Reactions in Aqueous Solutions

1
Reactions in Aqueous Solutions

1. Water as a universal solvent
2. Aqueous solutions - solutions with water as
   solvent
3. Solubility of ionic compounds in water
4. Solubility of molecular compounds in water
5. Strong, weak, and non-electrolyte solutions
6. Solution compositions - concentrations
7. Types of reactions in aqueous solutions
8. Chemical equations for reactions in aqueous
   solutions
9. Stoichiometry for reaction in aqueous solution
10. Redox reactions in aqueous solutions.

2
Water as a Universal Solvent

• Water molecule is very polar
• Water interacts strongly with ionic and polar
  molecules
• Strong interactions enable water to dissolve many
  solutes ionic and nonionic
• Solubility of ionic compounds depends on the
  relative strength of ion-dipole interactions
  between ions and water molecules and ionic bonds
  within the compounds
• Many ionic compounds dissolve in water because of
  strong ion-dipole interactions
• Polar nonionic compounds dissolve in water due to
  strong dipole-dipole interactions or hydrogen
  bonding

3
Electrolytes and Nonelectrolytes

• Electrolytes solutions capable of conducting
  electric current - contain ions that move freely
• Nonelectrolytes solutions not capable of
  conducting electric current - contains neutral
  molecules only
• Strong electrolytes ionic compounds, strong
  acids and strong bases they dissociate
  completely when dissolved in water, producing a
  lot of free ions
• Weak electrolytes weak acids or weak bases
  they only dissociate (ionize) partially when
  dissolved in water solutions contain mostly
  neutral molecules and very little free ions.

4
Strong and Weak Electrolytes

• Examples of strong electrolytes they ionize
  completely
• NaCl(aq) ? Na(aq) Cl-(aq)
• H2SO4(aq) ? H(aq) HSO4-(aq)
• Ca(NO3)2(aq) ? Ca2(aq) 2NO3-(aq)
• Examples of weak electrolytes they do not
  ionize completely
• HC2H3O2(aq) ?? H(aq) C2H3O2-(aq)
• NH4OH(aq) ?? NH4(aq) OH-(aq)
• Mg(OH)2(s) ?? Mg2(aq) 2 OH-(aq)

5
Nonelectrolytes

• Substances that do not ionize in aqueous solution
  are nonelectrolytes
• Most organic compounds are nonelectrolytes
• Solutions containing such substances cannot
  conduct electricity, because they do not have
  freely moving ions.
• Examples of nonelectrolytes
• C6H12O6, C12H22O11, CH3OH, C2H5OH, C3H7OH,
  HOC2H4OH, etc.

6
Solution Concentrations

• The concentration of a solution may be expressed
  in
• Percent by mass, percent by volume, or molarity
• Percent (by mass) (Mass of solute/Mass of
  solution) x 100
• Percent (by volume) (Vol. of solute/Vol. of
  solution) x 100
• Molarity (Mol of solute/Liter of solution)
• Mol of solute Liters of solution x Molarity

7
Percent by Mass

• Example
• A sugar solution contains 25.0 g of sugar
  dissolved in 100.0 g of water. What is the mass
  percent of sugar in solution?
• Percent sugar 25.0 g/(25.0 g 100.0 g) x
  100
• 20.0 (by mass)

8
Calculation of Mass from Percent

• Example
• Seawater contains 3.5 (by mass) of NaCl. How
  many grams of sodium chloride can be obtained
  from 5.00 gallons of seawater? (1 gall. 3.785
  L assume density of seawater 1.00 g/mL)
• Mass of seawater
• 5.00 gall x (3785 mL/gall.) x (1.00 g/mL)
  18925 g
• Mass of NaCl 18925 g sw x (3.5/100) 662 g

9
Percent by Volume

• Example
• A solution is prepared by mixing 150. mL of
  methanol, 100. mL of acetone, and 250. mL of
  water. What is the volume percent of methanol and
  acetone in solution?
• Percent methanol (150. mL/500. mL) x 100
• 30.0 (by
  volume)
• Percent acetone (100. mL/500. mL) x 100
• 20.0 (by volume)

10
Molar Concentration

• Example
• 4.0 g of sodium hydroxide, NaOH, is dissolved in
  enough water to make a 100.-mL of solution.
  Calculate the molarity of NaOH.
• Mole of NaOH 4.0 g NaOH x (1 mole/40.0 g)
• 0.10 mole
• Molarity of NaOH 0.10 mol/0.100 L 1.0 M

