is the maximum amount of solute in a solvent at a given temperature saturated solution, [ ]max...
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Transcript of is the maximum amount of solute in a solvent at a given temperature saturated solution, [ ]max...
• is the maximum amount of solute in a solvent at a given temperature
• saturated solution, [ ]max
• equilibrium between:
solid crystals dissolved ions
eg. AgNO3(s) Ag1+(aq) + NO3
1-(aq)
1st - H2O needs
to separate- H-bonding
needs to be overcome
- dissolving
- NRG input
1st
3rd
3rd
2nd
2nd - H2O removes
ions from the lattice- dissociation
- NRG input
3rd - hydration of ions
- Ion – dipole force
- with H2O
- NRG output
•solubility is a contest between:
• lattice energies vs hydration energies
• ionic bond strengths vs ion-dipole forces
• usually exothermic with increased disorder
• if undissolved solid is present, then it is a mixture with a saturated sol’n phase
If: AxBy(s) xAm+(aq) + yBn-
(aq)
Then:ym+ x n-[A ] [B ]
Ksp = [A B ]x y
and [A B ] is constant (solid)x y
ym+ x n- Ksp = [A ] [B ]
If: Q = Ksp then it is a saturated solution
Q > Ksp there is noticeable precipitate
Q < Ksp unsaturated
• refers to the amount of solid that can dissolve not how much is in solution
• solubility units are g/100g or g/100 mL or g/L
To find Concentration from Solubility:
[ ]max = solubility Msolute and V
adjustment
0.591 mol/ L=
eg. Solubility of Mg(OH)2 is 3.45 g/100 mL.
0.591molsol Mg(OH) = x 2 L
58.33g x
mol
100mL
0.1L
= molar solubility
To find Solubility from Concentration:
Solubility = molar solubility x Msolute and V change
3.45g[Mg(OH) ] x 2 100mL
mol x
58.33g
0.1L
100mL
= 3.45 g/ 100 mL
1) Ksp from solubility
eg. Calculate the Ksp of Ag2CO3 given its
solubility of 0.0014 g/100g.
2 30.0014 g
[Ag CO ] = x100 g
1st - calculate the [ ] from solubility:100 g
x 0.1 L
1 mol
275.75 g
-5= 5.1 x 10 mol/ L
2nd - create ICE table.
- Since dealing with a solid, the equilibrium is simplified
+ 2-2 3 3Ag CO 2 Ag + CO ;
I
C
E
+ 2 2-sp 3K = [Ag ] [CO ]
0 0
-5and x = 5.1 x 10 mol/ L (from solubility)
+2 x + x+2 x + x
Q – not necessary
100 Rule – not necessary
+ 2 2-sp 3K = [Ag ] [CO ]
-5 2 -5= [2(5.1 x 10 )] (5.1 x 10 )
-13= 5.3 x 10
2) Solubility from Ksptypes: solubility, molar solubility, [ion]eq, amount
of solid that will dissolve
- all from calculating “x”
eg. How much PbI2 at SATP will dissolve in 1.00 L
of water? Give the solute ion concentration,
[Pb2+]eq, solubility (g/100mL) and molar
solubility.
Ksp of PbI2 from textbook is 8.5 x 10-9
- when not given the Ksp and not asked to calculate it, find it in the reference table or the textbook
2+ 1-2PbI Pb + 2 I ; 2+ 1- 2
sp
-9
K = [Pb ][I ]
= 8.5 x 10
- as the solid is not included and the [ion]i = 0, we
can write ICE horizontally instead of vertically
ICE +x +2x
2 -9(x)(2x) = 8.5 x 10
2+ 1- 2 -9spK = [Pb ][I ] = 8.5 x 10
-3x = 1.3 x 10 mol/ L
3 -94 x = 8.5 x 102+
eq= [Pb ]
= the molar solubility
-3
21.3 x 10 mol
solubility of PbI = x L
461.00 g x
mol
= 0.060 g/ 100 mL
0.10 L
100 mL
Predicting Precipitation:
• Used to determine precipitation when mixing 2 sol’ns just like in double displacement rxns
Don’t by 100 - it is a unit 100 mL
1st – write down each reaction and calculate the V and C values
1+ 2-2 4 4Na CrO 2 Na + CrO
1 1V = 0.0050 L, C = 0.030 M
1+ 1-3 3AgNO Ag + NO
2 2V = 0.0010 L, C = 0.0050 M
+ 2-2 4 4Ag CrO 2 Ag + CrO ;
2nd – write down the equilibrium equation, calculate the new [ ] in the combined sol’n
+ 2 2-sp 4
-12
K = [Ag ] [CrO ]
= 1.1 x 10 1 1
total
C V
V
= 0.00083
2 2
total
C V
V
0.030M x 0.0050L
0.0060L
0.0050M x 0.0010L
0.0060L
= 0.025
3rd – calculate Q and compare with Ksp2Q = (0.025) (0.00083) -7= 5.2 x 10
a precipitate will form
-12Ksp (1.1 x 10 ) >