Chapter 5 Chemical Reactions

Transcription

Aqueous Solubility of Ionic Compounds
John W. Moore
Conrad L. Stanitski
Peter C. Jurs
• Not all compounds dissolve in water.
• Solubility varies from compound to compound.
http://academic.cengage.com/chemistry/moore
Chapter 5
Chemical Reactions
Soluble ionic
compounds
dissociate.
Stephen C. Foster • Mississippi State University
Soluble Ionic Compounds (Table 5.1)
Ions are
solvated
Insoluble Ionic Compounds (Table 5.1)
ALL
ALL
• ammonium and group 1A (Na+, K+… and NH4+ salts)
• phosphates (PO43-)
• nitrates (NO3-)
Molecular compounds
usually stay associated.
• carbonates (CO32-)
• acetates (CH3COO-)
• oxalates (C2O42-)
• chlorates (ClO3-)
• oxides (O2-)
• perchlorates (ClO4-)
• sulfides
MOST
Except grp 1A, NH4+
(S2-)
 MgS, CaS & BaS are slightly soluble.
• chlorides, bromides and iodides
• hydroxides (OH-)
 (not: AgX, Hg2X2, and PbX2 ; X = Cl-, Br-, I-).
 Sr, Ba & Ca are slightly soluble
• sulfates (SO42-)
 (not: CaSO4, SrSO4, BaSO4, and PbSO4)
Aqueous Solubility of Ionic Compounds
(a) Nitrates
(soluble)
Three types of exchange reactions:
• Form a precipitate
 an insoluble ionic compound
 AgNO3(aq) + KCl(aq) → KNO3(aq) + AgCl
AgCl(s)
(b) Hydroxides
(insoluble)
AgNO3 Cu(NO3)2
Exchange Reactions: Precipitation
Cu(OH)2 AgOH
• Form a molecular compound
 Often water
 HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(ℓ)
(c) Sulfides
• Form a molecular gas
CdS
Sb2S3
PbS
(NH4)2S
soluble
 2 HCl(aq) + Na2S(aq) → 2 NaCl(aq) + H2S(g)
insoluble
1
Precipitation Reactions
Precipitation Reactions
When ionic solutions mix, a precipitate may form:
AgNO3(aq) + NaCl(aq)
AgCl(s) + NaNO3(aq)
KNO3(aq) + NaCl(aq)
No reaction
Na2SO4(aq) and Ba(NO3)2(aq) are
mixed. Will they react?
Products?
Na+
Ba2+
SO42NO3-
NaNO3 and BaSO4 . Insoluble?
A reaction occurs if a product is insoluble.
The solubility rules help predict reactions.
Yes.
Yes BaSO4 – A reaction occurs:
Na2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2 NaNO3(aq)
Net Ionic Equations
Net Ionic Equations
Soluble ionic compounds fully dissociate:
AgNO3(aq)
Ag+(aq) + NO3-(aq)
+
KCl(aq)
K (aq) + Cl-(aq)
On mixing:
Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) →
AgCl(s) + K+(aq) + NO3-(aq)
K+ and NO3- appear on both sides
 NO3- and K+ are spectator ions.
 They are not directly involved in the reaction.
Net ionic equation:
Ag+(aq)
+
Cl-(aq)
General rules:
1. Write a balanced equation.
2. Solubility?
3. Dissociate soluble compounds.
4. Write the complete ionic equation.
5. Cancel ions appearing on both sides (spectators).
6. Check the charges are balanced.
→ AgCl(s)
Net Ionic Equations
Aqueous solutions of (NH4)2S and Hg(NO3)2 react to
give HgS and NH4NO3.
(a) Write the overall balanced equation
(b) Name each compound
(c) Write the net ionic equation
(d) What type of reaction is this?
(NH4)2S(aq) + Hg(NO3)2(aq) → HgS + NH4NO3
not balanced
(NH4)2S(aq) + Hg(NO3)2(aq) → HgS + 2 NH4NO3
Net Ionic Equations
(NH4)2S(aq) + Hg(NO3)2(aq) → HgS
+ 2 NH4NO3
Product Solubility
HgS – insoluble
(All sulfides – except group 1A, 2A and NH4+ salts)
NH4NO3 – soluble
(All ammonium salts; all nitrates)
(b) name each compound
ammonium sulfide
mercury(II) nitrate
mercury(II) sulfide
ammonium nitrate
balanced
2
Net Ionic Equations
Acids
Increase the concentration of H+ ions in water.
