Ksp : Solubility Product |
Key Concepts
| Condition |
Nature of Solution |
Possibility of Precipitation |
| ion product < Ksp |
unsaturated solution |
no precipitation occurs |
| ion product = Ksp |
saturated solution |
saturated solution at equilibrium |
| ion product > Ksp |
supersaturated solution |
precipitation occurs |
Example : Calculating the solubility of an ionic compound (MA).
Calculate how much silver bromide will dissolve in 1 L of water given Ksp = 5.0 x 10-13 at 25oC.
- Write the equation for dissolving AgBr in water:
AgBr(s) Ag+(aq) + Br-(aq)
- Write the equilibrium expression:
Ksp = [Ag+][Br-] = 5.0 x 10-13
- Determine the relative concentrations of each ion:
at equilibrium [Ag+] = [Br-] (from the balanced chemical equation)
So, [Ag+] = [Br-] = x
- Substitute these vales into the equilibrium expression:
5.0 x 10-5 = [x]2
- Solve for x:
x = √5.0 x 10-5 = 7 x 10-7M
So, [Ag+] = [Br-] = 7 x 10-7 mol L-1
The solubility of AgBr at 25oC is 7 x 10-7 M
7 x 10-7 moles AgBr will dissolve in 1L of water at 25oC.
Example : Calculating the solubility of an ionic compound (MA2).
Calculate how much strontium fluoride will dissolve in 1 L of water given Ksp = 2.5 x 10-9 at 25oC.
- Write the equation for dissolving SrF2 in water:
SrF2(s) Sr2+(aq) + 2F-(aq)
- Write the equilibrium expression:
Ksp = [Sr2+][F-]2 = 2.5 x 10-9
- Determine the relative concentrations of each ion:
at equilibrium [Sr2+] = x and [F-] = 2x (from the balanced chemical equation)
- Substitute these vales into the equilibrium expression:
2.5 x 10-9 = [x][2x]2 = 4x3
- Solve for x:
2.5 x 10-9 ÷ 4 = x3 = 6.25 x 10-10
x = 3√6.25 x 10-10 = 8.5 x 10-4M
The solubility of SrF2 at 25oC is 8.5 x 10-4 M
8.5 x 10-4 moles of SrF2 will dissolve in 1L of water at 25oC.
Example : Deciding whether a precipitate will form
Will a precipitate form if 25.0 mL of 1.4 x 10-9 M NaI and 35.0 mL of 7.9 x 10-7 M AgNO3 are mixed? (Ksp for AgI at 25oC is 8.5 x 10-17)
- Write the net ionic equation for the dissolving of the precipitate:
AgI Ag+ + I-
- Write the expression for the ion product:
ion product = [Ag+][I-]
- Calculate the concentration each reactant after mixing:
[I-] :    
M1 = [I-] before mixing = 1.4 x 10-9 M (assuming full dissociaiton of NaI)    
V1 = initial volume = 25.0 mL = 25.0 x 10-3 L    
V2 = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive)    
M2 = [I-] after mixing = M1 x V1 ÷ V2 = (1.4 x 10-9 x 25.0 x 10-3) ÷ (60.0 x 10-3) = 5.8 x 10-10 M
[Ag+]:    
M1 = [Ag+] before mixing = 7.9 x 10-7 M (assuming full dissociaiton of AgNO3)    
V1 = initial volume = 35.0 mL = 35.0 x 10-3 L    
V2 = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive)    
M2 = [Ag+] after mixing = M1 x V1 ÷ V2 = (7.9 x 10-7 x 35.0 x 10-3) ÷ (60.0 x 10-3) = 4.6 x 10-7
- Calculate the ion product:
   
ion product = [Ag+][I-] = 5.8 x 10-10 x 4.6 x 10-7 = 2.7 x 10-16
- Decide whether a precipitate forms:
   
If ion product > Ksp a precipitate will form    
If ion product < Ksp a precipitate will not form    
In this case, 2.7 x 10-16 > Ksp (8.6 x 10-17) so a precipitate will form
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| Practice Questions |
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For AUS-e-TUTE members:
- Click on the Ksp drill link:
Ksp drill
- Enter your username and password if prompted.
- Click the "New Question" button to begin the drill.
- Worked solutions are provided if you need some help!
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