Precipitation Reactions and Acid-Base Reactions

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Precipitation Reactions andAcid-Base Reactions
Lecture 11

• Chemistry 142 B
• James B. Callis, Instructor
• Autumn Quarter, 2004

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The Solubility of Ionic Compounds in Water
The solubility of Ionic Compounds in water
depends upon the relative strengths of the
electrostatic forces between ions in the ionic
compound and the attractive forces between the
ions and water molecules in the solvent. There
is a tremendous range in the solubility of ionic
compounds in water! The solubility of so called
insoluble compounds may be several orders of
magnitude less than ones that are
called soluble in water, for example
Solubility of NaCl in water at 20oC 365
g/L Solubility of MgCl2 in water at 20oC 542.5
g/L Solubility of AlCl3 in water at 20oC 699
g/L Solubility of PbCl2 in water at 20oC 9.9
g/L Solubility of AgCl in water at 20oC 0.009
g/L Solubility of CuCl in water at 20oC 0.0062
g/L
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Compare with Table 4.1 of Zumdahl
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Precipitation Reactions A solid product is formed
When ever two aqueous solutions are mixed, there
is the possibility of forming an insoluble
compound. Let us look at some examples to see
how we can predict the result of adding two
different solutions together.
Pb(NO3) (aq) 2 NaI(aq)
When we add these two solutions together, the
ions can combine in the way they came into the
solution, or they can exchange partners. In this
case we could have lead nitrate and sodium
iodide, or lead iodide and sodium nitrate formed,
to determine which will happen we must look
at the solubility rules to determine what could
form. Rule 3 in Table 4.1 (lecture) indicates
that lead iodide will be insoluble, so a
precipitate will form! The remainder (sodium
nitrate) is soluble.
Pb(NO3)2 (aq) 2 NaI(aq)
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Precipitation Reactions Will a precipitate form?
If we add a solution containing potassium
chloride to a solution containing ammonium
nitrate, will we get a precipitate?
KCl(aq) NH4NO3(aq) K (aq)
Cl-(aq) NH4(aq) NO3-(aq)
By exchanging cations and anions we see that we
could have potassium chloride and ammonium
nitrate, or potassium nitrate and ammonium
chloride. In looking at the solubility rules,
Table 4.1
If we mix a solution of sodium sulfate with a
solution of barium nitrate, will we get a
precipitate? Table 4.1 shows that barium sulfate
is insoluble - therefore we expect
Na2SO4(aq) Ba(NO3)2(aq)
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Problem 11-1 Predicting Whether a Precipitation
Reaction Occurs (see Table 4.1) Writing
Equations
a) calcium nitrate and sodium sulfate solutions
are added together.
Molecular Equation
Ca(NO3)2(aq) Na2SO4(aq)
Total Ionic Equation
Ca2(aq) 2 NO3-(aq) 2 Na(aq) SO42-(aq)

