Download Chapter 4 Reactions in Aqueous Solution 4.1 Aqueous Solutions

4.1 Aqueous Solutions
Chapter 4
Reactions in Aqueous
Solution
Electrolytes
• Electrolyte – substance that gives an
aqueous solution that conducts electricity
• Nonelectrolyte – substance that does
not produce conducting solution when
dissolved in water
• Mobile ions conduct electricity
– electrolytes break apart (dissociate or ionize)
into ions when dissolved in water
• Solution – homogeneous mixture of
2 or more substances
• Solute – the substance present in a
smaller amount (usually solid in
Chap. 4)
• Solvent – the substance present in
the larger amount (usually liquid in
Chap. 4)
• Aqueous solution – solvent is water
Electrical Conduction by
Electrolyte Solution
e-
e-
M+
X-
Cations (positive
ions) migrate to
cathode (negative
electrode)
Anions (n egative
ions) migrate to
anode (positive
electrode)
Electrochemistry at electrodes (Chap. 19)
Strong Electrolytes
• Dissociate completely into ions in H2O
• Salts like NaCl, KNO3, MgSO4
KNO3(s)
H 2O
K+ (aq) + NO3 - (aq)
•Strong acids like HCl, HNO3
HCl(g)
H2O
H + (aq) + Cl-(aq )
Weak Electrolytes
• Ionize partially into ions in H2O
• Weak acids (HF)
HF(aq)
H2 O
H + (aq) + F-(aq)
•Weak bases (NH3)
H2O
NH 3(aq)
NH 4+ (aq) + OH - (aq)
•Double arrow indicates that the reaction
is reversible, i.e., runs both ways
• Ionic compounds are strong electrolytes
• When KCl dissolves:
KCl(s) d K+(aq) + Cl-(aq)
• Ions are surrounded by polar molecules
– in general, solvation
– in water, hydration
+ end of dipole
points to anion
- end of dipole
points to cation
H2O
NH3(aq)
NH 4+(aq) + OH-(aq)
• Reversible reaction - can occur in both
directions
• Reactants form products as soon as
reaction begins
• Once products are formed, they in turn
react to re-form reactants
• Chemical Equilibrium
- when reactants form products as fast as
products form reactants, no further net
change in concentrations
Reversible Ionization
of a Weak Base
Principal Types of Reactions
• Precipitation Reactions (4.2)
• Acid - Base Reactions (4.3)
• Oxidation-Reduction Reactions
(4.4)
•Both NH3 and NH4 + exist in rapid equilibrium
= 1% is NH 4+ (Chap. 15)
4.2 Precipitation Reactions
• Reactions that result in the
formation of precipitate,
insoluble solid that separates
from the solution
• Often involve ionic compounds
AgNO3(aq) + 2NaBr(aq) d AgBr(s) + NaNO 3(aq)
Precipitate
Solubility
• Maximum amount of solute that will
dissolve in a given quantity of
solvent at a specific temperature
• Table 4.2 classifies substances in
one of 3 categories
–insoluble
–slightly soluble
–soluble
Simple solubility rules
Soluble salts
Exceptions
Alkal i meta l (Li + toC s+),
ammonium ( NH4+)
Using the solubility rules, classify
as soluble or insoluble.
• Ag2SO4
N itrate ( NO 3-),
bicar bon ate ( HCO 3-),
sulf ate ( SO 42- ), etc.
Sulfates of Ag +, C a2+, Sr2+,
Ba 2+, Pb 2+
H alide (Cl- , Br- , I -)
Halides of Ag +, Hg 22+, Pb 2+
insoluble
• Li2S
soluble
• Pb(NO3)2
Insoluble salts
Exceptions
C arbonat e (CO32- ),
Alkal i metal (Li + to Cs+),
phosphat e (PO 43- ), sulfide ammoni um ( NH4+)
(S 2-), etc.