11
Calculation of Solute Mass in Solution

• Example
• How many grams of NaOH are present in 35.0 mL of
  6.0 M NaOH solution?
• Mole of NaOH (6.0 mol/L) x 35.0 mL x (1 L/1000
  mL)
• 0.21 mol
• Mass of NaOH 0.21 mol x (40.0 g/mol)
• 8.4 g NaOH

12
Preparing Solutions from Pure Solids

1. From the volume (in liters) and molarity of
   solution, calculate the mole and mass of solute
   needed
2. Weigh the mass of pure solute accurately
3. Transfer solute into a volumetric flask of
   appropriate size
4. Add deionized water to the volumetric flask, well
   below the narrow neck, and shake well to dissolve
   the solute.
5. When completely dissolved, add more distilled
   water to fill the flask to the mark and mix the
   solution well.

13
Preparing Solution from Solid

• Example
• Explain how you would prepare 1.00 L of 0.500 M
  NaCl solution.
• Calculate mass of NaCl needed
• Mole of NaCl 1.00 L x (0.500 mol/L) 0.500 mol
• Mass of NaCl 0.500 mol x (58.44 g/mol) 29.2 g
• Preparing the solution
• Weigh 29.2 g of NaCl accurately and transfer into
  1-liter volumetric flask. Fill the flask half way
  with distilled water, shake well until all solid
  has dissolved. Fill the flask to the 1-liter mark
  with more distilled water and mix the solution
  well by inverting the flask back and forth
  several times.

14
Preparing Solution from Stock

• Calculate volume of stock solution needed using
  the formula MiVi MfVf (i initial f
  final)
• Measure accurately the volume of stock solution
  and carefully transfer to a volumetric flask of
  appropriate size
• Dilute stock solution with distilled water to the
  required volume (or to the mark on volumetric
  flask)
• Mix solution well.
• (Note if diluting concentrated acid, place some
  distilled water in the flask, add the
  concentrated acid, and then add more distilled
  water to the required volume.)

15
Preparing Solution from Stock

• Example
• Explain how you would prepare 1.0 L of 3.0 M
  H2SO4 solution from concentrated H2SO4, which is
  18 M.
• Calculate volume of concentrated H2SO4 needed
• Vol. of conc. H2SO4 (1.0 L x 3.0 M/18 M) 0.17
  L 170 mL
• Preparing the solution
• Place some distilled water in the 1-liter
  volumetric flask (that would fill the flask to
  about a quarter full). Measure accurately 170 mL
  of conc. H2SO4 and transfer carefully to the
  volumetric flask that already contains some
  distilled water. Then fill the flask to the
  1-liter mark with more distilled water and mix
  the solution well by inverting the flask back and
  forth several times.

16
Reactions in Aqueous Solution

1. Double-Displacement Reactions
2. Precipitation reactions
3. Neutralization (or Acid-Base) reactions
4. Oxidation-Reduction (Redox) Reactions
5. Combination reactions
6. Decomposition reactions
7. Combustion reactions
8. Single-Replacement reactions
9. Reactions involving strong oxidizing reagents

17
Precipitation Reactions

• Reactions that produce insoluble products (or
  precipitates) when two aqueous solutions are
  mixed.
• Examples
• 1) AgNO3(aq) KBr(aq) ? AgBr(s)
  KNO3(aq)
• 2) Pb(NO3)2(aq) K2CrO4 ? PbCrO4(s)
  2KNO3(aq)
• 3) BaCl2(aq) H2SO4(aq) ? BaSO4(s)
  2NaCl(aq)
• 4) 3Hg(NO3)2 (aq) 2Na3PO4 (aq) ? Hg3(PO4)2(s)
  6NaNO3(aq)

18
Solubility Rules for Predicting Solid Products

• Soluble salts
• All compounds of alkali metals and NH4
• All compounds containing nitrate, NO3-, and
  acetate. C2H3O2-, except silver acetate, which is
  sparingly soluble
• Most chlorides, bromides, and iodides, except
  AgX, Hg2X2, PbX2, and HgI2 where X Cl-, Br-,
  or I-.
• Most sulfates, except CaSO4, SrSO4, BaSO4, PbSO4
  and Hg2SO4.
• Insoluble or slightly soluble salts
• Most hydroxides (OH-), sulfides (S2-), carbonates
  (CO32-), chromates (CrO42-), and phosphate
  (PO43-), except those associated with the Group
  1A metals or NH4.

