Chapter 4
Reactions in Aqueous Solutions 

4.1 General Properties of Aqueous Solutions

•	Solution - a homogeneous mixture 
   
–	Solute: the component that is dissolved ‏
–	Solvent: the component that does the dissolving
Generally, the component present in the greatest quantity is considered to be the solvent.  Aqueous solutions are those in which water is the solvent.

•	Electrolytes and Nonelectrolytes
–	Electrolyte:  substance that dissolved in water produces a solution that  conducts electricity
•	Contains ions

NaCl(s)  (+ H2O)  →  Na+(aq)  +  Cl-(aq)

–	Nonelectrolyte: substance that dissolved in water produces a solution that  does not conduct electricity
•	Does not contain ions

C12H22O11(s) (+ H2O)  →  C12H22O11(aq)

•	Dissociation - ionic compounds separate into constituent ions when dissolved in solution 

NaCl(s)  (+ H2O)  →  Na+(aq)  +  Cl-(aq)

•	Ionization - formation of ions by molecular compounds when dissolved

HCl(g)  (+H2O)  →  H+(aq)  +  Cl-(aq)

Normally the water is simply written on the reaction arrow when it only dissoles the solutes but does not react

NH3(g)  +  H2O(l)  <=>   NH4+(aq)  +  OH-(aq)

•	Strong and weak electrolytes
–	Strong Electrolyte: 100% dissociation
•	All water soluble ionic compounds, strong acids and strong bases
–	Weak electrolytes
•	Partially ionized in solution
•	Exist mostly as the molecular form in solution 
•	Weak acids and weak bases

Table 4.1: Ionization Equations for acids

•	Examples of weak electrolytes

–	Weak acids 				
	    HC2H3O2(aq)   <=>        C2H3O2 (aq) + H+ (aq)‏

–	Weak bases 
	     NH3 (aq) +  H2O(l)  <=>  	NH4+ (aq) + OH (aq) 

  (Note: double arrows indicate a reaction that occurs in both directions - a state of dynamic equilibrium exists) 

Figure: Method to decide the kind of electrolyte

Classify the following as nonelectrolyte,
weak electrolyte or strong electrolyte 
–	NaOH
                    strong electrolyte
–	CH3OH
                    nonelectrolyte
–	H2CO3
                              weak electrolyte

4.2 Precipitation Reactions

•	Precipitation (formation of a solid from two aqueous solutions) occurs when product is insoluble 
•	Produce insoluble ionic compounds 
•	Double replacement (or metathesis reaction)
•	Solubility is the maximum amount of a solid that can dissolve in a given amount of solvent at a specified temperature  
•	Prediction based on solubility rules 

Table 4.2 and 4.3: Solubility rules

•	Hydration:  process by which water molecules remove and surround  individual ions from the solid. 

Figure on hydration

Identify the Precipitate
Pb(NO3)2(aq) + 2NaI(aq)  ® 2NaNO3 (aq)      +  PbI2  (s)   
 
All nitrates are soluble, lead iodide is insoluble

Figure: Mixing solutions of lead nitrate and sodium chloride

Classify the following as soluble or
insoluble in water
–	Ba(NO3)2
			           soluble
–	AgI
                       insoluble
–	Mg(OH)2
                                   insoluble

•	Molecular equation: shows all compounds represented by their chemical formulas

Na2SO4(aq)  +  Ba(OH)2(aq)  →  2 NaOH(aq)  +   BaSO4(s)

•	Ionic equation: shows all strong electrolytes as ions and all other substances (non- electrolytes, weak electrolytes, gases) by their chemical formulas
 Na2SO4(aq)  →  2 Na+(aq)  +  SO42-(aq)

Ba(OH)2(aq)  → Ba2+(aq)  +  2 OH-(aq)

NaOH(aq)  →  Na+(aq)  +  OH-(aq)

Ionic Equation:

	2 Na+(aq)  + SO42-(aq)  +  Ba2+(aq)  + 2 OH-(aq)  →  2 Na+(aq)  +  

+ 2 OH-(aq)  + BaSO4(s)
           
•	Net Ionic equation: shows only the reacting species in the chemical equation
–	Eliminates spectator ions
2 Na+(aq)  + SO42-(aq)  +  Ba2+(aq)  + 2 OH-(aq)  →  2 Na+(aq)  +  
cancel                                                     cancel              cancel    

+ 2 OH-(aq)  + BaSO4(s)
           
            cancel  

Net ionic equation:

Ba2+(aq)  +  SO42-(aq)  →  BaSO4(s)

•	Steps in writing a net ionic equation 
–	Write the balanced molecular equation.
•	Predict products by exchanging cations and anions in reactants.
–	Separate strong electrolytes into ions.
–	Cancel spectator ions.
–	Use the remaining species to write the net ionic equation.  

