Transcript No Slide Title
Reactions in Aqueous Solution
Chapter 4
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A
solution
is a homogenous mixture of 2 or more substances The
solute
is(are) the substance(s) present in the smaller amount(s) The
solvent
is the substance present in the larger amount Solution Soft drink (
l
) Air (
g
) Soft Solder (
s
) Solvent H 2 O N 2 Pb Solute Sugar, CO 2 O 2 , Ar, CH 4 Sn 4.1
An
electrolyte
is a substance that, when dissolved in water, results in a solution that can conduct electricity.
A
nonelectrolyte
is a substance that, when dissolved, results in a solution that does not conduct electricity.
nonelectrolyte weak electrolyte strong electrolyte 4.1
Conduct electricity in solution?
Cations (+) and Anions (-) Strong Electrolyte – 100% dissociation NaCl (
s
) H 2 O Na + (
aq
) + Cl (
aq
) Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO (
aq
) + H + (
aq
) 4.1
CH 3
Ionization
of acetic acid COOH CH 3 COO (
aq
) + H + (
aq
) A
reversible
reaction. The reaction can occur in both directions.
Acetic acid is a
weak electrolyte
because its ionization in water is incomplete.
4.1
Hydration
is the process in which an ion is surrounded by water molecules arranged in a specific manner.
d d + H 2 O 4.1
Nonelectrolyte does not conduct electricity?
No cations (+) and anions (-) in solution C 6 H 12 O 6 H 2 O (
s
) C 6 H 12 O 6 (
aq
) 4.1
Precipitation Reactions
Precipitate – insoluble solid that separates from solution precipitate PbI 2 Pb(NO 3 ) 2 (
aq
) + 2NaI (
aq
) PbI 2 (
s
)
molecular equation
+ 2NaNO 3 (
aq
) Pb 2+ + 2NO 3 + 2Na + + 2I PbI 2 (
s
) + 2Na + + 2NO 3 -
ionic equation
Pb 2+ + 2I PbI 2 (
s
)
net ionic equation
Na + and NO 3 are
spectator
ions 4.2
PbI 2
Precipitation of Lead Iodide
Pb 2+ + 2I PbI 2 (
s
) 4.2
Solubility
is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.
4.2
Writing Net Ionic Equations
1. Write the balanced molecular equation.
2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.
3. Cancel the spectator ions on both sides of the ionic equation 4. Check that charges and number of atoms are balanced in the net ionic equation Write the net ionic equation for the reaction of silver nitrate with sodium chloride.
AgNO 3 (
aq
) + NaCl (
aq
) AgCl (
s
) + NaNO 3 (
aq
) Ag + + NO 3 + Na + + Cl Ag + + Cl AgCl (
s
) + Na + + NO 3 AgCl (
s
) 4.2
Chemistry In Action:
An Undesirable Precipitation Reaction Ca 2+ (
aq
) + 2HCO 3
-
(
aq
) CaCO 3 (
s
) + CO 2 (
aq
) + H 2 O (
l
) CO 2 (
aq
) CO 2 (
g
) 4.2
Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid.
Cause color changes in plant dyes.
React with certain metals to produce hydrogen gas.
2HCl (
aq
) + Mg (
s
) MgCl 2 (
aq
) + H 2 (
g
) React with carbonates and bicarbonates to produce carbon dioxide gas 2HCl (
aq
) + CaCO 3 (
s
) CaCl 2 (
aq
) + CO 2 (
g
) + H 2 O (
l
) Aqueous acid solutions conduct electricity.
4.3
Bases
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Cause color changes in plant dyes.
Aqueous base solutions conduct electricity.
4.3
Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH in water 4.3
Hydronium ion
, hydrated proton, H 3 O + 4.3
A
Br ønsted acid
is a proton donor A
Br ønsted base
is a proton acceptor base acid acid base A Br ønsted acid must contain at least one ionizable proton!
