chapter 4: chemical reactions
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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 PresentationTRANSCRIPT
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Vanessa Prasad-PermaulValencia Community College
CHM 1045
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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
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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
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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
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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)
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SOLUBILITY RULES FOR IONIC COMPOUNDS
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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
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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
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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)
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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)
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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)
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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)
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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.
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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)
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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.
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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)
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COMMON ACIDS AND BASES
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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)
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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)
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Brønsted Acid: Can donate protons (H+) to another substance.
Brønsted Base: Can accept protons (H+) from another substance. (NH3)
20
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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
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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)
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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)
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Other Weak bases – trimethyl ammonia N(CH3)3, C5H5N pyridine, ammonium hydroxide NH4OH,
H2O water
STRONG ACIDS AND BASES
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EXERCISE 4.4Label each of the following as a strong or weak acid or base:a)H3PO4
b)HClOc)HClO4
d)Sr(OH)2
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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.
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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
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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).
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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)
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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
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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)
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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)
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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.
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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)
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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
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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
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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
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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
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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
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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
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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
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Some Common Oxidation-Reduction Reactions:
1.Combination Reaction
2.Decomposition Reaction
3.Displacement Reaction
4.Combustion Reaction
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1.Combination Reaction: two substances combine to form a third substance:
2Na(s) + Cl2(g) 2NaCl(s)
2Sb(s) + 3Cl2(g) 2SbCl3(s)
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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
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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)
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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)
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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)
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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)
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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)
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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)]
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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)
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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)
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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
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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?
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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
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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?
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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
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Mi x Vi = Mf x Vf
Where Mi = initial concentration (molarity)
Vi = initial volume Mf = final concentration (molarity)
Vf = final volume
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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?
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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