Aqueous solutions of silver nitrate and
sodium sulfate are mixed. Write the net
ionic reaction. 
Step 1: 

2AgNO3(aq)+Na2SO4(aq) ® 2NaNO3(aq)+Ag2SO4(s)

Step 2: Use solubility table for s and aq denotations on the left: all nitrates are soluble but silver sulfate is insoluble 

2Ag+(aq) + 2NO3(aq) + 2Na+(aq) + SO42(aq)

              ® 2Na+(aq)  + 2NO3(aq) + Ag2SO4(s) 

Step 3: Cancel spectators 
 
2Ag+(aq) + 2NO3(aq) + 2Na+(aq) + SO42(aq)
                       cancel            cancel        
		® 2Na+(aq)  +  2NO3(aq) + Ag2SO4(s)
                   cancel        cancel

Step 4: Write the net ionic reaction 
 
		2Ag+(aq) + SO42(aq) ® Ag2SO4(s) 

4.3 Acid-Base Reactions

•	Termed neutralization reactions.
•	Involve an acid and a base.
•	Double replacement (or metathesis) reaction 
•	A molecular compound (water) is a common product along with a salt (ionic compound).

                   Figure: Common Acids and Bases

Table 4.4: strong acids and bases

All the other acids and bases are weak 
electrolytes (important for net ionic 
equations). 

•	Definitions of acids and bases
–	Arrhenius acid - produces H+ in solution
–	Arrhenius base - produces OH in solution
–	More inclusive definitions: 
•	Brّønsted acid - proton donor
•	Brّønsted base - proton acceptor 

–	Examples of a weak base and weak acid
•	Ammonia with water: 

NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq) 

•	Hydrofluoric acid with water: 

HF(aq) + H2O(l) <=> H3O+(aq) + F-(aq)

Figure: Examples of a weak base and weak acid

•	Types of acids
–	Monoprotic: one ionizable hydrogen

         HCl + H2O  ® H3O+  + Cl		

–	Diprotic: two ionizable hydrogens
 
          H2SO4 + H2O  ® H3O+  + HSO4  
          HSO4 + H2O  ® H3O+  + SO42  

–	Triprotic: three ionizable hydrogens 

             H3PO4 + H2O ® H3O+ + H2PO4 
             H2PO4 + H2O ® H3O+  + HPO42 
             HPO42 + H2O ® H3O+  + PO43 

–	Polyprotic: generic term meaning more than one ionizable hydrogen 

•	Types of bases
–	Monobasic: One OH group

                KOH ®  K+   +  OH 

–	Dibasic: Two OH groups 

               Ba(OH)2  ®  Ba2+   +  2OH 

Acid-Base Neutralization 
•	Neutralization:  Reaction between an acid and a base
Acid + Base ®  Salt   +  Water 

Molecular equation:
HCl(aq) + NaOH(aq) ® NaCl(aq) + H2O(l)‏

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)‏

Solutions of acetic acid and lithium 
hydroxide are mixed. Write the net ionic
reaction. 

HC2H3O2(aq)  +  OH-(aq)  →  C2H3O2-(aq)  +  H2O(l)

4.4 Oxidation-Reduction Reactions
•	Often called “redox” reactions 
•	Electrons are transferred between the reactants 
–	One substance is oxidized, loses electrons
•	Reducing agent
–	Another substance is reduced, gains electrons
•	Oxidizing agent
•	Oxidation numbers change during the reaction 

–	Example
    Zn(s)  + CuSO4(aq) ® ZnSO4(aq)  + Cu(s) 

        Zn(s)  + Cu2+(aq) ® Zn2+(aq)  + Cu(s) 
–	Zinc is losing 2 electrons and oxidized.
•	Reducing agent
•	Zn(s)  ® Zn2+(aq)  + 2e
–	Copper ions are gaining the 2 electrons.
•	Oxidizing agent
•	Cu2+(aq) + 2e ® Cu(s)‏

Figure: Reaction of copper and zinc cations

•	Rules for assigning oxidation numbers 
–	Elements (uncombined) are 0.
	Al, N2, He, Zn, Ag, Br2, O2, O3
2.  Oxidation numbers must sum to the overall charge of the species. 
		SO42  = 2 (O is usually 2 so….)‏
		     ? + 4(2) = 2    
	Solve: ?  8 = 2         ? = + 6 (S) 

Table 4.5: Guidelines for Assigning Oxidation Numbers

Guidelines for Assigning Oxidation Numbers 

Assign oxidation numbers for all elements
in each species 
MgBr2
                          Mg  +2,  Br  1
ClO2
					 Cl  +1 ,   O  2 								

•	Displacement reactions
–	A common reaction: active metal replaces (displaces) a metal ion from a solution
 
      Mg(s) + CuCl2(aq)  ® Cu(s) + MgCl2(aq) 

–	The activity series of metals is useful in order to predict the outcome of the reaction. 