Net Ionic Equation
Spectator Ions are
b) ammonium sulfate and magnesium chloride are
added together.
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Testing the Reactivity of the Cations Ag, Ba2,
and Fe3 with the Anions Cl-, SO42-, and OH-
Cation NaCl(aq) (Cl-) Na2SO4(aq) (SO42-) NaOH(aq) (OH-)
Ag White ppt (AgCl) No Reaction White ppt that turns brown (AgOH -gt Ag2O)
Ba2 No Reaction White ppt (BaSO4) No Reaction
Fe3 Yellow solution, but no solid No Reaction Reddish Brn ppt Fe(OH)3
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A Scheme for Selective Precipitation
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Acid - Base Reactions Neutralization Rxns.
An Acid is a substance that produces H (H3O)
ions when dissolved in water. A Base is a
substance that produces OH- ions when dissolved
in water. Acids and Bases are electrolytes, and
their strength is categorized in terms of their
degree of dissociation in water to make hydronium
or hydroxide ions, resp. Strong acids and bases
dissociate completely, and are strong
electrolytes. Weak acids and bases dissociate
only to a small extent (ltlt100) and are weak
electrolytes.
The generalized reaction between an Acid and a
Base is
HX(aq) MOH(aq)
MX(aq) H2O(l)
Acid Base
Salt(aq) Water
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Selected Acids and Bases
Acids
Bases
Strong (100 of H)
Strong (100 of OH-) Hydrochloric, HCl
Sodium hydroxide, NaOH
Hydrobromic, HBr
Potassium hydroxide, KOH Hydroiodoic, HI
Calcium hydroxide,
Ca(OH)2 Nitric acid, HNO3
Strontium hydroxide, Sr(OH)2 Sulfuric
acid, H2SO4 Barium
hydroxide, Ba(OH)2 Perchloric acid,
HClO4 Weak (low of H)
Weak (low yield of OH-) Hydrofluoric, HF
Ammonia, NH3
Phosphoric acid, H3PO4 Acetic acid, CH3COOH
(or HC2H3O2)
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Problem 11-2 Writing Balanced Equations for
Neutralization Reactions - I
Problem Write balanced chemical reactions
(molecular, total ionic, and
net ionic) for the following Chemical reactions
a) calcium hydroxide(aq) and hydroiodic
acid(aq) b) lithium hydroxide(aq) and
nitric acid(aq) c) barium
hydroxide(aq) and sulfuric acid(aq) Plan These
are all strong acids and bases, therefore they
will make water and the corresponding
salts. Solution
a) Ca(OH)2(aq) 2HI(aq)
CaI2(aq) 2H2O(l) Ca2(aq) 2 OH -(aq)
2 H(aq) 2 I-(aq)

Ca2(aq)
2 I-(aq) 2 H2O(l) 2 OH -(aq) 2
H(aq) 2 H2O(l)
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Problem 11-2 Writing Balanced Equations for
Neutralization Reactions - II
b) LiOH(aq) HNO3(aq)
c) Ba(OH)2(aq) H2SO4(aq)

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Finding the Concentration of Acid from an
Acid - Base Titration
Volume (L) of base (difference in buret readings)
needed to titrate
M (mol/L) of base
Moles of base needed to titrate
molar ratio
Moles of acid which were titrated
volume (L) of acid
M (mol/L) of original acid soln.
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Potassium Hydrogenphthalate KHC8H4O4
O
C
K
O
O
C
H
O
KHP a common acid used to titrate bases
(M 204.2 g/mol)
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Problem 11-3 Finding the Concentration of Base
from an Acid - Base Titration - I
Problem A titration is performed between sodium
hydroxide and KHP (204.2 g/mol) to standardize
the base solution, by placing 50.00 mg of solid
potassium hydrogenphthalate in a flask with a
few drops of an indicator. A buret is filled with
the base, and the initial buret reading is 0.55
mL at the end of the titration the buret
reading is 33.87 mL. What is the concentration
of the base? Plan Use the molar mass of KHP to
calculate the number of moles of the acid, from
the balanced chemical equation, the reaction is
equal molar, so we know the moles of base, and
from the difference in the buret readings, we
can calculate the molarity of the base. Solution
The rxn is KHP(aq) OH-(aq) --gt KP-(aq)
H2O(l) or
HKC8H4O4(aq) OH-(aq)
KC8H4O4-(aq) H2O(l)
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Problem 11-3 Finding the Concentration of Base
from an Acid - Base Titration - II
50.00 mg KHP

moles KHP
Volume of base Final buret reading - Initial
buret reading
one mole of H one mole of OH- therefore
____________ moles of KHP will titrate
_____________ moles of NaOH.
moles L
molarity of NaOH

molarity of base
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Answers to Problems in Lecture 11

• (a) Net ionic equation Ca2(aq) SO42-(aq)
  CaSO4 (s) (b) No reaction occurs.
• (b) Total ionic reaction Li(aq) OH-(aq)
  H(aq) NO3-(aq) Li (aq) NO3-(aq)
  H2O(l)
• (c) Total ionic reaction Ba2(aq) 2 OH-(aq)
  2 H(aq) SO42-(aq) BaSO4(s) 2 H2O(l)
• 3. 0.07349 M