19
Predicting Precipitation Reactions

• Complete and balance the following reactions in
  aqueous solution and identify the precipitate, if
  formed.
• a) CaCl2(aq) Na2CO3(aq) ? ?
• b) NH4NO3(aq) MgCl2(aq) ? ?
• c) Pb(NO3)2(aq) KI(aq) ? ?
• d) AgNO3(aq) Na3PO4(aq) ? ?

20
Equations for Precipitation Reactions

• Molecular equation
• Pb(NO3)2(aq) K2CrO4(aq) ? PbCrO4(s)
  2KNO3(aq)
• Total Ionic equation
• Pb2 2NO3- 2K CrO42- ? PbCrO4(s) 2K
  2NO3-
• (K and NO3- are spectator ions)
• Net ionic equation
• Pb2(aq) CrO42-(aq) ? PbCrO4(s)

21
Acid-Base (Neutralization) Reactions

• Acid a compound that produces hydrogen ions
  (H) when dissolved in aqueous solution
• Base a compound that produces hydroxide ions
  (OH-) in aqueous solutions.
• Some examples of acids and strong bases
• Acids HCl, HClO4, HNO3, H2SO4, H3PO4, and
  HC2H3O2
• Bases NaOH, KOH, Ba(OH)2, and NH3.

22
Acid-Base Reactions

• Some example of acid-base reactions
• HCl(aq) NaOH(aq) ? H2O(l) NaCl(aq)
• H2SO4(aq) KOH(aq) ? H2O(l) K2SO4(aq)
• HC2H3O2(aq) NaOH(aq) ? H2O(l) NaC2H3O2(aq)
• 2HClO4(aq) Ba(OH)2(aq) ? 2 H2O(l)
  Ba(ClO4)2(aq)

23
Equations for Acid-Base Reactions

• An example of strong acid and strong base
  reaction
• Molecular equation
• HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
• Total ionic equation
• H(aq) Cl-(aq) Na(aq) OH-(aq) ? Na(aq)
  Cl-(aq) H2O(l)
• Net ionic equation
• H(aq) OH-(aq) ? H2O(l)

24
Equations for Acid-Base Reactions

• An example of week acid and strong base reaction
  
• Molecular equation
• HC2H3O2(aq) NaOH(aq) ? NaC2H3O2(aq) H2O(l)
• Total ionic equation
• HC2H3O2(aq) Na(aq) OH-(aq) ? Na(aq)
  C2H3O2-(aq) H2O(l)
• Net ionic equation
• HC2H3O2(aq) OH-(aq) ? C2H3O2-(aq) H2O(l)

25
Equations for Acid-Base Reactions

• An example of strong acid and weak base reaction
  
• Molecular equation
• HCl(aq) NH3(aq) ? NH4Cl(aq)
• Total ionic equation
• H(aq) Cl-(aq) NH3(aq) ? NH4(aq)
  Cl-(aq)
• Net ionic equation
• H(aq) NH3(aq) ? NH4(aq)

26
Stoichiometry in Aqueous Solution

• Example-1 Precipitation reaction
• How many grams of BaSO4 will be formed when 100.0
  mL of 0.100 M BaCl2 is reacted with 150.0 mL of
  0.100 M Na2SO4?
• Reaction BaCl2(aq) Na2SO4(aq) ? BaSO4(s)
  2NaCl(aq)
• Solution-1
• Find the limiting reactant by calculating the
  mole of each reactant, where the moles can be
  calculated by multiplying the volume (in liters)
  of the solution with the molarity.
• Mole of BaCl2 0.1000 L x 0.100 mol/L 0.0100
  mole
• Mole of Na2SO4 0.1500 L x 0.100 mol/L 0.0150
  mole

27
Stoichiometry in Aqueous Solution

• Solution-1 (continued)
• BaCl2 is the limiting reactant
• Therefore, mole of BaSO4 expected 0.0100 mole
• Mass of BaSO4 formed 0.0100 mol x 233.39 g/mol
• 2.33 g
• If 2.05 g of barium sulfate was actually
  obtained, what is the percent yield?
• Percent yield (2.05 g/2.33 g) x 100 88.0

28
Acid-Base Titration
29
Acid-Base Titration

• Example-1
• In a titration experiment, 32.40 mL of 0.2560 M
  NaOH solution was required to neutralize acetic
  acid in a 10.00-mL sample of vinegar. Calculate
  the molarity of acetic acid in the vinegar. If
  the density of vinegar is 1.0 g/mL, calculate the
  mass percent of acetic acid in the vinegar.
• Reaction HC2H3O2(aq) NaOH(aq) ? H2O(l)
  NaC2H3O2(aq)
• Solution-1
• Calculate mole of NaOH using its volume and
  molarity
• According to the equation, mole of acetic acid
  mole of NaOH