Table 4.6: Oxidation Series

•	Balancing redox reactions
–	Electrons (charge) must be balanced as well as number and types of atoms 
–	Consider this net ionic reaction: 
	       Al(s) + Ni2+(aq) ® Al3+(aq)  + Ni(s) 
–	The reaction appears balanced as far as number and type of atoms are concerned, but look closely at the charge on each side. 

        Al(s) + Ni2+(aq) ® Al3+(aq)  + Ni(s) 

–	Divide reaction into two half-reactions
                    Al(s) ® Al3+(aq)  + 3e 
            Ni2+(aq) + 2e ®  Ni(s) 

–	Multiply by a common factor to equalize electrons (the number of electrons lost must equal number of electrons gained)‏
                      2 [Al(s) ® Al3+(aq)  + 3e ]
         3 [Ni2+(aq) + 2e ®  Ni(s) ]

–	Cancel electrons and write balanced net ionic reaction 
                  2Al(s) ® 2Al3+(aq)  + 6e 
      3Ni2+(aq) + 6e ®  3Ni(s) 

	2Al(s) + 3Ni2+(aq) ® 2Al3+(aq) + 3Ni(s)‏

Predict whether each of the following will
occur. For the reactions that do occur,
write a balanced net ionic reaction for 
each. 
-  Copper metal is placed into a solution of silver nitrate 
    Silver below copper, so silver is reduced to the metal, and metallic copper oxidized to the doubly charged cation 

    Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)
-	A gold ring is accidentally dropped into a solution of hydrochloric acid 
        No reaction occurs, gold is below
        hydrogen on the activity series.

•	Combination Reactions
–	Many combination reactions may also be classified as redox reactions 
–	Consider: 
         Hydrogen gas reacts with oxygen
         gas 

			2H2(g) + O2(g) ® 2H2O(l) 

        Identify the substance oxidized and
        the substance reduced. 

•	Decomposition reactions
–	Many decomposition reactions may also be classified as redox reactions 
–	Consider: 
       Potassium chlorate is strongly heated  
		     2KClO3(s) ® 2KCl(s)   +  3O2(g) 

       Identify substances oxidized and
        reduced. 

•	Disproportionation reactions
–	One element undergoes both oxidation and reduction 
–	Consider: 

2 H2O2(aq)         →    2 H2O(l)         +    O2(g)

per atom:
   +1 –1                           +1 –2                  0
per molecule
    -2 –2                            +2 –2                  0

Oxygen in hydrogenperoxid is oxidized from –1 to 0 in the element and reduced from –1 to –2 in water

•	Combustion reactions
–	Common example, hydrocarbon fuel reacts with oxygen to produce carbon dioxide and water 
–	Consider: 

CH4(g)     +     2 O2(g)      →      CO2(g)                   +                       2 H2O(l)

per atom:
+4 –1                  0                     +4 –2                                                   +1 –2
per molecule:
+4 –4                  0                     +4 –4                                                   +2 –2

Figure: Oxidation Numbers on the Periodic Table       

4.5 Concentration of Solutions

•	Concentration is the amount of solute dissolved in a given amount of solvent.
•	Qualitative expressions of concentration
–	Concentrated – higher ratio of solute to solvent
–	Dilute - smaller ratio of solute to solvent

Figure: Comparison of a Concentrated and Dilute Solution

•	Quantitative concentration term 
–	Molarity is the ratio of moles solute per liter of solution 

Molarity = (moles of solute)/(liters of solution)

–	Symbols:  M  or [  ] 
–	Different forms of molarity equation

M = mol/L              L = mol/M                    mol = M x L

Calculate the molarity of a solution
prepared by dissolving 45.00 grams of
KI into a total volume of 500.0 mL.

(45.00 g KI)/(500.0 mL) x (1 mol KI)/(166.0 g KI) x 

x (1000 mL)/(1 L)  = 0.5422 M

How many milliliters of 3.50 M NaOH can 
be prepared from 75.00 grams of the
solid?

75.00 g NaOH x (1 mol NaOH/(40.00 g NaOH) x  

x (1 L )/(3.50 mol NaOH) x (1000 mL )/(1 L) = 536 mL

•	Dilution
–	Process of preparing a less concentrated solution from a more concentrated one.

moles of solute before dilution = moles of solute after dilution

moles of solute = (liters of solution) x (moles of solute)/(liters of solution)

                                       Mc x Vc (in L) = Md x Vd (in L)

For the next experiment the class will
need 250. mL of  0.10 M CuCl2. There is
a bottle of 2.0 M CuCl2. Describe how to
prepare this solution. How much of the
2.0 M solution do we need?