H ydroxide (OH- )
soluble
• AgCl
insoluble
Alkal i metal (Li + to Cs+),
Ba 2+
Ionic equations
• “Molecular” equation: write full formula
of each species
AgNO3(aq) + NaBr(aq) d AgBr(s) + NaNO3(aq)
•Ionic equation: write dissolved species
as free ions
Net ionic equations
• Spectator ions are not involved in the
overall reaction: Na+(aq) and NO3-(aq)
Ag+(aq) + NO3-(aq) + Na+(aq) + Br-(aq)
d AgBr(s) + Na+(aq) + NO3-(aq)
•Net ionic equation: write only
species that take part in reaction
Ag+(aq) + NO3-(aq) + Na+(aq) + Br-(aq)
d AgBr(s) + Na+(aq) + NO3-(aq)
Ag+(aq) + Br-(aq) d AgBr(s)
Predicting and writing
precipitation reactions
PbCl2 + Na2SO4
1 Write “molecular” equation
2 Dissociate electrolytes d ionic equation
3 Use solubility rules to predict precipitate
4 Cancel spectator ions d net ionic
equation
1. Write “molecular” equation, using solubility
rules to predict precipitate(s)
PbCl2(aq) + Na2SO 4(aq) d PbSO4(s) + 2NaCl (aq)
2. Dissociate electrolytes d ionic equation
Pb2 +(aq) + 2Cl-(aq) + 2Na +(aq) + SO42 -(aq) d
PbSO4(s) + 2Na +(aq) + 2Cl-(aq)
3. Cancel spectator ions d net ionic equation
Pb2 +(aq) + SO4 2-(aq) d PbSO4(s)
More examples (on your own)
• K2SO4 + AgNO 3
4.3 Acid-Base Reactions
• Acid-base reactions are very important
in chemistry
• Long history
• (NH4)2S + FeSO 4
– many models for acids and bases
– details in Chaps. 15 and 16
Empirical criteria
Empirical criteria
• Acid
– sour taste
– corrodes metals, often producing H2(g)
H2SO4(aq) + Fe(s) d FeSO4(aq) + H2(g)
– turns litmus (plant dye) red
– gives carbon dioxide (g) w/ carbonates
2HCl(aq) + CaCO 3(s)
H2O(l) + CO2(g)
• Base (alkali)
– bitter taste (NaHCO3 = baking soda)
– feels slippery (soap)
– turns litmus (plant dye) blue
– electrolyte
d CaCl2(aq) +
– electrolyte
Arrhenius concept (1887)
• Acid = Proton (H+) donor in H2O
HCl(aq) d H+(aq) + Cl-(aq)
H2SO4(aq) d H+(aq) + HSO4-(aq)
• Base = Hydroxide (OH-) donor in H2O
NaOH(aq) d Na +(aq) + OH-(aq)
MgO(s) + H2O(l) d Mg2+(aq) + 2 OH -(aq)
• Neutralization
H+(aq) + OH-(aq) d H2O(l)
Acid + Base d Water
Brønsted-Lowry concept (1923)
• Brønsted Acid = Proton donor
HCl(g) + HF(l) qeH2F+ (HF) + Cl-(HF)
• Brønsted Base = Proton acceptor
NH3(aq) + H2O(l) qeNH4+(aq) + OH-(aq)
• Neutralization
NH3 + HCl qeNH4+ + ClBase1 + Acid2 qeAcid1 + Base2
• Not necessarily in water
Dissociation of HCl in water
Acid1
Base 2
Acid2
Base 1
H3O+ = Hydronium (oxonium) ion
+
H (aq) + H2O(l)
+
H 3O (aq)
• H+ associates with
several H2O molecules
• e.g. H7 O3+ or
[H(H2O) n]+ (n = 3)
• Often abbreviated
H+(aq)
H3O+ = Hydronium ion
Weak base in water
NH3(aq) + H2O(l) qeNH4+(aq) + OH-(aq)
Base
Acid
• Very little NH3 (=1%) is ionized
– typical weak base in water
• H2O acts as a Brønsted acid in this reaction
• Some substances can act as either acid or
base depending on reaction
Acid - Base Neutralization Reaction
• Neutralization reaction - reaction
between acid and base to produce a
salt and water
• Salt - ionic compound w/ cation
besides H+
• acid + base d salt + water
HBr(aq) + KOH(aq) d KBr(aq) + H2O(l)
net: H+(aq) + OH-(aq) d H2O(l)