30
Acid-Base Titration

• Solution-1 (continued)
• Mole of NaOH used 0.03240 L x 0.2560 mol/L
• 0.008294 mol
• Mole of HC2H3O2 reacted 0.008294 mole
• Molarity of HC2H3O2 in vinegar 0.008294
  mol/0.01000 L
• 
  0.8294 M
• Mole of HC2H3O2 in 100.0 mL 0.1000 L x 0.8294
  mol/L
• 
  0.08294 mol
• Mass of HC2H3O2 in 100.0 mL 0.08294 mol x
  60.05 g/mol
• 
  4.981 g

31
Acid-Base Titration

• Solution-1 (continued)
• Mass of 100.0 mL vinegar 100.0 mL x 1.0 g/mL
• 1.0
  x 102 g
• Percent of acetic acid in vinegar (4.981 g/ 1.0
  x 102 g) x 100
• 
  5.0 (by mass)

32
Acid-Base Reactions

• Example-2
• How many milliliters of 0.2765 M NaOH solution
  will be required to neutralize 20.00 mL of 0.1500
  M H2SO4?
• Reaction H2SO4(aq) 2NaOH(aq) ? 2H2O(l)
  Na2SO4(aq)
• Solution-2
• Mole of H2SO4 present 0.02000 L x 0.1500 mol/L
  0.003000 mol
• Mole of NaOH needed 0.003000 mol x 2 0.006000
  mol
• Volume of 0.2765 M NaOH needed to neutralize the
  acid
• (0.006000 mol)/(0.2765 mol/L)
  0.02170 L
• 
  21.70 mL

33
Acid-Base Reactions

• Example-3
• A 4.00-mL sample of sulfuric acid is diluted to
  100.0 mL. 20.00 mL of the dilute acid is then
  titrated with 0.2750 M NaOH solution. If 35.60 mL
  of the base were required to neutralize the acid,
  calculate the molarity of the original sulfuric
  acid solution.
• Reaction H2SO4(aq) 2NaOH(aq) ? 2H2O(l)
  Na2SO4(aq)
• Solution-3
• Mole of NaOH used 0.03560 L x (0.2750 mol/L)
  0.009790 mole
• Mole of H2SO4 titrated ½ (0.009790 mol)
  0.004895 mole

34
Acid-Base Reactions

• Solution-3 (continued)
• Molarity of dilute acid (0.004895
  mol)/(0.02000 L) 0.24475 M
• Molarity of undiluted acid 0.24475 M x 100.0
  mL/4.00 mL
• 
  6.12 M

35
Reactions That Produce Gas

• Reactions producing CO2 gas
• CaCO3(s) 2HCl(aq) ? CaCl2(aq) H2O(l)
  CO2(g)
• NaHCO3(s) HCl(aq) ? NaCl(aq) H2O(l)
  CO2(g)
• Reaction that produces SO2 gas
• Na2SO3(s) 2HCl(aq) ? 2NaCl(aq) H2O(l)
  SO2(g)
• Reaction that produces H2S gas
• Na2S(aq) 2HCl(aq) ? 2NaCl(aq) H2S(g)

36
Oxidation-Reduction Reactions

• Oxidation ? loss of electrons and increase in
  oxidation number
• Reduction ? gain of electrons and decrease in
  oxidation number
• Oxidation-reduction (or Redox) reaction ? one
  that involves transfer of electrons from one
  reactant to the other
• Oxidizing agent ? the reactant that gains
  electrons and got reduced
• Reducing agent ? the reactant that loses
  electrons and got oxidized.

37
Types of Redox Reactions

• Reactions between metals and nonmetals
• Combustion reactions (reactions with molecular
  oxygen)
• Single replacement reactions
• Decomposition reactions that form free elements
• Reactions in aqueous solution involving oxidizing
  and reducing agents
• Disproportionation reactions.