Concentrated: 2.0 M  use  ? mL (Ld)
Dilute: 250. mL of 0.10 M 
			            
                  McLc  = MdLd

       (2.0 M) (Lc)  =  (0.10 M) (250.mL) 
                     Lc  = 12.5 mL
               
   12.5 mL of the concentrated solution are needed; add enough distilled water to prepare 250. mL of the solution. 

•	Solution Stoichiometry
–	Soluble ionic compounds dissociate completely in solution. 
–	Using mole ratios we can calculate the concentration of all species in solution. 
         NaCl dissociates into Na+ and Cl 
         Na2SO4 dissociates into 2Na+ and SO42
         AlCl3 dissociates into Al3+ and 3Cl

Find the concentration of all species in a
0.25 M solution of MgCl2

		         MgCl2   ®  Mg2+  +  2Cl

Given: MgCl2  = 0.25 M 

           [Mg2+ ] = 0.25 M (1:1 ratio) 
              [Cl ] = 0.50 M (1:2 ratio) 

Using the square bracket notation, 
express the molar concentration for all
species in the following solutions

    0.42 M Ba(OH)2
         [Ba2+ ] = 0.42 M (1:1 ratio)
         [OH ]  = 0.84 M (2:1 ratio) 

   1.2 M NH4Cl 
         [NH4+ ] = 1.2 M (1:1 ratio)
         [Cl ]     = 1.2 M (1:1 ratio) 

4.6 Aqueous Reactions and Chemical Analysis

•	Types of quantitative analysis
–	Gravimetric analysis (mass analysis)
•	Example:  precipitation reaction 
–	Volumetric analysis (volume analysis)
•	Example:  titration

•	Gravimetric Analysis
–	One form: isolation of a precipitate 
–	Typical steps: 
•	Determine mass of unknown solid 
•	Dissolve unknown in water
•	Combine with excess amount of known substance to form a precipitate (excess drives reaction to completion) 
•	Filter, dry and weigh the precipitate 
•	Use formula and mass of ppt to find % of ion in unknown solid

A 0.825 g sample of an ionic compound
containing chloride ions and an unknown
metal is dissolved in water and treated
with excess silver nitrate. If 1.725 g of
AgCl precipitate forms, what is the
percent by mass of Cl in the original
sample? 

Steps in solution: 
•	Find the % of Cl in AgCl 
•	Multiply the % of Cl by the mass of the precipitate to obtain the Cl in the sample
•	Divide the mass of Cl in sample by total mass of sample (multiply by 100 for %) 

% Cl = 100 %  x (35.45 g Cl)/(143.35 g AgCl) = 24.7 %

0.247 x 1.725 g AgCl ppt = 0.427 g Cl in sample

% Cl in unknown = 100 %  x (0.427 g Cl)/(0.825 g sample) 
= 51.7 % Cl

•	Volumetric analysis
–	Commonly accomplished by titration 
•	Addition of a solution of known concentration (standard solution) to another solution of unknown concentration.
–	Standardization is the determination of the exact concentration of a solution.
–	Equivalence point represents completion of the reaction.
–	Endpoint is where the titration is stopped.
–	An indicator is used to signal the endpoint.

Figure: Apparatus for a Titration

A student measured exactly 15.0 mL of 
an unknown monoprotic acidic solution
and placed in an Erlenmeyer flask. An indicator
was added to the flask. At the end of the
titration the student had used 35.0 mL of 0.12 M
NaOH to neutralize the acid. Calculate the
molarity of the acid.

0.035 L NaOH x 0.12 (mol NaOH/1 L) x ( 1 mol acid/1 mol base) = 0.0042 mol acid

M = (0.0042 mol)/(0.015 L) = 0.28 M acid 

Calculate the molarity of 25.0 mL of a 
monoprotic acid if it took 45.50 mL of 0.25 M
KOH to neutralize the acid. 

0.25 (mol/L) x 0.04550 L x (1 mol acid/1 mol KOH) = 0.01338 mol acid

(0.001338 mol acid)/(0.0250 L) = 0.455 M

Key Points
•	Electrolytes (strong, weak, and non) 
•	Precipitation reactions
–	Solubility rules  
•	Molecular, ionic, and net ionic reactions 
•	Acid-base neutralization reactions 
•	Oxidation-reduction reactions

Key Points 
–	Balancing redox reactions by the half reaction method 
–	Various types: decomposition, combination
•	Molarity 
•	Solution stoichiometry 
–	Gravimetric analysis
–	Volumetric analysis