(c) Write a net ionic equation
(NH4)2S(aq) + Hg(NO3)2(aq) → HgS(s) + 2 NH4NO3(aq)
2 NH4+(aq) + S2-(aq) + Hg2+(aq) + 2 NO3-(aq)
→ HgS(s) + 2 NH4+(aq) + 2 NO3-(aq)
• Protons (H+) always combine with
water to form H3O+ (hydronium
hydronium ion).
ion
• Sour tasting.
• Change the color of pigments (indicators)
§ Litmus, phenolphthalein…
S2-(aq) + Hg2+(aq) → HgS(s)
Strong acids dissociate (>99%) in water (strong
electrolytes).
(d) Type of reaction?
Exchange reaction
Precipitation reaction
Weak acids partially dissociate (weak electrolytes).
Acids
Bases
Hydrochloric acid (HCl) ≈ 100% dissociated in H2O
• Very few HCl molecules exist in solution.
HCl(aq) + H2O(ℓ)
H3O+(aq) + Cl-(aq)
• HCl is a strong acid.
Increase the concentration of OH- (hydroxide ion) in
water. Bases:
Acetic acid (CH3COOH) is 5% dissociated in H2O.
• 95% intact.
CH3COOH(aq) + H2O(ℓ)
H3O+(aq) + CH3COO-(aq)
• Acetic acid is a weak acid.
• Counteract an acid (neutralize an acid).
• Change an indicator’s color (phenolphthalein…).
• Have a bitter taste.
• Feel slippery.
Bases can be “strong” or “weak”.
H2O
NaOH(s)
Na+(aq) + OH-(aq)
NH3(aq) + H2O(ℓ)
Common Acids and Bases
Strong Acid
HCl
HBr
HI
HNO3
H2SO4
HClO4
Hydrochloric acid
Hydrobromic acid
Hydroiodic acid
Nitric acid
Sulfuric acid
Perchloric acid
Strong Base
LiOH
NaOH
KOH
Ca(OH)2
Ba(OH)2
Sr(OH)2
Weak Acid
HF
H3PO4
CH3COOH
H2CO3
HCN
HCOOH
C6H5COOH
Hydrofluoric acid
Phosphoric acid
Acetic acid
Carbonic acid
Hydrocyanic acid
Formic acid
Benzoic acid
Lithium hydroxide
Sodium hydroxide
Potassium hydroxide
Calcium hydroxide
Barium hydroxide
Strontium hydroxide
NH4
+(aq)
+
OH-(aq)
strong
weak
Neutralization Reactions
acid + base  salt + water.
salt = ionic compound made from an acid anion and
base cation.
HX(aq) + MOH(aq)
MX
MX(aq) + H2O(ℓ)
HBr(aq) + KOH(aq)
KBr(aq) + H2O(ℓ)
KBr
Weak Base
NH3
CH3NH2
Ammonia
Methylamine
H3PO4(aq) + 3 NaOH(aq)
Na3PO4(aq) + 3 H2O(ℓ)
Neutralizations are exchange reactions.
3
Net Ionic Equations for Acid-Base Reactions
Strong Acid + Strong Base
Overall:
HX + MOH → MX + H2O
Full ionic:
ionic
H+(aq) + X-(aq) + M+(aq) + OH-(aq) →
M+(aq) + X-(aq) + H2O(ℓ)
net ionic:
H+(aq) + OH-(aq) → H2O(ℓ)
Net Ionic Equations for Acid-Base Reactions
Weak Acid + Strong Base
Similar:
HA + MOH → MA + H2O
but the weak-acid remains undissociated:
Full ionic
HA(aq) + M+(aq) + OH-(aq) →
M+(aq) + A-(aq) + H2O(ℓ)
Net ionic:
Gas-Forming Exchange Reactions
Metal carbonate/acid exchange
Carbonic acid
(unstable)
Overall:
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2CO3(aq)
Tums®
HA(aq) + OH-(aq) → A-(aq) + H2O(ℓ)
Metal hydrogen carbonate/acid exchange
Overall:
NaHCO3(aq) + HCl(aq)→ NaCl(aq) + H2O(ℓ) + CO2(g)
H2CO3(aq) → H2O(ℓ) + CO2(g)
Alka-Seltzer ®
often written:
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(ℓ) + CO2(g)
Net ionic:
ionic
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2O(ℓ) + CO2(g)
Metal sulfite/acid exchange
Sulfurous acid
(unstable)
Similar. Overall:
Na2SO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2SO3(aq)
H2SO3(aq) → H2O(ℓ) + SO2(g)
or
Na2SO3(aq) + 2 HCl(aq) →
2 NaCl(aq) + H2O(ℓ) +SO2(g)
Net ionic
HCO3-(aq) + H+(aq) → H2O(ℓ) + CO2(g)
Metal sulfide/acid exchange
Overall:
Na2S(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2S(g)
net ionic: S2-(aq) + 2 H+(aq) → H2S(g)
net ionic: SO32-(aq) + 2 H+(aq) → H2O(ℓ) + SO2(g)
4
Oxidation-Reduction Reactions
Oxidation
Originally: add oxygen.
