1. Reactions in Aqueous Solution
Lecture Presentation
Chapter 4
Reactions in Aqueous Solution
James F. Kirby
Quinnipiac University
Hamden, CT

2. Solutions
Solutions are defined as homogeneous mixtures of two or more pure substances.
The solvent is present in greatest abundance.
All other substances are solutes.
When water is the solvent, the solution is called an aqueous solution.

3. Aqueous Solutions Substances can dissolve in water by different ways:
Ionic Compounds dissolve by dissociation, where water surrounds the separated ions.
Molecular compounds interact with water, but most do NOT dissociate.
Some molecular substances react with water when they dissolve.

4. Electrolytes and Nonelectrolytes
An electrolyte is a substance that dissociates into ions when dissolved in water.
A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

5. Electrolytes
A strong electrolyte dissociates completely when dissolved in water.
A weak electrolyte only dissociates partially when dissolved in water.
A nonelectrolyte does NOT dissociate in water.

6. Strong Electrolyte – 100% dissociation
NaCl (s) Na+ (aq) + Cl- (aq)
H2O
Weak Electrolyte – not completely dissociated
CH3COOH CH3COO- (aq) + H+ (aq)
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7. Ionization of acetic acid
CH3COOH CH3COO- (aq) + H+ (aq)
A reversible reaction. The reaction can occur in both directions.
Acetic acid is a weak electrolyte because its ionization in water is incomplete.
4.1

8. Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. d+ d-
H2O
4.1

9. Nonelectrolyte does not conduct electricity?
No cations (+) and anions (-) in solution
C6H12O6 (s) C6H12O6 (aq)
H2O
4.1

10. Solubility of Ionic Compounds
Not all ionic compounds dissolve in water.
A list of solubility rules is used to decide what combination of ions will dissolve.

11. Electrolytes vs. Solubility
Some substances are not very soluble in water, but the amount that does dissolve dissociates almost completely, therefore making it a strong electrolyte.
Ex. Ca(OH)2
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12. Precipitation Reactions
When two solutions containing soluble salts are mixed, sometimes an insoluble salt will be produced. A salt “falls” out of solution, like snow out of the sky. This solid is called a precipitate.

13. Metathesis (Exchange) Reactions
Metathesis comes from a Greek word that means “to transpose.”
It appears as though the ions in the reactant compounds exchange, or transpose, ions, as seen in the equation below.
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)

14. Completing and Balancing Metathesis Equations
Steps to follow
Use the chemical formulas of the reactants to determine which ions are present.
Write formulas for the products: cation from one reactant, anion from the other. Use charges to write proper subscripts.
Check your solubility rules. If either product is insoluble, a precipitate forms.
Balance the equation.

15. Ways to Write Metathesis Reactions
Molecular equation
Complete ionic equation
Net ionic equation

16. AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
Molecular Equation
The molecular equation lists the reactants and products without indicating the ionic nature of the compounds.
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)

17. Complete Ionic Equation
In the complete ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions.
This more accurately reflects the species that are found in the reaction mixture.
Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq) 
AgCl(s) + K+(aq) + NO3−(aq)

18. Net Ionic Equation
To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
The ions crossed out are called spectator ions, K+ and NO3−, in this example.
The remaining ions are the reactants that form the product—an insoluble salt in a precipitation reaction, as in this example.
Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq) 
AgCl(s) + K+(aq) + NO3−(aq)

19. Writing Net Ionic Equations
Write a balanced molecular equation.
Dissociate all strong electrolytes.
Cross out anything that remains unchanged from the left side to the right side of the equation.
Write the net ionic equation with the species that remain.

20. Acids
The Swedish physicist and chemist S. A. Arrhenius defined acids as substances that increase the concentration of H+ when dissolved in water.
Both the Danish chemist J. N. Brønsted and the British chemist T. M. Lowry defined them as proton donors.

21. Acids HCl H2SO4 HNO3 H2CO3 HF HI H3PO4 Monoprotic Acids
Polyprotic Acids
Diprotic:
H2SO4
H2CO3
Triprotic:
H3PO4
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22. Bases
Arrhenius defined bases as substances that increase the concentration of OH− when dissolved in water.
Brønsted and Lowry defined them as proton acceptors.

23. Strong or Weak?
Strong acids completely dissociate in water; weak acids only partially dissociate.
Strong bases dissociate to metal cations and hydroxide anions in water; weak bases only partially react to produce hydroxide anions.

24. Acid-Base Reactions
In an acid–base reaction, the acid (H2O above) donates a proton (H+) to the base (NH3 above).
Reactions between an acid and a base are called neutralization reactions.
When the base is a metal hydroxide, water and a salt (an ionic compound) are produced.

25. Neutralization Reactions
When a strong acid (like HCl) reacts with a strong base (like NaOH), the net ionic equation is circled below:
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: H+(aq) + OH−(aq)  H2O(l)

26. Gas-Forming Reactions
Many other bases react with H+ to form molecular compounds. Two of these are
Sulfide (S2-)
Carbonate (CO32-)
These anions react with acids to form gases with low solubilities
Ex. 2 HCl (aq) + Na2S (aq) → H2S (g) + 2NaCl (aq)
Net Ionic: 2H+ (aq) + S2- (aq) → H2S (g)

27. Gas-Forming Reactions
When a carbonate or bicarbonate reacts with an acid, the products are a salt, carbon dioxide, and water.
HCl(aq) + NaHCO3 (aq) NaCl(aq) + H2CO3 (aq)
Carbonic acid is unstable, and decomposes to H2O(l) and CO2(g). The overall reaction is written as:
HCl(aq) + NaHCO3(aq) NaCl(aq) + CO2(g) + H2O(l)
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28. Oxidation-Reduction Reactions
Loss of electrons is oxidation.
Gain of electrons is reduction.
One cannot occur without the other.
The reactions are often called redox reactions.

