Chapter 4 Page 1

CHAPTER FOUR: CHEMICAL REACTIONS

Part One: Ions in Aqueous Solution A. Electrolytes and Non-electrolytes.

1. Pure water does not conduct electric current appreciably. It is the ions dissolved in the

water that conduct charge.

2. Electrolyte = substance that dissolves in water forming ions

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3. Classification of water-soluble substances.

Water-soluble Substances

Electrolytes Nonelectrolytes -dissociate to some extent -do not dissociate into in water to form ions ions in water -solution conducts electricity -solution is non-conductive Strong Weak Electrolytes Electrolytes -dissociate to large extent -dissociate to small extent -conduct well -conduct slightly HAc (acetic acid) strong strong most soluble acids bases salt HCl, HNO3, NaOH NaCl, CaCl2 H2SO4

a. Dissociation = process in which a solid ionic compound separates into ions in

solution.

NaCl H2O → Na+ +Cl −

b. Ionization = process in which a molecular compound separates to form ions.

HCl H2O → H+ + Cl− c. When one of the ions is H+, the substance is an acid.

HCl H2O → H+ + Cl−

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d. When OH- are produced, substance is a base.

NaOH H2O → Na+ +OH−

e. When neither ion is H+ or OH-, substance is a salt.

KBr H2O → K+ +Br −

B. Strong and weak acids. 1. Strong acids ionize completely in dilute aqueous solution. Example: HCl is a gas when pure, dissolves to large extent in water, and fully ionizes.

HCl(g) H2O → HCl(aq) H2O → H+ (aq) +Cl−(aq)

2. Know these:

3. H+ ions exist in aqueous solution attached to H2O (i.e., hydrated).

H+ (aq) really means H3O+ H + H O hydronium ion H

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4. Weak acids ionize only slightly. Example: acetic acid

CH3COOH(aq) H+ (aq) + CH3COO-(aq)

dominant form ≈ 99.5% ≈ 0.5% Abbreviation: HAc(aq) H+ (aq) + Ac- (aq)

5. Know these:

6. Note that we have introduced the idea of reversible reactions. ( ) Reaction is continually happening to appreciable extent in both directions to produce a

balance in which both Reactants and Products are present.

C. Bases = most commonly metal hydroxides. (Mx(OH)y)

1. Strong soluble bases:

NaOH(s) H2O → Na+(aq) +OH−(aq) dissociate completely

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a. Insoluble bases: metal hydroxides which have low solubility in H2O, produce low

amount of OH-(aq).

Cu(OH)2, Zn(OH)2, Fe(OH)2, Fe(OH)3

b. Weak bases: soluble in H2O but ionize only slightly. Example:

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

molecular compound

D. Solubility rules for ionic compounds in aqueous solution.

1. Compounds of Group IA and NH4

+ are soluble. 2. Nitrates, NO3

-; acetates, CH3COO-; chlorates, ClO3-; and perchlorates, ClO4

-, are soluble.

3. Chlorides, Cl-, are soluble EXCEPT AgCl, Hg2Cl2, and PbCl2. Bromides, Br-, and

iodides, I-, similar to chlorides, with some exceptions. As these halide ions (Cl-, Br-, I-) increase in size, solubilities decrease.

4. Sulfates, SO4

2-, are soluble EXCEPT PbSO4, BaSO4, HgSO4, CaSO4 , SrSO4 and Ag2SO4

5. Carbonates, CO32-, are insoluble EXCEPT those of the Group IA metals and NH4

+. MgCO3 is moderately soluble.

6. Phosphates, PO4

3-, and arsenates, AsO43-, are insoluble EXCEPT those of the Group

IA metals and NH4+.

7. Sulfides, S2-, are insoluble EXCEPT Group IA and IIA metals and NH4

+. 8. Metal hydroxides, OH-, are insoluble EXCEPT Group IA metals and Group IIA

metals below Mg.

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E. Writing Reactions in Aqueous Solution. (Section 4.2) 1. Three ways of writing equations. (illustrated by neutralization of Mg(OH)2 with HCl

in aqueous solution. a. Molecular equations = show complete formulas for all compounds. (best for

stoichiometric calculations)

2HCl(aq) + Mg(OH)2(s) → MgCl2 (aq) + 2H2O (insoluble) (soluble)

b. Complete ionic equations = show predominant form of compounds as ions if they

strongly dissociate. (the bridge to the net ionic equation)

2[H+(aq) + Cl-(aq)] + Mg(OH)2(s) → [Mg2+ (aq) + 2Cl-(aq)] + 2H2O

c. Net ionic equation = complete ionic equation after “spectator ions” have been canceled out.

