Transcript Chapter 4 – Reactions in Aqueous Solutions

Reactions in Aqueous Solutions
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Water as a universal solvent
Aqueous solutions - solutions with water as solvent
Solubility of ionic compounds in water
Solubility of molecular compounds in water
Strong, weak, and non-electrolyte solutions
Solution compositions - concentrations
Types of reactions in aqueous solutions
Chemical equations for reactions in aqueous solutions
Stoichiometry for reaction in aqueous solution
Redox reactions in aqueous solutions.
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;
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.
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)
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.
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
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)
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
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)
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
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
Preparing Solutions from Pure Solids
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From the volume (in liters) and molarity of solution,
calculate the mole and mass of solute needed;
Weigh the mass of pure solute accurately;
Transfer solute into a volumetric flask of appropriate
size;
Add deionized water to the volumetric flask, well below
the narrow neck, and shake well to dissolve the solute.
When completely dissolved, add more distilled water to
fill the flask to the mark and mix the solution well.
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.
Preparing Solution from Stock
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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.)
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.
Reactions in Aqueous Solution
1. Double-Displacement Reactions
1. Precipitation reactions
2. Neutralization (or Acid-Base) reactions
2. Oxidation-Reduction (Redox) Reactions
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Combination reactions
Decomposition reactions
Combustion reactions
Single-Replacement reactions
Reactions involving strong oxidizing reagents
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)
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+.
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)  ?
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)
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.
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)
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)
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)
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)
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;
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%
Acid-Base Titration
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
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
–
– Mass of HC2H3O2 in 100.0 mL
–
= 0.1000 L x 0.8294 mol/L
= 0.08294 mol
= 0.08294 mol x 60.05 g/mol
= 4.981 g
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)
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
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
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
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);
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.
Types of Redox Reactions
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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.
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);
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);
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);
Guidelines for Determining Oxidation Numbers of Elements
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Atoms in the free elemental form are assigned oxidation number zero (0);
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);
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;
Hydrogen is assigned an o.n. of +1 in compounds or polyatomic ions with
nonmetals, and an o.n. of –1 in metal hydrides;
In compounds and polyatomic ions, oxygen is assigned an o.n. of -2, except in
peroxides, in which its o.n. is –1;
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.
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);
•
Cr2O72-(aq) + 3C2H5OH(l) + 2H+(aq)  2Cr3+(aq) + 3CH3COOH(aq) + 4H2O(l)
6.
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.
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 halfequation with all atoms balanced will look like this:
–
MnO4-(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l)
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?)
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);
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-;
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);
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.
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-;
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);
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);