4.3
Monoprotic
acids HCl H + + Cl HNO 3 H + + NO 3 CH 3 COOH H + + CH 3 COO Strong electrolyte, strong acid Strong electrolyte, strong acid Weak electrolyte, weak acid
Diprotic
acids H 2 SO 4 H + + HSO 4 HSO 4 H + + SO 4 2-
Triprotic
acids H 3 PO 4 H 2 PO 4 HPO 4 2 H + + H 2 PO 4 H + H + + HPO 4 2 + PO 4 3 Strong electrolyte, strong acid Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid 4.3
Identify each of the following species as a Br ønsted acid, base, or both. (a) HI, (b) CH 3 COO , (c) H 2 PO 4 HI (
aq
) H + (
aq
) + I (
aq
) Br ønsted acid CH 3 COO (
aq
) + H + (
aq
) CH 3 COOH (
aq
) Br ønsted base H 2 PO 4 (
aq
) H + (
aq
) + HPO 4 2 (
aq
) H 2 PO 4 (
aq
) + H + (
aq
) H 3 PO 4 (
aq
) Br ønsted acid Br ønsted base 4.3
Neutralization Reaction
acid + base salt + water HCl (
aq
) + NaOH (
aq
) NaCl (
aq
) + H 2 O H + + Cl + Na + + OH Na + + Cl + H 2 O H + + OH H 2 O 4.3
Oxidation-Reduction Reactions
(electron transfer reactions) 2Mg 2Mg 2+ + 4e -
Oxidation
half-reaction (lose e ) O 2 + 4e 2O 2 2Mg + O 2 + 4e -
Reduction
2Mg + O 2 2Mg 2+ half-reaction (gain e ) + 2O 2MgO 2 + 4e 4.4
4.4
Zn (
s
) + CuSO 4 (
aq
) ZnSO 4 (
aq
) + Cu (
s
) Zn Zn 2+ + 2e Zn is oxidized Zn is the
reducing agent
Cu 2+ + 2e Cu Cu 2+ is reduced Cu 2+ is the
oxidizing agent
Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu (
s
) + 2AgNO 3 (
aq
) Cu(NO 3 ) 2 (
aq
) + 2Ag (
s
) Cu Cu 2+ + 2e Ag + + 1e Ag Ag + is reduced Ag + is the oxidizing agent 4.4
Oxidation number
The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation number of zero.
Na, Be, K, Pb, H 2 , O 2 , P 4 = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion.
Li + , Li = +1 ; Fe 3+ , Fe = +3 ; O 2 , O = -2 3. The oxidation number of oxygen is and O 2 2 it is –1 .
usually
–2 . In H 2 O 2 4.4
4. The oxidation number of hydrogen is +1
except
when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1 .
5. Group IA metals are +1 , IIA metals are +2 always –1 .
and fluorine is 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.
7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O 2 , is ½ .
HCO 3 Oxidation numbers of all the elements in HCO 3 ?
O = -2 H = +1 3x( -2) + 1 + ?
= -1 C = +4 4.4
The oxidation numbers of elements in their compounds 4.4
Oxidation numbers of all the elements in the following ?
NaIO Na = +1 I = +5 3 O = -2 3x( -2 ) + 1 + ?
= 0 IF 7 F = -1 7x( -1 ) + ?
= 0 I = +7 K 2 Cr 2 O 7 O = -2 K = +1 7x( -2 ) + 2x( +1 ) + 2x( ?) = 0 Cr = +6 4.4
Types of Oxidation-Reduction Reactions
Combination Reaction A + B C 0 0 2Al + 3Br 2 +3 -1 2AlBr 3 Decomposition Reaction C A + B +1 +5 -2 2KClO 3 +1 -1 0 2KCl + 3O 2 4.4
Types of Oxidation-Reduction Reactions
Combustion Reaction A + O 2 0 0 S + O 2 B +4 -2 SO 2 0 0 2Mg + O 2 +2 -2 2MgO 4.4
Types of Oxidation-Reduction Reactions
Displacement Reaction A + BC AC + B 0 +1 +2 Sr + 2H 2 O Sr(OH) 2 0 + H 2 +4 TiCl 4 0 0 +2 + 2Mg Ti + 2MgCl 2 Hydrogen Displacement Metal Displacement 0 Cl 2 -1 -1 0 + 2KBr 2KCl + Br 2 Halogen Displacement 4.4
The Activity Series for Metals
Hydrogen Displacement Reaction M + BC AC + B M is metal BC is acid or H 2 O B is H 2 Ca + 2H 2 O Ca(OH) 2 + H 2 Pb + 2H 2 O Pb(OH) 2 + H 2 4.4
The Activity Series for Halogens F
2
> Cl
2
> Br
2
> I
2 Halogen Displacement Reaction 0 Cl 2 -1 -1 0 + 2KBr 2KCl + Br 2 I 2 + 2KBr 2KI + Br 2 4.4
Types of Oxidation-Reduction Reactions
Disproportionation Reaction Element is simultaneously oxidized and reduced.
