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Vanessa Prasad-Permaul Valencia Community College CHM 1045

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Chapter 4: CHEMICAL REACTIONS. Vanessa Prasad- Permaul Valencia Community College CHM 1045. IONS IN AQUEOUS SOLUTION. IONIC THEORY OF SOLUTIONS AND SOLUBILITY RULES CERTAIN SUBSTANCES PRODUCE FREELY MOVING IONS WHEN THEY DISSOLVE IN WATER - PowerPoint PPT Presentation

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Page 1: Chapter 4: CHEMICAL REACTIONS

Vanessa Prasad-PermaulValencia Community College

CHM 1045

Page 2: Chapter 4: CHEMICAL REACTIONS

IONIC THEORY OF SOLUTIONS AND SOLUBILITY RULES

CERTAIN SUBSTANCES PRODUCE FREELY MOVING IONS WHEN THEY DISSOLVE IN WATER

THOSE IONS CONDUCT AN ELECTRICAL CURRENT IN AQUEOUS SOLUTION

Page 3: Chapter 4: CHEMICAL REACTIONS

ELECTROLYTE: A SUBSTANCE THAT DISSOLVES IN WATER TO GIVE AN ELECTRICALLY CONDUCTING SOLUTION

In general, ionic solutions are electrolytesNaCl(aq) Na+(aq) + Cl-(aq)

some electrolytes are molecular substancesHCl (aq) H+(aq) + Cl-(aq)

NONELECTROLYTE: A SUBSTANCE THAT DISSOLVES IN WATER TO GIVE A NONCONDUCTING OR VERY POORLY CONDUCTING SOLUTION

Page 4: Chapter 4: CHEMICAL REACTIONS

STRONG ELECTROLYTE: AN ELECTROLYTE THAT EXISTS IN SOLUTION ALMOST ENTIRELY AS IONS

MOSTLY IONIC SUBSTANCES

NaCl(aq) Na+(aq) + Cl-(aq)

WEAK ELECTROLYTE: AN ELECTROLYTE THAT DISSOLVES IN WATER TO GIVE A RELATIVELY SMALL PERCENTAGE OF IONS

GENERALLY MOLECULAR SUBSTANCES

Page 5: Chapter 4: CHEMICAL REACTIONS

WEAK ELECTROLYTES:

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

NH4+(aq) + OH-(aq) NH3(aq) + H2O(l)

NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

Page 6: Chapter 4: CHEMICAL REACTIONS

SOLUBILITY RULES FOR IONIC COMPOUNDS

Page 7: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.1Determine whether the following compounds

aresoluble or insoluble in water:

A) NaBrB) Ba(OH)2

C) CaCO3

D) Hg(NO3)2

E) AgCl

Page 8: Chapter 4: CHEMICAL REACTIONS

a. According to the chart, all compounds that contain sodium, Na+ are soluble, so NaBr is soluble in water.b. According to Table 4.1, most compounds that contain hydroxides, OH, are insoluble in water. However, Ba(OH)2 is listed as one of the exceptions to this rule, so it is soluble in water.c. According to the chart, most compounds that contain

carbonate, CO32−, are insoluble. CaCO3 is

not one of theexceptions, so it is insoluble in water.

d. Mercuric nitrate or Hg(NO3)2 is soluble because all forms of nitrates are soluble.e. AgCl is insoluble. According to the chart, all Cl- containing compounds are soluble except Ag, Hg & Pb

Page 9: Chapter 4: CHEMICAL REACTIONS

MOLECULAR & IONIC EQUATIONS

Molecular equation: a chemical equation in which allcompounds are written as molecules (even if ions):

Ca(OH)2(aq) + Na2CO3(aq) CaCO3(s) + 2NaOH(aq)

Complete Ionic Equations: a chemical equation in which

strong electrolytes are written as separate ions:Ca2+(aq) + 2OH-(aq) + 2Na+(aq) + CO3

2-(aq) CaCO3(s) + 2Na+(aq) +

2OH-(aq)

Page 10: Chapter 4: CHEMICAL REACTIONS

Spectator Ion: an ion that does not take part in an ionic

equation:Ca2+(aq) + 2OH-(aq) + 2Na+(aq) + CO3

2-(aq) CaCO3(s) + 2Na+(aq) +

2OH-(aq)

