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Exp 4A: Conductivity Of Aqueous Solutions

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Title: Exp 4A: Conductivity Of Aqueous Solutions


1
Exp 4A Conductivity Of Aqueous Solutions
  • Purpose
  • Study conductivity of a series of solutions to
    determine the difference between strong
    electrolytes, weak electrolytes and
    nonelectrolytes
  • Use conductivity to distinguish between strong
    and weak acids and strong and weak bases
  • Use conductivity to study the effects of ion
    concentration

2
  • Conductivity of Solutions
  • The conductivity (or specific conductance) of an
    electrolyte solution is a measure of its ability
    to conduct or allow the passage of electricity.
  • Conductivity units
  • The SI unit of conductivity is siemens per meter
    (S/m)
  • Conductivity vs Conductance
  • Consider a piece of wire.
  • Electrical conductivity is a property of the
    material of the wire, and its value changes only
    with temperature (it decreases linearly with
    temperature).
  • On the other hand conductance is the measure of
    the ease with which current can flow in the wire.
    It depends on the physical parameters of the wire
    (length,area of cross section) as well as the
    conductivity of the material of the wire.
    conductanceconductivity x area of
    cross section / length

3
  • APPLICATIONS and CONCEPT CONNECTIONS
  • Conductivity measurements are used routinely in
    many industrial and environmental applications as
    a fast, inexpensive and reliable way of measuring
    the ionic content in a solution. For example, the
    measurement of product conductivity is a typical
    way to monitor and continuously trend the
    performance of water purification systems.
  • For water quality- Conductivity is linked
    directly to the total dissolved solids (T.D.S.).
  • High quality deionized water has a
    conductivity of about 5.5 µS/m,
  • Typical drinking water in the range of 5-50
    mS/m
  • Sea water about 5 S/m (one million times
    higher than deionized water).
  • Conductivity is traditionally determined by
    measuring the AC (alternating currect) resistance
    of the solution between two electrodes.
  • Dilute solutions follow Kohlrausch's Laws of
    concentration dependence and additivity of ionic
    contributions. Onsager gave a theoretical
    explanation of Kohlrausch's law by extending the
    DebyeHückel theory.

4
Exp 4A Conductivity Of Aqueous Solutions
  • Electrolytes
  • Aqueous solutions of ionic compounds
  • Ionic compounds dissolve and dissociate in water
  • NaCl(s) ? Na(aq) Cl-(aq)
  • Formation of positive and negative ions in
    solution
  • Solution conducts electricity
  • Strong electrolytes conduct electricity easily
  • Strong electrolytes are completely dissociated
    (100)
  • Weak electrolytes conduct electricity poorly
  • Weak electrolytes are only partially dissociates
    (lt100)
  • Mostly undissociated molecular form
  • CH3COOH(l) H(aq) CH3COO-(aq)
  • Nonelectrolytes
  • Compounds that do not conduct electricity in
    solution
  • Compounds that do not form ions in aqueous
    solution

5
Exp 4A Conductivity Of Aqueous Solutions
  • NaCl(s) H2O(l) ? Na(aq) Cl-(aq)
  • 1 positive charge 1 negative charge
  • 2 ionic charges
  • MgCl2(s) H2O(l) ? Mg2(aq) 2 Cl-(aq)
  • 1 ion with 2 charge, 2 ions with 1 charge
  • 4 ionic charges
  • Fe(NO3)3 H2O(l) ? Fe3(aq) 3 NO3-(aq)
  • 1 ion with 3 charge, 3 ions with 1 charge
  • 6 ionic charges
  • Conductivity depends on
  • Concentration of ions
  • Charge of ions
  • (Size of ions (mobility in solution large ions
    move more slowly))

6
Exp 4A Conductivity Of Aqueous Solutions
Electrolytes conduct electricity
http//nvcicourse.accd.edu8900/SCRIPT/1020720051/
scripts/student/serve_page.pl/1020720051/chapter4/
animations.html?11187622481859016556OFFolc/dl/1
71024/4_1_Stg_Wk_Nonelelytes.swf
7
Exp 4A Conductivity Of Aqueous Solutions
  • Strong electrolytes
  • Conduct electricity easily
  • Electrolyte (almost) completely dissociated in
    solution
  • NaCl(s) H2O(l) ? Na(aq) Cl-(aq)
  • Weak electrolytes
  • Poorly conducting solutions
  • Electrolytes mostly in molecular form with few
    ions
  • NaH2PO4(s) H2O(l) Na(aq) H2PO4-(aq)
  • Magnitude of conductance of a solution
  • proportional to the number and type of ions in
    solution
  • More ions (higher ion concentration) more
    conductivity
  • More ionic charges more conductivity

8
Exp 4A Conductivity Of Aqueous Solutions
  • Strong Acids
  • HCl(g) H2O(l) ? H(aq) Cl-(aq)
  • HNO3(l) H2O(l) ? H(aq) NO3-(aq)
  • Strong Bases
  • KOH(s) H2O(l) ? K(aq) OH-(aq)
  • Weak Acids
  • CH3COOH(l) H2O(l) H(aq) CH3COO-(aq)
  • Weak Bases
  • NH3(g) H2O(l) NH4(aq) OH-(aq)

9
Exp 4A Conductivity Of Aqueous Solutions
  • Strong and Weak electrolytes

Strong Electrolyte
Weak Electrolyte
10
Exp 4A Conductivity Of Aqueous Solutions
  • Mixtures of electrolytes
  • Conductance shows additive effect if electrolytes
    do not react with each other
  • more ions, more charges, more conductivity
  • If a chemical reaction occurs between the
    electrolytes, the properties of the new
    substance(s) determine conductivity
  • typically reaction between a weak acid and a base
    or a weak base and an acid
  • NH3(aq) HCl(aq) ? NH4(aq) Cl-(aq)