29. Oxidation Numbers
To determine if an oxidation–reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.

30. Rules to Assign Oxidation Numbers
Elements in their elemental form have an oxidation number of zero. (ex. C, Ne, H2)
The oxidation number of a monatomic ion is the same as its charge. (ex. Na+, Mg2+, Br-, O2-)

31. Rules to Assign Oxidation Numbers
Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.
Oxygen has an oxidation number of −2, except in the peroxide ion (O22-), in which it has an oxidation number of −1.
Hydrogen is −1 when bonded to a metal (ex. LiH), +1 when bonded to a nonmetal (ex, HBr).

32. Rules to Assign Oxidation Numbers
Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.
Fluorine always has an oxidation number of −1.
The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, most notably in oxyanions (ex. BrO3-).

33. Rules to Assign Oxidation Numbers
The sum of the oxidation numbers in a neutral compound is zero.
Ex. CO2
The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
Ex. PO43-

34. Displacement Reactions
In displacement reactions, ions oxidize an element.
In this reaction, silver ions oxidize copper metal:
Copper was oxidized
Silver was reduced
(silver takes electrons from copper)
Cu(s) + 2 AgNO3(aq) → Cu(NO3)2 (aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
The reverse reaction does NOT occur. Why not?

35. Activity Series
Elements higher on the activity series are more reactive.
They are more likely to exist as ions.

36. Practice Problem #1:
Will an aqueous solution of iron (II) chloride oxidize magnesium metal? If so, write the balanced molecular and net ionic equations for the reaction.
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37. Practice Problem #2
Does a reaction occur when an aqueous solution of NiCl2(aq) is added to a test tube containing strips of zinc metal? If so write the molecular and net ionic equations.
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38. Practice Problem #3
Will a reaction occur when solid magnesium is added to a solution of hydrochloric acid? If so, write the molecular and net ionic equations.
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39. Metal/Acid Displacement Reactions
The elements above hydrogen will react with acids to produce hydrogen gas.
The metal is oxidized to a cation.

40. volume of solution in liters
Molarity
The quantity of solute in a solution can matter to a chemist.
We call the amount dissolved its concentration.
Molarity is one way to measure the concentration of a solution:
moles of solute
volume of solution in liters
Molarity (M) =

41. Mixing a Solution
To create a solution of a known molarity, weigh out a known mass (and, therefore, number of moles) of the solute.
Then add solute to a volumetric flask, and add solvent to the line on the neck of the flask.

42. Practice Problems
Calculate the molarity of a solution made by dissolving 23.4g of Na2SO4 in enough water to make 125 mL of solution.
What is the molar concentration of each ion present in a M aqueous solution of calcium nitrate?
What volume of 0.30 M HNO3 solution is required to supply 2.0 mol of HNO3?
How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4 solution.
How many milliliters of 0.50 M Na2SO4 solution are needed to provide mol of the salt?

43. Dilution One can also dilute a more concentrated solution by
using a pipet to deliver a volume of the solution to a new volumetric flask, and
adding solvent to the line on the neck of the new flask.

44. Dilution
The molarity of the new solution can be determined from the equation
Mc  Vc = Md  Vd,
where Mc and Md are the molarity of the concentrated and dilute solutions, respectively, and Vc and Vd are the volumes of the two solutions.

45. Practice Problems
1) How many milliliters of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M H2SO4?
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46. Using Molarities in Stoichiometric Calculations

47. Practice Problems
What volume of 0.250M HCl is required to react completely with mL of M NaOH?
How many grams of silver chromate will precipitate when mL of M silver nitrate are added to mL of M potassium chromate?
2 AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2 KNO3(aq)

48. Titration
A titration is an analytical technique in which one can calculate the concentration of a solute in a solution.

49. Titration
A solution of known concentration, called a standard solution, is used to determine the unknown concentration of another solution.
The reaction is complete at the equivalence point.

50. Titration Practice
If 45.7 mL of M H2SO4 is required to neutralize 20.0 mL of NaOH solution, what is the concentration of the NaOH?
What is the molarity of an HCl solution if 27.3 mL of it neutralizes mL of M Ba(OH)2?
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51. Gravimetric Analysis Dissolve unknown substance in water
React unknown with known substance to form a precipitate
Filter and dry precipitate
Weigh precipitate
Use chemical formula and mass of precipitate to determine amount of unknown ion
4.6

52. % Cl = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑋𝐶𝑙 𝑥 100
A g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with excess AgNO3. If g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound? XCl(aq) + AgNO3 (aq) → AgCl + XNO3
g g
% Cl = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑋𝐶𝑙 𝑥 100
Given: mass of XCl
Find: mass of Cl
What do we know?
The only source of chloride ions is the original compound
These ions end up in the precipitate
2. What do we need to know?
a. mass of chloride ions in the precipitate, AgCl

53. 3. How can we figure this out?
a. Find the % of Cl in AgCl and calculate the mass of chloride ions in the g AgCl precipitate
b. This will be the mass of chloride ions in the original ionic compound.
c. Use this and the mass of the ionic compound ( g) to calculate the percent.
% Cl in AgCl = 𝑔 𝐶𝑙 𝑔 𝐴𝑔𝐶𝑙 x 100
= 24.72%
Mass of Cl in g of AgCl = x g
= g Cl
This will also be the mass of Cl in “XCl”
% Cl in “XCl” = 𝑔 𝐶𝑙 𝑔 𝑋𝐶𝑙 x 100 = 47.51%