Net Ionic Equation: an ionic equation from which thespectator ions have been cancelled:

Ca2+(aq) + CO32-(aq) CaCO3(s)

Page 11: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.2Write the complete net ionic equation for each

of thefollowing:2HNO3(aq) + Mg(OH)2(s) 2H2O(l) +

MgNO3(aq)*nitric acid is a strong electrolytePbNO3(aq) + Na2SO4(aq) PbSO4(s) +

2NaNO3(aq)

2HClO4(aq) + Ca(OH)2(aq) Ca(ClO4)2(aq) + 2H2O(l)

*perchloric acid is a strong electrolyte

HC2H3O2(aq) + NaOH(aq) NaC2H3O2(aq) + H2O(l)

Page 12: Chapter 4: CHEMICAL REACTIONS

The resulting complete ionic equation is:2H+(aq) + 2NO3

−(aq) + Mg(OH)2(s) 2H2O(l) + Mg2+(aq) + 2NO3

−(aq)

The corresponding net ionic equation is:2H+(aq) + Mg(OH)2(s) 2H2O(l) + Mg2+(aq)

The resulting complete ionic equation is:Pb2+(aq) + 2NO3

−(aq) + 2Na+(aq) + SO42−(aq)

PbSO4(s) + 2Na+(aq) + 2NO3

−(aq)

The corresponding net ionic equation is:Pb2+(aq) + SO4

2−(aq) PbSO4(s)

Page 13: Chapter 4: CHEMICAL REACTIONS

PRECIPITATION REACTIONS: Two ionic solutions mix and a solid ionic substance (precipitate) forms.

ACID-BASE REACTIONS: an acidic substance reacts with as basic substance. This reaction involves the transfer of a proton between reactants.

Neutralization Acid-Base with Gas Formation

OXIDATION-REDUCTION REACTIONS: a reaction in which there is a transfer of electrons between the reactants.

Page 14: Chapter 4: CHEMICAL REACTIONS

PRECIPITATION REACTIONS: Two ionic solutions mix and a solid ionic substance (precipitate) forms.

Molecular Equation:MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) +

Mg(NO3)2(aq)

Complete Ionic Equation:Mg2+(aq) + 2Cl-(aq) + 2Ag+(aq) + 2NO3

- (aq)

2AgCl(s) + Mg2+(aq) + 2NO3

-(aq)

Net Ionic Equation:

2Cl-(aq) + 2Ag+(aq) 2AgCl(s)

Page 15: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.3• For each of the following, decide whether a

precipitation reaction occurs. If it does, write the balanced molecular equation and then the net ionic equation. If no precipitation reaction occurs, write NPR.

• Aqueous solution of sodium chloride and iron(II) nitrate are mixed

• Aqueous solution of aluminum sulfate and sodium hydroxide are mixed

• You mix an aqueous solution of sodium iodide and lead (II) acetate. If a reaction occurs, write the balanced molecular equation, the complete ionic equation and the net ionic equation.

Page 16: Chapter 4: CHEMICAL REACTIONS

The formulas of the compounds are NaI which is soluble and Pb(C2H3O2)2 is also soluble. Exchanging anions, you get sodium acetate, NaC2H3O2 which is soluble, and lead(II) iodide, PbI2 which is insoluble and will form a precipitate.

The balanced molecular equation is:Pb(C2H3O2)2(aq) + 2NaI(aq) PbI2(s) + 2NaC2H3O2(aq)

To get the net ionic equation, first get the complete ionic equation:Pb2+(aq) + 2C2H3O2

-(aq) + 2Na+(aq) + 2I-(aq) PbI2(s) + 2Na+(aq) +

C2H3O2-(aq)

The final result is:Pb2+(aq) + 2I−(aq) PbI2(s)

Page 17: Chapter 4: CHEMICAL REACTIONS

COMMON ACIDS AND BASES

Page 18: Chapter 4: CHEMICAL REACTIONS

Acid–Base Neutralization: A process in which an acid reacts with a base to yield water plus an ionic compound called a salt.