2 Cu(s) + O2(g)
2 CuO(s)
2 CO(g) + O2(g)
2 CO2(g)
Cu and CO are oxidized. O2 is the oxidizing agent for both.
Oxidation-Reduction Reactions
In all cases:
• If something is oxidized, something must be reduced.
• Ox
Oxidation-red
reduction = redox.
redox
• Redox reactions move e-.
+2 e-
2 Ag+(aq) + Cu(s)
Reduction
2 Ag(s) + Cu2+(aq)
-2 e-
Originally: reduce ore to metal; reverse of oxidation.
CuO(s) + H2(g)
Cu(s) + H2O
SnO2(s) + 2 C(s)
Sn(s) + 2 CO(g)
CuO and SnO2 are reduced. H2 and C are the reducing agents
(note: H2 is oxidized, so CuO is an oxidizing agent)
Redox Reactions and Electron Transfer
Here:
• Cu changes to Cu2+.
• Cu loses 2 e-; each Ag+ gains one e• Ag+ is reduced (ore turned to metal)
• Gain of e- = reduction
(so, loss of e- = oxidation)
Redox Reactions and Electron Transfer
Loss of electrons is oxidation
Gain of electrons is reduction
Leo says ger
Oxidation is loss
Reduction is gain
e-
M
X
M loses electron(s)
X gains electron(s)
M is oxidized
X is reduced
M is a reducing agent
X is an oxidizing agent
Oil rig
Common Oxidizing and Reducing Agents
Oxidizing Agent
O2 (oxygen)
H2O2 (hydrogen peroxide)
F2, Cl2, Br2, I2 (halogen)
HNO3 (nitric acid)
Cr2O7- (dichromate ion)
MnO4- (permanganate ion)
Reaction Product
O2- (oxide ion)
H2O(ℓ)
F-, Cl-, Br-, I- (halide ion)
nitrogen oxides (NO, NO2..)
Cr3+ (chromium(III) ion)
Mn2+ (manganese(II) ion)
Reducing Agent
H2 (hydrogen)
C
M (metal: Na, K, Fe…..)
Reaction Product
H+ or H2O
CO and CO2
Mn+ (Na+, K+, Fe3+…..)
Oxidation Numbers & Redox Reactions
Oxidation number
Compares the charge of an uncombined atom with its
actual or relative charge in a compound.
General rules:
 Pure element = 0.
 Monatomic ion = charge of ion.
 Some elements have the same oxidation
number in almost all their compounds…
 The sum of the oxidation numbers of all atoms in
any species = the charge of the species.
5
Element
Find the oxidation number for all elements in SO32-
Ox#
Exceptions
F
−1
None
O
−2
Metal peroxides (MO2) = -1
Sulfur?
Halogen oxides (OXn) (varies)
Sum of the oxidation numbers = charge = -2
(ox. no. for S) + 3(ox. no. for O) = -2
(ox. no. for S) + 3(-2) = -2
(ox. no. for S) – 6 = -2
Sulfur = +4
O = -2 ( not a peroxide, not bonded to halogen)
Cl, Br, I
−1
Interhalogens (strongest = -1)
H
+1
Metal hydrides (e.g. NaH) = -1
Compound
SO2
SO42NH4+
NO2NO3OF2
ClF5
KMnO4
H2O2
Known
Unknown
O = -2
O = -2
H = +1
S = +4
S = +6
N = -3
O = -2
O = -2
N = +3
N = +5
O = +2
Cl = +5
F = -1
F = -1
K = +1 O = -2
Mn = +7
H = +1
O = -1
A more complex example:
0
+1
+5 −2
Cu(s) + 4 H+(aq) + 2 NO3-(aq)
Cu2+ (aq) + 2 NO2 (g) + 2 H2O(ℓ)
+2
+4 −2
+1 −2
Ox. numbers always change during redox reactions
Increase ox. number = Oxidation
Decrease ox. number = Reduction
If an element reacts to form a compound, it is a redox
reaction.