38
Types of Redox Reactions

• Reactions between metals and nonmetals
• 4Al(s) 3 O2(g) ? 2Al2O3(s)
• 3Mg(s) N2(g) ? Mg3N2(s)
• Combustion reactions
• CH4(g) 2 O2(g) ? CO2(g) 2H2O(g)
• 2C8H18(l) 25 O2(g) ? 16CO2(g) 18H2O(g)
• C2H5OH(l) 3 O2(g) ? 2CO2(g) 3H2O(g)

39
Types of Redox Reactions

• Single-Replacement Reactions
• Mg(s) 2HCl(aq) ? MgCl2(aq) H2(g)
• Cu(s) 2AgNO3(aq) ? Cu(NO3)2(aq) 2Ag(s)
• Cl2(aq) 2KBr(aq) ? 2KCl(aq) Br2(aq)
• Decomposition Reactions
• 2HgO(s) ? 2Hg(l) O2(g)
• (NH4)2Cr2O7(s) ? Cr2O3(s) N2(g) 4H2O(g)

40
Types of Redox Reactions

• Reactions in aqueous solutions that involve
  strong oxidizing reagents
• MnO4-(aq) 5Fe2(aq) 8H(aq) ? Mn2(aq)
  5Fe3(aq) H2O(l)
• Cr2O72-(aq) 3H2O2(aq) 8H(aq) ? 2Cr3(aq)
  7H2O(l) 3 O2(g)
• 2Cr(OH)4-(aq) 3H2O2(aq) 2 OH-(aq) ?
  2CrO42-(aq) 8H2O(l)
• Disproportionation reaction
• Cl2(g) 2NaOH(aq) ? NaOCl(aq) NaCl(aq)
  H2O(l)
• 3Br2(aq) 6NaOH(aq) ? NaBrO3(aq)
  NaBr(aq) H2O(l)

41
Guidelines for Determining Oxidation Numbers of
Elements

• 1. Atoms in the free elemental form are
  assigned oxidation number zero (0)
• 2. The sum of oxidation number in neutral
  molecules or formula units is 0
• the sum of oxidation number (o.n.) of atoms
  in a polyatomic ion is equal to the net charge of
  the ion (in magnitude and sign)
• 3. In their compounds, each Group IA metal is
  assigned an o.n. of 1 each Group IIA metal an
  o.n. of 2 boron and aluminum each an o.n. of
  3, and fluorine an o.n. of 1
• 4. Hydrogen is assigned an o.n. of 1 in
  compounds or polyatomic ions with nonmetals, and
  an o.n. of 1 in metal hydrides
• 5. In compounds and polyatomic ions, oxygen is
  assigned an o.n. of -2, except in peroxides, in
  which its o.n. is 1
• 6. In binary compounds with metals, chlorine,
  bromine, and iodine each has an o.n. of -1
  sulfur, selenium, and tellurium each has an o.n.
  of -2.

42
Oxidation-Reduction Reactions

• In the following equations, identify all
  reactions that are redox reactions
• 1. 2KMnO4(aq) 16HCl(aq) ? 2MnCl2(aq)
  2KCl(aq) 5Cl2(aq) 8H2O(l)
• 2. 2KClO3(s) ? 2KCl(s) 3 O2(g)
• 3. CaCO3(s) ? CaO(s) CO2(g)
• 4. Mg(OH)2(s) ? MgO(s) H2O(g)
• 5. Mg(s) ZnSO4(aq) ? MgSO4(aq) Zn(s)
• 6. Cr2O72-(aq) 3C2H5OH(l) 2H(aq) ?
  2Cr3(aq) 3CH3COOH(aq) 4H2O(l)

43
Balancing Redox Reactions by Half-Equation Method

• Example-1 Balance the following
  oxidation-reduction reaction in aqueous solution
• MnO4-(aq) Fe2(aq) ? Mn2(aq)
  Fe3(aq)
• Solution-1
• Note the above equation is both not balanced and
  not complete. It only shows the components
  (reactants) that undergoes changes is oxidation
  numbers
• Redox reactions in acidic solution means that you
  need to add H ion in the equation, which
  produces water as one of the products.

44
Balancing Redox Reactions by Half-Equation Method

• Solution-1 (continued)
• Balancing the equation first step, break up the
  equation into two half equations oxidation and
  reduction half-equations
• MnO4-(aq) ? Mn2(aq)
• In this case, all the four oxygen in MnO4- will
  become water in acidic solution. So, add enough
  H ion to form H2O with the four oxide ions in in
  MnO4-. The half-equation with all atoms balanced
  will look like this
• MnO4-(aq) 8H(aq) ? Mn2(aq)
  4H2O(l)

45
Balancing Redox Reactions by Half-Equation Method

• Solution-1 (continued)
• Next step, check the total charges on both side
  of the half-equation note that they are not
  equal. Add enough electrons on the side that has
  more positive charges (or less negative charges),
  so that the total charges on both sides become
  equal (in magnitude and sign).
• MnO4-(aq) 8H(aq) 5e- ? Mn2(aq)
  4H2O(l)
• Now we have a balanced reduction half-equation.
• (How do you know it is reduction half-equation?)