The driving force of this reaction is the formation of the stable water molecule.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Page 19: Chapter 4: CHEMICAL REACTIONS

Arrhenius Acid: A substance which dissociates in water to form hydrogen ions (H+).

HNO3(aq) H+(aq) + NO3-(aq)

Arrhenius Base: A substance that dissociates in (or reacts with) water to form hydroxide ions (OH–).

NaOH(s) Na+(aq) + OH-(aq)

Limitations: Has to be an aqueous solution and doesn’t account for the basicity of substances like NH3.NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

Page 20: Chapter 4: CHEMICAL REACTIONS

Brønsted Acid: Can donate protons (H+) to another substance.

Brønsted Base: Can accept protons (H+) from another substance. (NH3)

20

Page 21: Chapter 4: CHEMICAL REACTIONS

NH3(aq) + H2O(l) NH4+(aq) + OH-

(aq)

H+

acidbase

HNO3(aq) H+(aq) + NO3-(aq)

HNO3(aq) + H2O(l) H3O+(aq) + NO3-(aq)

HNO3(aq) + H2O(l) H3O+(aq) + NO3-(aq)

H+

acid base

Page 22: Chapter 4: CHEMICAL REACTIONS

Strong acid: strong electrolyte - almost completely dissociates in water:

HCl, H2SO4, HNO3, HClO4, HI HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)

Weak acid: weak electrolyte - does not dissociate well in water:

HF, HCN, CH3CO2H HF(aq) + H2O(l) H3O(aq) + CN-(aq)

Page 23: Chapter 4: CHEMICAL REACTIONS

Strong base: strong electrolyte - almost completely dissociates in water:

Metal hydroxides NaOH(s) Na+(aq) + OH-(aq)

Weak base: does not dissociate well in water:

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Page 24: Chapter 4: CHEMICAL REACTIONS

Other Weak bases – trimethyl ammonia N(CH3)3, C5H5N pyridine, ammonium hydroxide NH4OH,

H2O water

STRONG ACIDS AND BASES

Page 25: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.4Label each of the following as a strong or weak acid or base:a)H3PO4

b)HClOc)HClO4

d)Sr(OH)2

Page 26: Chapter 4: CHEMICAL REACTIONS

a. H3PO4 is not listed as a strong acid in the table so it is a weak acid.

b. Hypochlorous acid, HClO, is not one of the strong acids listed in the table, so we assume that HClO is a weak acid.

c. As noted in the table, HClO4 is a strong acid.

d. As noted in the table, Sr(OH)2 is a strong base.

Page 27: Chapter 4: CHEMICAL REACTIONS

NEUTRALIZATION REACTIONS: A reaction of an acid and a base that results in an ionic compound (salt)and possibly water:

2HCl(aq) + Ca(OH)2(aq) CaCl2(aq) + 2H2O(l)

HCN(aq) + KOH(aq) KCN(aq) + H2O(l)

acid base salt

acid base salt

Page 28: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.5•Write the molecular equation and the complete ionic equation and the net ionic equation for the neutralization of nitrous acid by sodium hydroxide, both in aqueous solution.

•Write the molecular equation and the complete ionic equation and the net ionic equation for the neutralization of hydrocyanic acid by lithium hydroxide (both in aqueous solution).

Page 29: Chapter 4: CHEMICAL REACTIONS

HCN(aq) + LiOH(aq) LiCN(aq) + H2O(l)

Note that LiOH (a strong base) and LiCN (a soluble ionic substance) are strong electrolytes; HCN is a weak electrolyte (it is not one of the strong acids in the table

HCN(aq) + Li+(aq) + OH-(aq) Li+(aq) + CN-(aq) + H2O(l)

After eliminating the spectator ions (Li+ and CN−), the net ionic equation is:

HCN(aq) + OH−(aq) H2O(l) + CN−(aq)

Page 30: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.6Write the molecular equation, the complete ionic equation and the net ionic equation for the successive neutralization of each of the acidic hydrogens of sulfuric acid with potassium hydroxide

Page 31: Chapter 4: CHEMICAL REACTIONS

The first step in the neutralization is described by the following molecular equation:

