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CH107 - Physical Chemistry
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  • Course
  • Chemistry: Principles and Practice 2 (CH107)
  • Institution
  • The University Of Strathclyde (UOS)
  • Book
  • Chemistry(3)

Following document contains EQUATIONS, graphs, definitions and example calculations from the following areas: - RMS - Mean free path - Equipartition theorem - Internal energy - Van der Waals' equation for real gases - Gas mixtures and equilibrium - Gases and liquids - Raoult's law - Colli...

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Document information

  • March 15, 2021
  • 25
  • 2020/2021
  • Class notes
  • Dr. nico bruns
  • All classes

Subjects

  • kinetic theory
  • rms speed of gas
  • degrees of freedom
  • definition of temperature
  • internal energy
  • heat capacity
  • partial pr
  • boyles law
  • charles law
  • avogadros hypothesis
  • van der waals equation

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  • The University of Strathclyde (UOS)
  • The University of Strathclyde
  • Chemistry: Principles and Practice 2 (CH107)
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Solubility Solubility of a solid in a solvent is the
maximum that will dissolve in a given
mass
volume of the solvent at a particular
temperature in the presence of excess
undissolved solid so that the solution is
saturated


Solubility rules Some empirical rules have been made
based on observation




solubility product constant
For insoluble PbCLz Salt with its ions in aqueous
solution

Ploch s T Pflag 2cL Cag

µ apbzt.CA Pf Ct
Apoch
Solubility product constant
Ksp Pf cc is calculated as any eg
constant

, When solving for equilibrium concentrations similar to
ice tables we obtain molar solubility s which
gives the number of moles of solute dissolved in 1dm
of a saturated solution

common ion effect reduction in solubility of a salt
occurs due to the presence in solution of one of the
ions from another source



Tfsolid Catz



Nat
we can use solubility product constant to determine
when a compound will precipitate

consider AgCLCs T
Ag Caq Ct Cag


We have reaction quotient Q Agt Ci
If Q2 Ksp solution is unsaturated
If Q Ksp solution is saturated
If Q Ksp Solution is supersaturated
solid precipitate will form

, s
e g c Pb Noda 0.1 moldm
Pb Noda 50mL

c NaCl 0.005moldm 3
NaCl 50mL

solution V 100mL
concentrations are halved

Pb NOD 2 NaCl PbCLzt 2 NANO
5
We have Ksp Ptc 1.17 10 M
since this is the solid
7
Q Pb Nos NaCl 0.05 0.0252 3.13 10 M

Thus Qi Ksp No precipitate

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