Solution Equilibrium exam outline
Solutions:
Concentration units:
molal (m)
moles/kg solvent
%(mass)
g/g solution
ppm
g/106g total or μg/g total
ppb
g/109g total or ng/g total
mole fraction(X)
moles/moles total
molarity (M) moles/L solution
Raoult’s Law Ptot = PA + PB = XAPºA +XBPºB Ideal solutions: no intermolecular forces.
Pressure just the sum of individual pressures times the fraction of molecules of that type.
Freezing point depression
For nonvolatile solutes: ∆Tf = kf i m
∆Tb = kb i m k’s property of solvents
i = # ions / molecule
Remember
Tf = Tºf - ∆Tf
Tb = Tºb + ∆Tb
Calculate new freezing or boiling temperature given solute and concentration or the i
factor given concentration and Tf.
Change of vapor pressure with temperature:
H vap
P
ln  1   
R
 P2 
1 1 
  
 T1 T2 
Calculate ∆H given P,T data or Calculate new P or T given ∆H and P,T data.
Osmosis
Π = M RT Osmotic pressure (Π ) increases with molarity of solute.
More concentrated side absorbs solvent, rises in U tube.
Concepts: Intermolecular forces (see solutions HW)
Ideal Solutions, what are the assumptions
Solubility and temperature
Enthalpy of solution
Equilibrium:
Know how to write the mass action expression for the equilibrium relation for a balanced
reaction.
[ A]a [ B]b
K
[C ]c [ D]d
aA + bB --> cC + dD
Solids and pure liquids not included in rate expression.
LeChatellier’s Principle and direction of reaction.
Calculate K from equilibrium concentrations:
Given [A], [B], [C], [D] plug these into the expression for K above
and calculate the value for K.
Relate Kc to Kp using P = (n/V) RT = [ ] RT So Kp = Kc (RT)(nprod-nreact)
Calculate the equilibrium concentrations given K and the beginning concentrations.
You can shift concentrations to product or reactant side and not affect the equilibrium
values. DON’T FORGET THE STOICHIOMETRY!
Examples of approximations when K is large or small:
Small K:
HCN (aq)  H+(aq) + CN-(aq)
K = 4.9 x10-10
Init.:
0
5x10-3
0.001
Mod. Init
.001
4x10-3
0
Change
-x
+x
+x
Equilib
.001-x
4x10-3+x
K tiny, shift to reactant side
x
K small, so x<<.001, drop from first two entries above since it is insignificant compared to
those.
So
K
Giving
[CN  ][ H  ]
x(4 103 )
 4.9 1010 
 4x
[ HCN ]
.001
[HCN] = .001M
[H+] = 4x10-3 M
x  1.2 1010
[CN-]= 1.2x10-10M
Large K:
Cu+
Init.:
+
.006
2CN-

.015
Cu(CN)20
Mod. Init
0
3x10-3
.006
Change
+x
+x
-x
Equilib
x
3x10-3+x
K=1.0 x 1016
K large, shift to product side
.006 - x
K large, so x<<.001, drop from last two entries above since it is insignificant compared to
those.
Approx:
So, K 
x
3x10-3
.006
[Cu (CN )2 ]
.006
 1.0 1016 

 2
[Cu ][CN ]
x(.003)2
Giving [Cu+] = 6.7 x 10-14 M
so x  6.7 1014
[CN-] = 3x10-3 M
[Cu(CN)2-] = .006 M
Suppose on the first example you do not shift to the reactant side.
HCN (aq)  H+(aq)
Init.:
0
Change
+x
Equilib
x
CN-(aq)
+
5x10-3
K = 4.9 x10-10
0.001
-x
K tiny, shift to reactant side
-x
5x10-3-x
.001-x
[CN  ][ H  ]
(.001  x)(5 103  x) 5 106  6 103 x  x 2
10
K
 4.9 10 

[ HCN ]
x
x
4.9 1010 x  5 106  6 103 x  x 2
x 2  (6 103  4.9 1010 ) x  5 106  0
Drop the 4.9 x10-10 added to 6x10-3 :
x 2  6 103 x  5 106  0  (.001  x)(.005  x) from above.
So, either from quadratic equation or looking back at the original equilibrium equation factors,
x = .001 or .005, the second is unphysical since it leaves [CN-] negative.
This gives: [HCN] = .001M
[H+] = 4x10-3M
[CN-] = 0 Oops!
Just put the nonzero concentrations into the equilibrium expression and put in a variable, y, for
the zero concentrations:
4.9 1010 
(.004) y
 4 y y  1.2 1010
.001
[HCN] = .001M
giving
[H+] = 4x10-3M
[CN-] = 1.2x10-10 M
So you can recover from this if you forget to approximate.
More examples on the links page.
Energy and Equilibrium:
ΔG =ΔH – TΔS = -RTln(Q) Q = K expression with actual concentrations substituted.
If ΔG <0 for a reaction, it is spontaneous, will proceed forward
If ΔG = 0 reaction is at equilibrium, concentrations won’t change
If ΔG >0 Reaction will proceed toward reactants, non-spontaneous.