H2SO4(aq) + KOH(aq) KHSO4(aq) + H2O(l)

The corresponding net ionic equation is:H+(aq) + OH−(aq) H2O(l)

The reaction of the acid salt KHSO4 is given by the following molecular equation:

KHSO4(aq) + KOH(aq) K2SO4(aq) + H2O(l)

The corresponding net ionic equation is:HSO4

−(aq) + OH−(aq) H2O(l) + SO42−(aq)

Page 32: Chapter 4: CHEMICAL REACTIONS

Acid-Base Reactions with Gas Formations

Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)

carbonate acid salt gas

Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2CO3(aq)

CO32-(aq) + 2H+(aq) H2O(l) + CO2(g)

Page 33: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.7

Write the molecular equation and the net ionic equation for the reaction of zinc sulfide with hydrochloric acid.

Write the molecular equation and the net ionic equation for the reaction of calcium carbonate with nitric acid.

Page 34: Chapter 4: CHEMICAL REACTIONS

First, write the molecular equation for the exchange reaction, noting that the products of the reaction would be soluble Ca(NO3)2 and H2CO3. The carbonic acid decomposes to water and carbon dioxide gas. The molecular equation for the process is:CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + H2O(l) +

CO2(g)

The corresponding net ionic equation isCaCO3(s) + 2H+(aq) Ca2+(aq) + H2O(l) +

CO2(g)

Page 35: Chapter 4: CHEMICAL REACTIONS

Oxidation-Reduction Reactions: reactions involving a transfer of electrons from one species to another (or in which atoms change oxidation states).

Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

Net ionic equation:Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)

Iron loses electrons = oxidizedCopper gains electrons = reduced

Page 36: Chapter 4: CHEMICAL REACTIONS

Oxidation Number: the actual charge of the atom if it exists as a monatomic ion (or a hypothetical charge assigned to the atom in the substance by simple rules).

2Ca(s) + O2(g) 2CaO(s) 0 0 +2 -2

2Ca(s) + O2(g) 2CaO(s)

*The concept of oxidation numbers (states) was developed as a simple way of keeping track of electrons in a reactions

Page 37: Chapter 4: CHEMICAL REACTIONS

1. Free elements are assigned an oxidation state of zero. 2. The sum of the oxidation states of all that atoms in a species

must be equal to the net charge on the species. 3. The alkali metals (Li, Na, K, Rb, and Cs) in compounds are

always assigned an oxidation state of +1.

4. Fluorine in compounds is always assigned an oxidation state of -1.

5. The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) and also Zn and

Cd in compounds are always assigned an oxidation state of +2.

6. Hydrogen in compounds is assigned an oxidation state of +1. 7. Oxygen in compounds is assigned an oxidation state of -2. 8. Halogen in compounds is assigned an oxidation state of -1.

Rules for Assigning Oxidation Numbers

Page 38: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.8Use the oxidation number rules to obtain

the oxidation number of the chlorine atom in each of the following:A)HClO4B)ClO3

-

Obtain the oxidation numbers of the atoms ineach of the following:A)Potassium dichromateB)Permanganate ion

Page 39: Chapter 4: CHEMICAL REACTIONS

A) For potassium dichromate, K2Cr2O7:2 x (oxidation number of K) + 2 x (oxidation number of Cr) + 7 x (oxidation number of O) = 0For oxygen, the oxidation number is −2 (rule 3), and for potassium ion, the oxidation number is +1 (rule 2)[2 x (+1)] + 2 x (oxidation number of Cr) + [7 x (−2)] = 0Therefore,2 x oxidation number of Cr = − [2 x (+1)] − [7 x (−2)] = +12or, oxidation number of Cr = +6.B) For the permanganate ion, MnO4

−:(Oxidation number of Mn) + 4 x (oxidation number of O) = −1For oxygen, the oxidation number is −2 (rule 3).(oxidation number of Mn) + [4 x (−2)] = −1Therefore,Oxidation number of Mn = −1 − [4 x (−2)] = +7

Page 40: Chapter 4: CHEMICAL REACTIONS

Redox reactions are those involving the oxidation and reduction of species.