0 Cl 2 + 2OH +1 ClO -1 + Cl + H 2 O Chlorine Chemistry 4.4
Classify the following reactions.
Ca 2+ + CO 3 2 NH 3 + H + CaCO 3 NH 4 + Precipitation Acid-Base Zn + 2HCl ZnCl 2 + H 2 Redox (H 2 Displacement) Ca + F 2 CaF 2 Redox (Combination) 4.4
Chemistry in Action: Breath Analyzer
+6 3CH 3 CH 2 OH + 2K 2 Cr 2 O 7 + 8H 2 SO 4 +3 3CH 3 COOH + 2Cr 2 (SO 4 ) 3 + 2K 2 SO 4 + 11H 2 O 4.4
Solution Stoichiometry
The
concentration
of a solution is the amount of solute present in a given quantity of solvent or solution.
M
=
molarity
= moles of solute liters of solution What mass of KI is required to make 500. mL of a 2.80
M
KI solution?
M
KI volume of KI solution
M
KI moles KI grams KI 500. mL 1 L x 1000 mL x 2.80 mol KI x 1 L soln 166 g KI 1 mol KI = 232 g KI 4.5
4.5
Dilution
is the procedure for preparing a less concentrated solution from a more concentrated solution.
Dilution Add Solvent Moles of solute before dilution (i)
M
i V i
=
= Moles of solute after dilution (f)
M
f V f 4.5
How would you prepare 60.0 mL of 0.200
M
HNO 3 from a stock solution of 4.00
M
HNO 3 ?
M
i V i =
M
f V f
M
i = 4.00
M
f = 0.200
V f = 0.06 L V i = ? L V i =
M
f V f
M
i = 0.200 x 0.06
4.00
= 0.003 L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution 4.5
Gravimetric Analysis
1. Dissolve unknown substance in water 2. React unknown with known substance to form a precipitate 3. Filter and dry precipitate 4. Weigh precipitate 5. Use chemical formula and mass of precipitate to determine amount of unknown ion 4.6
Titrations
In a
titration
a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point
– the point at which the reaction is complete
Indicator
– substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 4.7
What volume of a 1.420
M
NaOH solution is Required to titrate 25.00 mL of a 4.50
M
solution?
H 2 SO 4 WRITE THE CHEMICAL EQUATION!
M
volume acid acid H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 rx moles acid coef.
moles base
M
base volume base 25.00 mL x 4.50 mol H 2 SO 4 1000 mL soln x 2 mol NaOH 1 mol H 2 SO 4 1000 ml soln x 1.420 mol NaOH = 158 mL 4.7
Chemistry in Action: Metals from the Sea
CaCO 3 (
s
) CaO (
s
) + CO 2 (
g
) CaO (
s
) + H 2 O (
l
) Ca 2+ (
aq
-
aq
) Mg 2+ (
aq
-
aq
) Mg(OH) 2
(s
) Mg(OH) 2 (
s
) + 2HCl (
aq
) MgCl 2 (
aq
) + 2H 2 O (
l
) Mg 2+ + 2e 2Cl MgCl 2 (
aq
) Mg Cl 2 + 2e Mg (
s
) + Cl 2 (
g
)