Polyprotic acids
•Diprotic – supply 2 H+ in 2 steps
H2S(aq) qe H+(aq) + HS-(aq)
SH-(aq) qe H+(aq) + S2-(aq)
•Triprotic – supply 3 H+ in 3 steps
H3 PO4(aq) qe H+(aq) + H2 PO 4-(aq)
H2 PO4-(aq) qe H+(aq) + HPO42 -(aq)
HPO42-(aq) qe H+(aq) + PO43 -(aq)
•Each step is successively weaker
Examples of Neutralization Reactions
• HF(aq) + NaOH(aq) d NaF(aq) + H2 O(l)
• 2HNO3(aq) + Ba(OH) 2(aq) d Ba(NO3) 2(aq) + 2H2O(l)
• H2SO4(aq) + 2LiOH(aq) d Li2SO4(aq) + 2H2O(l)
4.4 Oxidation-Reduction Reactions
Redox mnemonic
• “Redox reactions”
– electron transfer reactions
• 2K(s) + Cl2(g) d 2KCl(s)
• K loses an electron to become K +
2K d 2K+ + 2e • Cl2 gains 2 electrons to become 2ClCl2 + 2e- d 2Cl • Each step is a half-reaction
• LEO
Loss of Electrons =
Oxidation
GER
• GER
Gain of Electrons =
Reduction
– explicitly shows electron transfer
Redox Definitions
• 2K(s) + Cl2(g) d 2KCl(s)
• K is oxidized to K +
– oxidation half-reaction
2K d 2K+ + 2e – K is the reducing agent (reductant)
-
• Cl2 is reduced to 2Cl
– reduction half-reaction
– Cl2 + 2e- d 2Cl – Cl2 is the oxidizing agent (oxidant)
Short Activity Series
•
•
•
•
•
•
•
•
•
Li
most reactive
K
Ba
Na
Zn
H
Cu
Hg
Au
least
reactive
LEO
The Activity Series for Metals
Displacement Reactions
M + n H2O
M + n H+
M(OH)n + n/2 H2
Mn+ + n/2 H2
Examples:
Ca + 2H2O
Ca(OH)2 + H2
Pb + 2H2O
Pb(OH) 2 + H2
Oxidation Number
• Reactions of molecular, not just ionic,
compounds are redox reactions
P4(s) + 5O2(g) d P4 O10 (s)
– There are no ionic charges shown, but it is a
redox reaction
• Oxidation number (state) – charge an
atom would have if e- were transferred
completely
P(+5) and O(-2)
– remember: Σ Ox# = charge on ion (molecule)
• Chap. 9 explains direction of transfer
Ox# Rules
Calculate Ox#s
1. In free elements each atom has Ox # = 0
2. For monatomic ions, Ox # = charge
3. Ox # of oxygen is usually -2
4. Ox # of hydrogen is usually +1
except metal hydrides, Ox # (H) = -1
5. Ox # of fluorine is always -1
6. Σ Ox # = charge on molecule or ion
7. Ox # can be fractional
Types of Redox Reactions
• Combination Reaction
– substances combine to form 1 product
– Fe(s) + O2(g) d Fe2O3(s)
• Decomposition Reaction
– breakdown of compound into components
– NH4 NO2(s) d N2(g) + H2O(l)
H3PO4
SO2
SF6
CO
Re2Cl82SH3+
Displacement Reactions
• Hydrogen displacement
Mg(s) + H2 O(l) d Mg(OH) 2(aq) + H2(g)
• Metal displacement
– more active metal displaces less active
metal (Chap. 19)
Al(s) + Fe 2O3(s) d Fe(l) + Al2O3(s)
• Halogen displacement
– reactivity F2 > Cl2 > Br 2 > I2
Cl2(g) + CaBr 2(s) d Br 2(l) + CaCl2(s)
Disproportionation Reaction
• Element in one oxidation state is
simultaneously oxidized and reduced
2CuCl(s) d Cu(s) + CuCl2(s)
• Cu(+1) goes to Cu(0) and Cu(+2)
4.5 Concentration of Solutions
• Amount of solute present in a given
quantity of solvent or solution
• Molarity (M) – the number of moles of
solute in 1 liter of solution
M = molarity =
mol solute
L solution
Conc. NaCl = 0.34 M
Calculating molarity
What is the molarity of a solution made by
dissolving 5.0 g of NaCl in enough
water to make 250.0mL of solution?
