Chapter 1 Introduction · Introduction Analytical Chemistry • qualitative analysis (Chem 101-103)...
Transcript of Chapter 1 Introduction · Introduction Analytical Chemistry • qualitative analysis (Chem 101-103)...
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Chapter 1
Introduction
Analytical Chemistry
• qualitative analysis (Chem 101-103)
• quantitative analysis
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Quantitative Analysis
• gravimetric
• volumetric
• electroanalytical
• spectroscopic
• chromatographic (if time allows)
Analytical Terminology
• heterogeneous
• homogeneous
• analyte
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Steps in a Chemical Analysis
• Sampling: representative
• Sample Preparation
• Analyzing the Sample
• Interpreting the results
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Grind to small pieces
Extraction
Decant, quantitative transfer, supernatant liquid
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Analyzing the Sample
Step 1. Obtain a representative bulk sample.
Step 2. Extract from the bulk sample asmaller, homogeneous laboratory sample.
Step 3. Convert the laboratory sample into aform suitable for analysis, a process thatusually involves dissolving the sample.
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Analyzing the Sample
Step 4. Remove or mask species that willinterfere with the chemical analysis.
Step 5. Measure the concentration of theanalyte in several aliquots.
Step 6. Interpret your results and drawconclusions.
SI Prefixes
especially useful in this course
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
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Classification of Analytical MethodsAccording to Size of Sample
<1Ultramicro
<501-10Micro
50-10010-100Semimicro
>100>100Meso
Sample Volume( L)
Sample Weight(mg)
Method
Constituents
<0.1%trace
0.1-1%minor
>1%major
ConstituentsMethod
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Solution Terminology
• solute
• solvent
• aqueous solution
• liter
• atomic weight
• molecular weight
Molarity
# moles AMolarity => M = -------------------
# liters solutionor
# millimoles AMolarity => M = -------------------------
# milliliters solution
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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)
Types of Solutions
• strong electrolyte
• weak electrolyte
• non-electrolyte
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Useful Algebraic Relationships
n = M ×V
n = W (g)FM(g /mol)
M = nV
(mol /L) Molarity
Formal Concentration
• used for systems which separate (ionize) insolution
• same form for equation as molarity,substitute formula weight for molecularweight for those substances which do notform molecules
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Molality => m# moles A
molality => m = ------------------------- # kilograms solvent
• this concentration unit is temperatureindependent as the mass does not changewith temperature whereas volume does
• used in freezing point depression/boilingpoint elevation
• not commonly used.
p-Functions
pX = - log10[X]
examples:
pH
pOH
pCl
pAg
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Percent Composition
wt of a solutew - w% = -------------------- 102
wt of solutionvol of a solute
v - v% = -------------------- 102
vol of solutionwt of a solute
w - v% = --------------------- 102
vol of solution
Parts per Million
wt of a solutecppm = ------------------- 106
wt of solution
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Parts per Billion
wt of a solutecppb = ------------------- 109
wt of solution
Preparing Solutions
EXAMPLE: Describe the preparation of 1.00L of 0.100 M NaOH solution (f.w.40.00) from reagent grade solid.
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EXAMPLE: Describe the preparation of 1.00L 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 1.00L 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)
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EXAMPLE: Describe the preparation of 1.00L 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 1.00L of 0.100 M NaOH solution (f.w. 40.00)from reagent grade solid.
(1.00)(0.100 mol NaOH)# g NaOH = ------------------------------
(1)
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EXAMPLE: Describe the preparation of 1.00L 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 1.00L 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)
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EXAMPLE: Describe the preparation of 1.00L 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 1.00L 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
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EXAMPLE: Describe the preparation of 1.00L 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 Lvolumetric flask, and dilute to the line.
Dilution
#moles solute in conc. soln
equals
#moles solut in dil. soln
therefore
Mconc Vconc = Mdil Vdil
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Examples
• 2. How can one prepare 500 mL 1.0 M HClsolution from concentrated HCl solution(12M)?
Examples• 3. What is the molarity of 1:6 (v:v) HNO3
solution?
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Examples
• 3. Calculate the molarity of concentratedsulfuric acid (Strength = 95.5-96.5%,Density = 1.84)?
• AAcetic Acid, Glacial, 100% 117 MolarAmmonia, 29% 115 MolarHydrochloric Acid, 37%, 12 MolarNitric Acid, 70%, 16 MolarPhosphoric Acid, 85%, 15 MolarPerchloric Acid, 71%, 11 MolarSodium Hydroxide, 50%, 19 MolarSulfuric Acid, 96%, 18 Molar (36 Normal)
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Comparison of Different AnalyticalMethods