46
Balancing Redox Reactions by Half-Equation Method

• Solution-1 (continued)
• Next, write the other half-equation and balance
  it
• Fe2(aq) ? Fe3(aq) e- (this is
  oxidation half-equation)
• To obtain the overall equation, we add the two
  balanced half-equations, but make sure the number
  of electrons on both half-equations are equal, so
  that they cancel out. The overall equation should
  not contain any electrons. In this case, we
  multiply the above oxidation half-equation by 5
  and obtain
• 5Fe2(aq) ? 5Fe3(aq) 5e-
• Now, adding the two half-equations yields the
  following balanced net ionic equation
• MnO4-(aq) 5Fe2(aq) 8H(aq) ? Mn2(aq)
  5Fe3(aq) 4H2O(l)

47
Balancing Redox Reactions by Half-Equation Method

• Example-2 Balance the following
  oxidation-reduction reaction in acidic solution
• Cr2O72-(aq) H2O2(aq) ? Cr3(aq)
  O2(g) H2O(l)
• Solution-2 Write the two half-equations and
  balance them.
• Reduction half-equation
• Cr2O72-(aq) 14H(aq) 6e- ? 2Cr3(aq)
  7H2O(l)
• Oxidation half-equation
• H2O2(aq) ? O2(g) 2H(aq) 2e-

48
Balancing Redox Reactions by Half-Equation Method

• Solution-2 (continued)
• Multiply the oxidation half-equation by 3 to make
  the number of electrons equal with that of the
  reduction-half equation.
• 3H2O2(aq) ? 3 O2(g) 6H(aq) 6e-
• Then add the two half-equations, canceling all
  the electrons on both side of the equation, all
  the H ions on the right-hand side, and the same
  number of H ions on the left-hand side. The
  final (balanced) net ionic equation will be as
  follows
• Cr2O72-(aq) 3H2O2(aq) 8H(aq) ? 2Cr3(aq)
  3O2(g) 7H2O(l)

49
Balancing Redox Reactions by Half-Equation Method

• Example-3 Balance the following
  oxidation-reduction reaction in basic solution
• Cr(OH)4-(aq) H2O2(aq) ? CrO42-(aq)
  H2O(l)
• Solution-3
• Write the two half-equations and balance them,
  adding electrons on either side as needed to
  balance the charges.
• Note that this reaction is in basic solution,
  which means the overall equation should contains
  OH- ion instead of H ions.

50
Balancing Redox Reactions by Half-Equation Method

• Solution-3 (continued)
• Oxidation half-equation
• Cr(OH)4-(aq) 4OH-(aq) ? CrO42-(aq)
  4H2O(l) 3e-
• Reduction half-equation
• H2O2(aq) 2e- ? 2OH-
• Multiply the oxidation half-equation by 2 and the
  reduction half-equation by 3 to make the number
  of electrons in both half-equations equal.
• 2Cr(OH)4-(aq) 8OH-(aq) ? 2CrO42-(aq)
  8H2O(l) 6e-
• 3H2O2(aq) 6e- ? 6OH-

51
Balancing Redox Reactions by Half-Equation Method

• Solution-3 (continued)
• Now add the two half-equations, canceling all the
  electrons on both sides of the equation, all the
  6OH- ions on the right-hand side, and the same
  number of OH- ions on the left-hand side.
• The overall balanced net ionic equation will
  appear as follows
• 2Cr(OH)4-(aq) 3H2O2(aq) 2OH-(aq) ?
  2CrO42-(aq) 8H2O(l)

52
Uses of Reactions in Aqueous Solutions

• 1. Dissolving Insoluble Compounds
• Fe2O3(s) 6HNO3(aq) ? 2Fe(NO3)3(aq)
  3H2O(l)
• Mg(OH)2(s) 2HCl(aq) ? MgCl2(aq) 2H2O(l)
• 2. Syntheses of Inorganic Compounds
• AgNO3(aq) NaBr(aq) ? AgBr(s) NaNO3(aq)
• Pb(NO3)2(aq) K2CrO4(aq) ? PbCrO4(s)
  2KNO3(aq)
• 3. Extraction of Metals form Solution
• Mg2(aq) Ca(OH)2(aq) ? Mg(OH)2(s)
  Ca2(aq)