Conc. = 0.34 M
• What is the conc. of Na + in solution?
[Na+] = 0.34 M
• What is the conc. of Cl- in solution?
[Cl-] = 0.34 M
Consider a 0.50 M BaCl2 solution.
• [Ba2+] = ?
0.50 M
• [Cl-] = ?
1.0 M
• Molarity (mol/L) can be used to convert
– mol solute to L solution
– L solution to mol solute
How many moles of HCl are in 500 mL of
0.30 M HCl?
0.50 L x (0.30 mol/ L) = 0.15 mol
•
a
b
c
Prepare 250 mL of 2.0 M KOH solution
• Need (2.0 mol KOH/ L)(0.25 L) = 0.50
mol KOH
• (0.50 mol KOH)(56 g /mol) = 28 g KOH
• Dissolve 28 g KOH in enough water to
make 250 mL of solution
Convert molarity to mass of solute needed
Transfer solute to volumetric flask
Add enough solvent to dissolve solute
Dilute to mark
Dilution
• Start with a concentrated stock solution
• Add more solvent to produced solution
of lower concentration
• M = mol/ L
• mol contained = M•V
Dilution Formula
Add solvent
How many mL H2O are required to dilute
205 mL of 1.15 M HCl solution to 0.81 M?
MiVi = MfVf
Vf = MiVi / Mf
Vf = (1.15 M)(0.205 L)/(0.81 M) = 0.291 L
• Same # mol solute in each beaker
• mol = M•V in each beaker
VH2O = Vtotal - Vin
VH2O = 0.291 L - 0.205 L = 0.086 L = 86 mL
• Mi Vi = M fVf
i = initial, f = final
Quantitative Analysis
•
•
Determination of the amount or
concentration of a substance in a sample
Two Quantitative Analysis Techniques
1. Gravimetric Analysis – based on the
measurement of mass (Section 4.6)
2. Titration - solution of known concentration is
added to solution of unknown concentration
until chemical reaction is complete (endpoint)
How many grams of NaCl are required to
precipitate all the Ag+ ions from 125 mL
of 0.0081 M AgNO3 solution?
NaCl (aq) + AgNO 3 (aq) d AgCl (s) +
NaNO3 (aq)
[work on blackboard]
0.059 g
a. Acid-Base Titrations (Section 4.7)
b. Redox Titrations (Section 4.8)
4.7 Acid-Base Titration
• Equivalence point – point at which acid
is completely neutralized by added base
(or vice versa)
• Equivalence point (or end point) is
signaled by an indicator – substance
that is different color in acid and base
22.5 mL of 0.383 M H2SO4 are required to
neutralize 20.0 mL of a KOH solution.
Calculate the molarity of the KOH solution.
H2SO4 + 2 KOH d K2SO4 + 2 H2O
M KOH = (0.0225 L H2SO4)(0.383 mol
H2SO4/L H2SO4)(2 mol KOH/ 1 mol
H2SO4)/ (0.0200 mL KOH) =
0.862 M KOH
4.8 Redox titrations
• Read on your own
• Won’t be on exam