3(4.5*10^-10)]^3=27(4.5*10^-10)^4
Ksp= ?

and this problem please the calculations:

Question: What are the concentrations of each of the ions in a saturated solution of PbBr2, given that the Ksp of PbBr2 is 2.1*10^-6? If 5 g of PbBr2 are dissolved in water to make 1L of solution and this is then brought to 25degrees C, would the solution be saturated, unsaturated, or supersaturated?

please work this out for me........

#### KappaSig

##### Future Dentist!
10+ Year Member
15+ Year Member
I realized my mistake on the 1st question... i forgot to raise the 4.5^4...

but the 2nd on i'm still having trouble with..
and still cant figure out how they derived with this

i dont get this part here how they got [OH^-]= 3[Fe^3+] where is this coming from

#### Doc Smile

##### senior member
10+ Year Member
1. Equilibrium equations
Fe(OH)3(s) <---> Fe^3+ (aq) + 3OH^- (aq)
H20 <---> H+ + OH-

2.Charge Balance gives: [OH-] = 3[Fe^3+] + [H+]
*this relates charges together, where the coefficeint equals the charge on the ion. It is how Ksp calculations where calulated before the ICE method (it is called the systematic approach to equilibrium and I will do the rest of the system not occording to ICE).
* The concentration of H+ from water is sooooooooooooo small it can be ignored (not 1.0x10-7 M in this equation bc of common ion)

3. Mass Balance Gives:
4.5x10^10 M = [Fe^3+]
4.5x10^10 M = 3[OH-]
[Fe^3+] = 3[OH-]

4. Equations
Ksp= [Fe^3+][OH^-]^3
Kw= [H+][OH-]

5. skip
6. solve
Substituting charge balance into Ksp you get
Ksp=[Fe^3+](3[Fe^3+])^3 = (x)(3(x)^3 = x*27x^3 = 27x^4
substitute x and solve, good luck w/o a calc

Comprende?

second question
What are the concentrations of each of the ions in a saturated solution of PbBr2, given that the Ksp of PbBr2 is 2.1*10^-6? If 5 g of PbBr2 are dissolved in water to make 1L of solution and this is then brought to 25degrees C, would the solution be saturated, unsaturated, or supersaturated?

Ksp= 2.1E-6 = [Pb2+][Br-]^2
5g PbBr2 / 1L = 5 M PbBr2

2.1E-6 = x(2x)^2 = 4x^3
x = 8.067E-3 = [Pb2+]
[Br-] = 1.6E-2

I am going to guess this is unsaturated i cant remember the exat definitions but supersatureated is used to make uber pure crystals, saturation is the point of percipitation and unsaturation is still dissolving.

##### Full Member
10+ Year Member
Think about it this way: For every Fe3+ molecule you get 3 OH-, so at equilibrium you will have 3 times teh concentration of OH- than Fe3+ that's where they get the [OH^-]= 3[Fe^3+]

Question: The solubility of Fe(OH)3 in an aqueous solution was determine to be 4.5*10^-10 mol/L. What is the vaule of the Ksp for Fe(OH)3?

so i got the part of :

Fe(OH)3(s) <---> Fe^3+ (aq) + 3OH^- (aq)

Ksp= [Fe^3+][OH^-]^3

i dont get this part here how they got [OH^-]= 3[Fe^3+] where is this coming from