Oxidation and reduction must both occur.
S8(s) + 8 O2(g)
S : 0 → +4 (oxidized: added O; lost e-; increased ox. no.)
O : 0 → -2 (reduced: lost O; gained e-; decreased ox. no.)
Exchange reactions are not redox
– no change in oxidation state occurs.
e.g.
+1
Cu is oxidized (ox. no. ↑; loss of
e-).
8 SO2(g)
+1 −1
+1
+1 −1
AgNO3(aq) + KCl(aq)  KNO3(aq) + AgCl (s)
H is unchanged.
O is unchanged.
NO3-: N is in a +5, and O is in a -2 ox. state.
N is reduced (ox. no. ↓; gain of e-).
6
Displacement Reactions
+
A
Redox:
Activity Series of Metals
Displace H2 from H2O (ℓ),
steam or acid
+
XZ
Fe(s) + CuSO4(aq)
0
+2
Cu(s) + 2 AgNO3(aq)
0
+1
AZ
X
FeSO4(aq) + Cu(s)
+2
0
Cu(NO3)2(aq) + 2 Ag(s)
+2
0
Displace H2 from steam or
acid
Mg
Al
Mn
Zn
Cr
Displace H2 from acid
Fe
Ni
Sn
Pb
No reaction with H2O,
steam or acid
Sb
Cu
Hg
Ag
Pd
Pt
Au
Not all metals can displace another from its salts:
Cu(s) + ZnSO4(aq)
An activity series was developed…
Powerful reducing agents at the top.
Higher elements displace lower ones:
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)
Metals at the bottom are unreactive.
• Coinage metals
• Their ions are powerful oxidizing agents.
Ease of
oxidation
increases
H
no reaction
Li
K
Ba
Sr
Ca
Na
Li
K
Ba
Sr
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Ni
Sn
Pb
H
Sb
Cu
Hg
Ag
Pd
Pt
Au
Solution Concentration
Potassium + water
Relative amounts of solute and solvent.
2 K(s) + 2 H2O(ℓ) 
2 KOH(aq) + H2(g)
 solute
– substance dissolved.
 solvent – substance doing the dissolving.
There are several concentration units.
Most important to chemists: Molarity
7
Molarity
Molarity
Molarity =
moles solute = mol
L
liters of solution
• V of solution not solvent.
• Shorthand:
[NaOH] =1.00 M
Calculate the molarity of sodium sulfate in a solution
that contains 36.0 g of Na2SO4 in 750.0 mL of
solution.
nNa2SO4 =
[Na2SO4 ] =
Brackets [ ] represent
“molarity of ”
36.0 g
= 0.2534 mol
142.0 g/mol
0.2534 mol
0.7500 L
Unit change!
(mL to L)
[Na2SO4 ] = 0.338 mol/L = 0.338 M
Molarity
6.37 g of Al(NO3)3 are dissolved to make a 250. mL
aqueous solution. Calculate (a) [Al(NO3)3] (b) [Al3+]
and [NO3-].
(a) Al(NO3)3
FM = 26.98 + 3(14.00) + 9(16.00)
= 213.0 g
mol
6.37 g
nAl(NO3)3 =
= 2.991 x 10-2 mol
213.0 g/mol
-2
[Al(NO3)3] = 2.991 x 10 mol = 0.120 M
0.250 L
Solution Preparation
Solutions are prepared either by:
1. Diluting a more concentrated solution.
or…
2. Dissolving a measured amount of solute and
diluting to a fixed volume.
Molarity
6.37 g of Al(NO3)3 in a 250. mL aqueous solution. Calculate (a) the molarity of the
Al(NO3)3 , (b) the molar concentration of Al3+ and NO3- ions in solution.
(b) Molarity of Al3+ , NO3-?
Al(NO3)3(aq) → Al3+(aq) + 3 NO3-(aq)
1 Al(NO3)3 ≡ 1 Al3+
1 Al(NO3)3 ≡ 3 NO3-
[Al3+] = 0.120 M Al(NO3)3
1 Al3+
= 0.120 M Al3+
1 Al(NO3)3
[NO3-] = 0.120 M Al(NO3)3
3 NO3- = 0.360 M NO 3
1 Al(NO3)3
Solution Preparation by Dilution
MconcVconc = MdilVdil
Example
Commercial concentrated sulfuric acid is 17.8 M. If
75.0 mL of this acid is diluted to 1.00 L, what is the
final concentration of the acid?