Oxidation and reduction must occur together. They cannot exist alone.

Fe2+ + Cu0 Fe0 + Cu2+

Half-reaction: one of two parts of an oxidation-reduction reaction. One part involves a loss of

electron, the other a gain of electrons:Reduced: Iron gained 2 electrons Fe2+ + 2 e

Fe0

Oxidized: Copper lost 2 electrons Cu0 Cu2+ + 2e

Page 41: Chapter 4: CHEMICAL REACTIONS

Fe2+ + Cu0 Fe0 + Cu2+

Fe2+ gains electrons, is reduced, and we call it an oxidizing agent Oxidizing agent is a species that can gain

electrons and this facilitates in the oxidation of another species. (electron deficient)

Cu0 loses electrons, is oxidized, and we call it a reducing agent Reducing agent is a species that can lose

electrons and this facilitates in the reduction of another species. (electron rich)

reducingagent

oxidizingagent

reduction

oxidation

Page 42: Chapter 4: CHEMICAL REACTIONS

Some Common Oxidation-Reduction Reactions:

1.Combination Reaction

2.Decomposition Reaction

3.Displacement Reaction

4.Combustion Reaction

Page 43: Chapter 4: CHEMICAL REACTIONS

1.Combination Reaction: two substances combine to form a third substance:

2Na(s) + Cl2(g) 2NaCl(s)

2Sb(s) + 3Cl2(g) 2SbCl3(s)

Page 44: Chapter 4: CHEMICAL REACTIONS

2. Decomposition Reaction: a single compound reacts to give two or more substances.

2HgO(s) 2Hg(l) + O2(g)

2KClO3(s) 2KCl(s) + 3O2(g)MnO2

Page 45: Chapter 4: CHEMICAL REACTIONS

3. Displacement Reaction (Single Replacement Reaction): an element reacts with a compound displacing another element from that compound

Cu(s) + 2 AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)

Zn(s) + 2H+ ZnCl2(aq) + H2(g)

Page 46: Chapter 4: CHEMICAL REACTIONS

4. Combustion Reaction: a substance reacts with oxygen, usually with a rapid release of heat to produce a flame.

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)

4Fe(s) + 3O2(g) 2Fe2O3(s)

Page 47: Chapter 4: CHEMICAL REACTIONS

Half-Reaction Method: Allows you to focus on the transfer of electrons. This is important when considering batteries and other aspects of electrochemistry.

The key to this method is to realize that the overall reaction can be broken into two parts, or half-reactions. (oxidation half and reduction half)

Page 48: Chapter 4: CHEMICAL REACTIONS

Balance for an acidic solution:MnO4

–(aq) + Br–

(aq) Mn2+(aq) + Br2(aq)

1. Determine oxidation and reduction half-reactions:

Oxidation half-reaction: Br–(aq) Br20(aq)

Reduction half-reaction: MnO4–(aq) Mn2+(aq)

2. Balance for atoms other than H and O:Oxidation: 2 Br–(aq) Br2(aq)Reduction: MnO4

–(aq) Mn2+(aq)

Page 49: Chapter 4: CHEMICAL REACTIONS

3. Balance for oxygen by adding H2O to the side with less oxygen

Oxidation: 2 Br–(aq) Br2(aq)Reduction: MnO4

–(aq) Mn2+(aq) + 4 H2O(l)

4. Balance for hydrogen by adding H+ to the side with less hydrogens

Oxidation: 2 Br–(aq) Br2(aq)Reduction: MnO4

–(aq) + 8 H+(aq) Mn2+(aq) + 4 H2O(l)

Page 50: Chapter 4: CHEMICAL REACTIONS

5. Balance for charge by adding electrons (e–):

Oxidation: 2 Br–(aq) Br2(aq) + 2 e–

Reduction: MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) +

4 H2O(l)

6. Balance for numbers of electrons by multiplying:

Oxidation: 5[2 Br–(aq) Br2(aq) + 2 e–]Reduction: 2[MnO4

–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4 H2O(l)]

