Quantities and Concentrations SI Base Units Physical quantityName of UnitsAbbreviation...
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Transcript of Quantities and Concentrations SI Base Units Physical quantityName of UnitsAbbreviation...
Quantities and Concentrations
SI Base UnitsPhysical quantity Name of Units Abbreviation
Mass kilogram kg
Length meter m
Time second s
Temperature kelvin K
Amount of substance mole mol
Electric current ampere A
Luminous intensity candela cd
SI Prefixes
especially useful in this course giga G 109
mega M 106
kilo k 103
centi c 10-2
milli m 10-3
micro 10-6
nano n 10-9
pico p 10-12
Quantities
• Mole
• Molar mass
• Mass
• Weight
• What is the difference between mass and weight?
Useful Algebraic Relationships
wt A (g) mol A = -----------------
fw A (g/mol)
mol A = V (L) x M (mol A/L soln)
or
wt A (mg) mmol A = -----------------
fw A (g/mol)
mmol A = V (mL) x M (mmol A/mL soln)
Solution Terminology
• solute
• solvent
• aqueous solution
• liter
• atomic weight
• molecular weight
Molar concentrtion-Molarity
no. moles AMolarity => M = -------------------
no. liters solutionor
no. millimoles AMolarity => M = -------------------------
no. milliliters solution
Analytical Molarity Equilibrium molarity
• What are analytical molarity and equilibrium molarity?
• What is the difference between them?
Percent Composition
wt of a solutew - w% = -------------------- × 100%
wt of solution
vol of a solutev - v% = -------------------- × 100%
vol of solution
wt of a solutew - v% = --------------------- × 100%
vol of solution
p-Functions
pX = - log10[X]
examples:
pH
pOH
pCl
pAg
Parts per Million / Billion
wt of a solutecppm = ------------------- × 106
wt of solution
wt of a solutecppb = ------------------- × 109
wt of solution
Empirical FormulasMolecular FormulasStructure Formulas
Chemical Stoichiometry
Preparing Solutions
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
(1.00 L soln) # g NaOH = ----------------
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
(1.00 L soln)(0.100 mol NaOH)
# g NaOH = --------------------------------------- (1 L soln)
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
(1.00 L soln)(0.100 mol NaOH)
# g NaOH = --------------------------------------- (1 L soln)
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
(1.00)(0.100 mol NaOH)# g NaOH = ------------------------------
(1)
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
(1.00)(0.100 mol)(40.00g NaOH)
# g NaOH = ---------------------------------------- (1) (1 mol)
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
(1.00)(0.100 mol)(40.00g NaOH)
# g NaOH = ---------------------------------------- (1) (1 mol)
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
(1.00)(0.100)(40.00g NaOH)# g NaOH = ----------------------------------------
(1) (1)
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
(1.00)(0.100)(40.00g NaOH)# g NaOH = ----------------------------------------
(1) (1)
= 4.00 g NaOH
EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.
(1.00)(0.100)(40.00g NaOH)# g NaOH = ----------------------------------------
(1) (1)
= 4.00 g NaOH
Weigh 4.00 g of NaOH, transfer to a 1.00 L volumetric flask, and dilute to the line.
Dilution
#moles solute in conc. soln
equals
#moles solut in dil. soln
therefore
Mconc Vconc = Mdil Vdil