Mconc = 17.8 M
Mdil = ?
Vconc = 75.0 mL
Vdil = 1000. mL
Mdil = MconcVconc = 17.8 M x 75.0 mL = 1.34 M
Vdil
1000. mL
8
Solution Preparation from Pure Solute
Prepare a 0.5000 M solution of potassium
permanganate in a 250.0 mL volumetric flask.
• Weigh exactly 19.75 g of pure KMnO4
• Transfer it to a volumetric flask.
Mass of KMnO4 required
nKMnO4 = [KMnO4] x V
= 0.5000 M x 0.2500 L (M ≡ mol/L)
= 0.1250 mol KMnO4
Mass of KMnO4 = 0.1250 mol x 158.03 g/mol
= 19.75 g
Use molar
mass of B
Use molar
mass of A
Moles of
A
Use mole ratio
Grams of
B
Molarity and Reactions in Aqueous Solution
What volume, in mL, of 0.0875 M H2SO4 is required
to neutralize 25.0 mL of 0.234 M NaOH?
H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(ℓ)
Moles of
B
Use solution
molarity of A
Use solution
molarity of B
Liters of
A solution
nNaOH = 0.0250 L
Liters of
B solution
nA = [ A ] x V
[product] = nproduct / (total volume).
H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(ℓ)
0.002925 mol 0.00585 mol
V?
25.0 mL
0.234 mol
L
= 5.850 x 10-3 mol
2 NaOH ≡ 1 H2SO4
nH2SO4 = 5.850 x 10-3 mol NaOH
1 H2SO4
2 NaOH
= 2.925 x 10-3 mol
A 4.554 g mixture of oxalic acid, H2C2O4 and NaCl
was neutralized by 29.58 mL of 0.550M NaOH. What
was the weight % of oxalic acid in the mixture?
H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(ℓ)
Vacid needed:
V =
• Rinse all the solid from the weighing
dish into the flask.
• Fill the flask ≈ ⅓ full.
• Swirl to dissolve the solid.
• Fill the flask to the mark on the neck.
• Shake to thoroughly mix.
Grams of
A
Solution Preparation from Pure Solute
mol H2SO4
1L
= 2.925 x 10-3 mol
[H2SO4]
0.0875 mol
nNaOH = 0.02958 L x 0.550 mol/L = 0.01627 mol
1 H2C2O4 ≡ 2 NaOH
Oxalic acid reacted =
= 0.0334 L = 33.4 mL
0.01627 mol NaOH
1 H2C2O4
= 8.135 x10-3 mol
2 NaOH
9
A 4.554 g H2C2O4 / NaCl mixture … Wt % of oxalic acid in the mixture?
25.0 mL of 0.234 M FeCl3 and 50.0 mL of 0.453 M
NaOH are mixed. Which reactant is limiting? How
many moles of Fe(OH)3 will form?
Mass of acid consumed, macid
= 8.135 x10-3 mol x (90.04 g/mol acid)
= 0.7324 g
FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s)
Weight % =
macid
sample mass x 100%
nFeCl3 = 0.0250 L x 0.234 mol/L = 0.005850 mol
Weight % =
0.7324 g
x 100% = 16.08%
4.554 g
nNaOH = 0.0500 L x 0.453 mol/L = 0.02265 mol
FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s)
0.005850 mol
0.01925 mol
0.00585 mol FeCl3
nFe(OH)3 ?
1 Fe(OH)3
= 0.00585 mol Fe(OH)3
1 FeCl3
Aqueous Solution Titrations
Titration = volume-based method used to determine
an unknown concentration.
A standard solution (known concentration) is added
to a solution of unknown concentration.
 Monitor the volume added.
0.02265 mol NaOH
1 Fe(OH)3
= 0.00755 mol Fe(OH)3
3 NaOH
FeCl3 is limiting; 0.00585 mol Fe(OH)3 produced.
 Add until equivalence is reached – stoichiometrically
equal moles of reactants added.
 An indicator monitors the end point.
Often used to determine acid or base concentrations.
Aqueous Solution Titrations
Titrant: Base
of known
concentration
Buret = volumetric glassware used for
titrations.
Slowly add standard solution.
End point: indicator changes
color.
Determine Vtitrant added.
Unknown acid + phenolphthalein
(colorless in acid)…
…turns pink
in base
10

Chapter 5 Chemical Reactions

Transcription

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