11
Exp 4A Conductivity Of Aqueous Solutions
  • Dilution
  • A certain amount of a solution added to an amount
    of solvent to lower the concentration
  • Add 10 mL of 1.0 M NaCl to 90 mL of H2O
  • Final volume 10 mL 90 mL 100 mL
  • Dilution 110
  • Concentration (10 mL x 1.0 M NaCl)/ 100 mL
    0.10 M NaCl
  • The concentration changes. Does the total amount
    of NaCl particles change?
  • Dilution formula
  • initial molarity (mol/L) x initial volume (L)
  • final molarity (mol/L) x final volume (L)
  • Mi x Vi Mf x Vf
  • mol/L x L mol

12
Exp 4A Conductivity Of Aqueous Solutions
  • Prelab Question 4a
  • How will you prepare 10 mL of 0.050 M HCl from
    0.10 M HCl?
  • Answer
  • Use dilution formula V1 x M1 V2 x M2
  • M1 0.10 M
  • M2 0.050 M
  • V2 10 mL
  • V1 V2 x M2/M1 10 mL x 0.050 M/0.10M 5.0 mL
    of 0.10 M HCl
  • Take 5.0 mL of 0.10 M hydrochloric acid in a
    10.0-mL graduated cylinder and dilute to 10 mL
    with dH2O

13
Exp 4A Conductivity Of Aqueous Solutions
  • Prelab Question 5a
  • How will you prepare 80 mL of 0.10 M CH3COOH from
    6.0 M acetic acid?
  • Answer
  • Use dilution formula V1 x M1 V2 x M2
  • M1 6.0 M
  • M2 0.10 M
  • V2 80 mL
  • V1 V2 x M2/M1 80 mL x 0.10 M/6.0M 1.3 mL of
    6.0 M acetic acid
  • Take 1.3 mL of 6.0 M acetic acid in a 5.0-mL
    graduated cylinder or use a 5.0 mL pipet. Add
    50 mL dH2O to a 100-mL graduated cylinder, then
    add the 1.3 mL of acetic acid. If you use a 5.0
    mL cylinder for the acetic acid, rinse it out
    with dH2O and add it to the 100-mL cylinder.
    Repeat this process 2 more times. Add dH2O to a
    total volume of 80 mL.
  • Prelab Question 5b
  • Same as 5a

14
Exp 4A Conductivity Of Aqueous Solutions
  • Part 1 Compare conductance of
  • 20 mL dH2O, 20 mL tap water, 20 mL ethanol in 50
    mL beaker
  • Why is there a difference in conductance, if any,
    between distilled water and tap water?
  • Part 2 Measure conductance of hydrochloric acid
    solutions
  • 20 mL 0.10 M HCl
  • 0.050 M HCl dilute 10 mL 0.10 M HCl 10 mL dH2O
  • 0.020 M HCl dilute 10 mL 0.050 M HCl 15 mL
    dH2O
  • How do you make dilutions?
  • ? See prelab assignment 4!

15
Exp 4A Conductivity Of Aqueous Solutions
  • Part 3 Measure conductivity in different
    solutions

Solution Concentration Conductance
1 10 mL 0.10 M HNO3 10 mL dH2O
2 10 mL 0.10 M KOH 10 mL dH2O
3 10 mL 0.10 M KCl 10 mL dH2O
4 10 mL 0.10 M KNO3 10 mL dH2O
5 10 mL 0.10 M Ca(NO3)2 10 mL dH2O
6 10 mL 0.10 M NH3 10 mL dH2O
7 10 mL 0.10 M HC2H3O2 10 mL dH2O
8 10 mL 0.10 M HCl 10 mL 0.10 M KNO3 Calculate new concentrations
9 10 mL 0.10 M HNO3 10 mL 0.10 M KCl Calculate new concentrations
10 10 mL 0.10 M HCl 10 mL 0.10 M KOH Calculate new concentrations
11 10 mL 0.10 M NH3 10 mL 0.10 M HC2H3O2 Calculate new concentrations
16
Exp 4A Conductivity Of Aqueous Solutions
  • Measure conductivity in different solutions

Solution Concentration Conductance
1 10 mL 0.10 M HNO3 10 mL dH2O 0.05
2 10 mL 0.10 M KOH 10 mL dH2O 0.05
3 10 mL 0.10 M KCl 10 mL dH2O 0.05
4 10 mL 0.10 M KNO3 10 mL dH2O 0.05
5 10 mL 0.10 M Ca(NO3)2 10 mL dH2O 0.05
6 10 mL 0.10 M NH3 10 mL dH2O
7 10 mL 0.10 M HC2H3O2 10 mL dH2O
8 10 mL 0.10 M HCl 10 mL 0.10 M KNO3
9 10 mL 0.10 M HNO3 10 mL 0.10 M KCl
10 10 mL 0.10 M HCl 10 mL 0.10 M KOH
11 10 mL 0.10 M NH3 10 mL 0.10 M HC2H3O2
17
Exp 4A Conductivity Of Aqueous Solutions
  • Next week
  • No Formal report.
  • Turn in the following, stapled in this order
  • On paper Type or write 1)a summary of the
    experiment and 2)your conclusions about the
    relationship between conductivity and
    concentration
  • Data and Calculations sheets for Exp 4A
    Conductivity Of Aqueous Solutions
  • Answers to post-lab questions on the lab manual
    sheets
  • Prelab Assignment for Exp 4B Ionic Reactions in
    Aqueous Solutions
  • Avoid getting definitions from Google or
    Wikipedia. Use a textbook or lab manual or a
    science dictionary
  • Read Prelab preparations and protocol
  • Answer Prelab questions 1a-d, 2a-h, 3, 4
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