unit 2: solids, liquids, equilibrium, solubility

Ca(NO3)2

YESAll nitrates are soluble!

Zn(OH)2

NOHydroxides are insoluble except for sodium and potassium ions!

K3PO4

YES Phosphates are insoluble except for sodium, potassium, and ammonium ions!

Ene

rgy

to b

oil (

kJ/m

ol)

1100

Ion-ion

700

500

Covalent bonds

100

60

Intermolecular forces

0.1

H-bonds

Polar bonds

Non-polar bonds

(intramolecular forces)

Why so strong?

-Small size of hydrogen

-High electronegativity of O, N, and F

What are they?

-A network between H and either N, O, or F molecules

NameName Simple CubicSimple Cubic Body-Body-Centered Centered CubicCubic

Face-Face-Centered Centered CubicCubic

# atoms/unit # atoms/unit cellcell

11 22 44

VolumeVolume

e = edge e = edge lengthlength

r = cell r = cell

radiusradius

ee33 = (2r) = (2r)33 = 8r = 8r3 3 ee33 = =

(4r/√3)(4r/√3)3 3

ee33 = =

(4r/√2)(4r/√2)33

For:

3A(aq) + B(s) ↔ 2C(g) + D(l) + E(aq)

** Only gaseous and dissolved particles are expressed in the equilibrium expression because their concentrations can vary (whereas solids and liquids cannot)

2

3

[ ] [ ]

[ ]c

C EK

A

If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

3A(aq) + B(s) ↔ 2C(g) + D(l) + E(aq)

If the concentration of A is raised, as the mixture returns to equilibrium a portion of all of the reactants are consumed, and as a result, the concentrations of C, D, and E will increase

N2 + O2 ↔ 2NOI

C

E

0.25 0.25 0.0042

-x -x +2x

0.25 0.25 0.0042 + 2x

Kc = 1.7 x 10-3 = (0.0042 + 2x)2 / (0.25)(0.25)

Eventually… x = 0.0030 M change

Would a precipitate form?

AgCl(s) ↔ Ag+(aq)+ Cl-(aq)

Ksp = 1.8 x 10-10

Q = [Ag+][Cl-]

25 mL of 0.1M AgNO3 solution is added to 100 mL of 0.0050M NaCl solution

If Q<Ksp, no precipitation

If Q>Ksp, precipiation

[Ag+] = 0.10M(25mL/125mL)

[Cl-] = 0.005M(100mL/125mL)

Therefore Q>Ksp, and this is expected to precipitate

Q = (0.100)(0.005) = (5.0 x 10-4)

Molar Concentration “molarity”

M = # moles/vol. of solution

Molal Concentration “molality”

m = # moles solute / mass of solvent (in kg)

Mass Fraction

= mass solute / mass total

Ppm (parts per million)

= mass solute (in mg) / mass total (in kg)

Mole Fractions

XA = molesA/molesA + molesB +…

Cd(OH)2 Ksp = 1.2 x 10-14

Ca(OH)2 Ksp = 7.9 x 10-6

Ca2+

Cd2+

0.10 M Cd2+

0.10 M Ca2+

At what concentration of OH- will one of the ions precipitate?

(OH-) is more attracted to Cd2+, because, as the equations are similar in structure (1 cation, 2 anions) they are comparable and Cd(OH)2 would precipitate first

Cd(OH)2 Cd2+(aq) + 2OH-

(aq)

0.10 ~0

+x +x

Ksp = 1.2*10-14 = (0.10)(x)2

X2 = 12*10-12

x = 3.5*10-6

Freezing Point Depression

ΔTf = iKfm

Boiling Point Elevation

ΔTb = iKbmi = # dissociated particles in empirical formula (Van’t Hoff factor)

Kf or Kb = boiling point elevation / freezing point depression constant

m = (n mol solute / mass solvent (kg))

Volatile SolutionPvaptotal = XAPvapA + XBPvapB +…

Non-volatile Solution

PvapA = XAPvapA + 0

volatile= changes to gas

“the solubility of a gas is proportional to the pressure of the gas”

Solg = KHenryPg

Osmotic Pressure

Π = (n/v)RT

R = 0.082057 L · atm / K · mol

C = (n / v) = (mol/L)

T = temperature in Kelvin