Page 51: Chapter 4: CHEMICAL REACTIONS

7. Combine and cancel to form one equation:

Oxidation: 10 Br–(aq) 5 Br2(aq) + 10 e–

Reduction: 2 MnO4–(aq) + 16 H+(aq) + 10 e– 2 Mn2+

(aq) + 8 H2O(l)

2 MnO4–(aq) + 10 Br–(aq) + 16 H+(aq) 2 Mn2+(aq) + 5

Br2(aq) + 8 H2O(l)

We will not be balancing in basic solutions!! (until CHM 1046)

Page 52: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.9Use the half reaction method to balance the equation:

Ca(s) + Cl2(g) CaCl2(s)

Identify the oxidation states of the elements.Break the reaction into two half-reactions, making sure that both mass and charge are balanced.

Ca Ca2+ + 2e−

Cl2 + 2e− 2Cl−Since each half-reaction has two electrons, it is not necessary to multiply the reactions by any factors to cancel them out. Adding the two half-reactions together and canceling out the electrons, you get:

Ca(s) + Cl2(g) CaCl2(s)

Page 53: Chapter 4: CHEMICAL REACTIONS

SOLUTE: The substance that dissolves in liquid.

SOLVENT: The liquid that the solute dissolves in.

MOLAR CONCENTRATION:

MOLARITY(M) = moles of solute liters of solution

Page 54: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.10: A sample of NaNO3 weighing 0.38-g is placed in a 50.0-mL volumetric flask. The flask is then filled with DI water to mark and carefully mixed. What is the molarity of the solution?

A sample of NaCl weighing 0.0678-g is placed in a 25.0-mL volumetric flask. Enough water is added to dissolve the NaCl, and the flask is filled to mark and carefully mixed. What is the molarity of the solution?

Page 55: Chapter 4: CHEMICAL REACTIONS

Convert mass of NaCl (molar mass, 58.44 g) to moles of NaCl.

Then divide moles of solute by liters of solution. Note that 25.0 mL = 0.0250 L.

0.0678 g NaCl x 1 mol NaCl = 1.160 x 10−3 mol NaCl

58.44g

Molarity = 1.160 x 10-3mol = 0.04641 = 0.0464 M NaCl

0.0250L

Page 56: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.11An experiment calls for the addition to the reaction vessel of 0.814-g of sodium hydroxide(aq). How many mL of 0.150M NaOH should be added?

How many mL of 0.163M NaCL are required to give 0.0958-g of NaCl?

EXERCISE 4.12How many moles of NaCl should be put in a 50.0mL volumetric flask to give 0.15M NaCl solution? How many grams of NaCl is this?

Page 57: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.11: Convert grams of NaCl (molar mass, 58.44 g) to moles NaCl and then to volume of NaCl solution.0.0958 g NaCl x mol NaCl x L NaCl x 1000mL = 10.06

58.44g 0.163mol 1L = 10.1 mL NaCl

EXERCISE 4.12: One (1) liter of solution is equivalent to 0.15 mol NaCl. The amount of NaCl in 50.0 mL of solution is:50.0 mL x 1L x 0.150mol = 0.00750 mol NaCl 1000mL 1LConvert to grams using the molar mass of NaCl (58.44 g/mol).

0.00750 mol NaCl x 58.44g NaCl = 0.438 = 0.44 g NaCl mol

Page 58: Chapter 4: CHEMICAL REACTIONS

Mi x Vi = Mf x Vf

Where Mi = initial concentration (molarity)

Vi = initial volume Mf = final concentration (molarity)

Vf = final volume

Page 59: Chapter 4: CHEMICAL REACTIONS

EXERCISE 4.13There is a solution of 14.8M NH3. How many mL of this solution is needed to give 100.0mL of 1.00M NH3 solution?

There is a solution of 1.5M sulfuric acid. How many mL of this acid is needed to prepare 100.0mL of 0.18M sulfuric acid?

Page 60: Chapter 4: CHEMICAL REACTIONS

Use the rearranged version of the dilution formula from the text to calculate the initial volume of 1.5 M sulfuric acid required:

Mi x Vi = Mf x Vf

1.5M H2SO4 x Vi = 0.18M H2SO4 x 100.0mL

Vi = 0.18M x 100.0mL 1.5M

Vi = 12.0 = 12 mL