Transcript Chapter 4 PPT - Richsingiser.com

Daniel L. Reger
Scott R. Goode
David W. Ball
http://academic.cengage.com/chemistry/reger
Chapter 4
Chemical Reactions in
Solution
Solutions
• Solvent: compound that has the same
physical state as the solution - frequently
a liquid.
• Solute: substance being dissolved.
• Aqueous solution: water is the solvent.
• Strong electrolyte: compound that
separates completely into ions in water.
• Weak electrolyte: molecule that only
partially ionizes when dissolved in water.
Solubility Experiments
NiCl2 Hg2Cl2 CoCl2 Fe(NO3)3 NaNO3 Cr(NO3)3
Solubility Rules
Solubility of Ionic Compounds
Predict the solubility of (a) (NH4)2SO4 and
(b) PbCl2.
Precipitation Reactions
• A precipitation reaction involves the
formation of an insoluble product or
products from the reaction of soluble
reactants.
• Example: Mixing AgNO3 and LiCl, both
of which are soluble, produces insoluble
AgCl.
AgNO3(aq) + LiCl(aq) AgCl(s) + LiNO3(aq)
Precipitation Reactions
Precipitation Reactions
What insoluble compound, if any, will
form when solutions of Pb(NO3)2 and
Na2SO4 are mixed? Write the
chemical equation.
Precipitation Reactions
Build a table of the reactants and possible
products; label each as soluble or insoluble.
Cations
Anions
Na+
Pb2+
NO3- soluble NaNO3 soluble Pb(NO3)2
SO42- soluble Na2SO4 insoluble PbSO4
PbSO4 is insoluble and the equation is:
Pb(NO3)2(aq) + Na2SO4(aq) 
PbSO4(s) + 2NaNO3(aq)
Complete Ionic Equation
• Complete ionic equation shows all
strong electrolytes as ions in solution
Overall equation:
Pb(NO3)2(aq) + Na2SO4(aq) 
PbSO4(s) + 2NaNO3(aq)
Complete ionic equation:
Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + SO42-(aq)
 PbSO4(s) + 2Na+(aq) + 2NO3-(aq)
Net Ionic Equation
• Net ionic equation shows only those
species in the solution that actually
undergo a chemical change.
Overall equation:
Pb(NO3)2(aq) + Na2SO4(aq) 
PbSO4(s) + 2NaNO3(aq)
Net ionic equation:
Pb2+(aq) + SO42-(aq)  PbSO4(s)
• Spectator ions are those that do not
participate in the chemical reaction.
Test Your Skill
• Write the net ionic equation for the
reaction of HCl(aq) and KOH(aq).
Molarity
• Molarity (M) is the number of moles
of solute in one liter of solution.
Molarity =
moles of solute
liters of solution
Example: Molarity of Solution
• What is the molar concentration of NaF
in a solution prepared by dissolving
2.51 g of NaF in enough water to form
200 mL of solution?
Molarity of Ions
• One mole of K2SO4 dissolves in water to
form two moles of K+ ions and one mole
of SO42- ions.
K2SO4(s) 
 2K+(aq) + SO42-(aq)
H2O
Dilution
• Solutions of lower concentration can
be prepared by dilution of more
concentrated solutions of known
molarity.
Volume (L)
of dilute
solution
Molarity
of dilute
solution
Molarity
Moles of
solute
of
concentrated
solution
Volume (L)
of concentrated solution
Dilution
• In a dilution problem: moles of solute in
dilute solution = moles of solute in the
concentrated solution
molarity(conc) x volume(conc) =
molarity(dil) x volume(dil)
• Use this formula only for dilution
problems, not for problems involving
equations.
Example: Dilution
• Calculate the volume of 6.00 M H2SO4
that is needed to prepare 2.00 L of a
0.200 M solution of H2SO4.
Test Your Skills
(a) Calculate the volume of 4.00 M K2SO4
that is needed to prepare 600 mL of a
0.0200 M solution of K2SO4.
(b) Calculate the molar concentration of
K+ ions in the 0.0200 M solution.
Solution Stoichiometry Calculations
• In stoichiometric calculations, molarity is used
to calculate moles from volume of solution
analogous to using molar mass from mass of
a solid.
Solution Stoichiometry Calculations
• Calculate the mass of lead(II) sulfate
formed in the reaction of 145 mL of
0.123 M lead(II) nitrate and excess
sodium sulfate.
Pb(NO3)2(aq) + Na2SO4(aq) 
PbSO4(s) + 2NaNO3(aq)
Test Your Skills
• Calculate the mass of magnesium
hydroxide need to react completely with
356 mL of 6.92 M H2SO4.
Titrations
• In a titration, the concentration and
volume of a solution of known
concentration is used to determine the
concentration of an unknown solution.
• Equivalence point: the point in a
titration where stoichiometrically
equivalent amounts of the two reactants
have been added.
Titrations
• An indicator is a compound that
changes color as an acidic solution
becomes basic or basic solution
becomes acidic.
• The indicator changes color at the end
point – the end point of the indicator
should match the equivalence point.
Titration with Phenolphthalein Indicator
• Left: acidic solution with indicator added
• Center: end point - very slight pink color
• Right: pink color after excess base added
Example: Titrations
• Calculate the molarity of an HCl solution
if 26.4 mL of the solution neutralizes 30.0
mL of 0.120 M Ba(OH)2.
Test Your Skills
• Calculate the molarity of an NaOH
solution if 33.4 mL of the solution is
neutralized by 16.0 mL of a 0.220 M
solution of H2 SO4.
Gravimetric Analysis
• Calculate the molarity of Cl- ions in a 250
mL solution if addition of excess silver
nitrate yielded 1.34 g of silver chloride.
Test Your Skill
• What mass of lead(II) chloride forms in
the reaction of 24.3 mL of 1.34 M lead(II)
nitrate and 38.1 mL of 1.22 M sodium
chloride?