2H+(aq) + Mg(OH)2(s) → Mg+ (aq) + 2H2O

[This is the essence of the reaction.] 2. To get the net ionic equation:

a. Write down the full molecular equation first. b. For each compound, ask: Is it a strong electrolyte? If not, keep full formula. For

example, AgCl(s). c. Cancel out ions that appear on both sides. These are “spectator ions.”

3. Problem: Write down all 3 types of equations for the reaction of sodium sulfide with silver nitrate to produce silver sulfide and sodium nitrate. a. Molecular equation:

Na2S(aq) + 2AgNO3(aq) → Ag2S(s) + 2NaNO3(aq)

↑ ↑ ↑ all Na salts all nitrates insoluble soluble soluble sulfide

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b. Complete ionic equation:

[2Na+(aq) + S2-(aq)] + 2[Ag+(aq) + NO3-(aq)] →

Ag2S(s) + 2[Na+(aq) + NO3-(aq)]

c. Net ionic equation:

S2-(aq) + 2Ag+(aq) → Ag2S(s)

(Simple formation of insoluble silver sulfide.)

Part Two: Types of Chemical Reactions There are millions. We will look at three important types: a) precipitation; b) acid-base; and c)

oxidation-reduction.

Three driving forces for ionic reaction to happen, all in which there is a removal of ions from solution: a. Formation of insoluble solid precipitate. (e.g. Ag2S) b. Formation of a very stable nonionized molecule. (e.g. H2O) c. Formation of a gas that escapes. (e.g., H2)

A. Precipitation Reactions. (Metathesis Reaction I) (Section 4.3)

1. An insoluble solid (precipitate) forms and then settles out of solution. 2. Driving force is strong attraction between cations and anions, resulting in removal of

ions from solution. 3. Example: addition of sodium phosphate solution to a calcium chloride solution. What

happens? Pictorially:

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Na+ Ca2+ Na+ Ca2+ + PO4

3- Cl- PO43- Cl-

Na+ Cl-

solid Ca3(PO4)2 Ca3(PO4)2 ppts!

2Na3PO4 (aq) + 3CaCl2(aq) → Ca3(PO4)2(s) + 6NaCl(aq) Net ionic equation:

2PO43- (aq) + 3Ca2+(aq) → Ca3(PO4)2(s)

4. Additional example: Potassium chromate solution added to lead nitrate solution. What

happens? K+ Pb2+ soluble insoluble CrO4

2- NO3-

K2CrO4(aq) + Pb(NO3)2(aq) → PbCrO4(s) + 2KNO3(aq)

Net ionic equation:

CrO42-(aq) + Pb2+(aq) → PbCrO4(s)

yellow solid

B. Acid/Base Reactions. (Metathesis Reaction II) (Section 4.4)

1. General form of neutralization between acid and base:

acid + base → salt + H2O

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2. Driving force here is formation of H2O from H+ and OH-. 3. Example - strong acid + strong base:

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

Total ionic reaction:

[H+(aq) + Cl-(aq)] + [Na+(aq) + OH-(aq)] → [Na+(aq) + Cl-(aq)] + H2O

Net ionic reaction for strong acid + strong base:

H+(aq) + OH-(aq) → H2O(l)

4. Example - weak acid + strong base:

HAc(aq) + NaOH(aq) → NaAc(aq) + H2O(l)

Total ionic reaction:

HAc(aq) + [Na+(aq) + OH-(aq)] → [Na+(aq) + Ac-(aq)] + H2O(l) Net ionic reaction for weak acid + strong base:

HAc(aq) + OH-(aq) → Ac-(aq) + H2O(l)

5. Example of Acid-Base reaction with Formation of a Gas (a Driving Force for Reaction)

reaction of insoluble carbonate salt with strong acid:

2HCl(aq) + CaCO3(s) → H2CO3(aq) + CaCl2(aq) limestone carbonic acid seashells (weak acid) chalk

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Net ionic reaction:

2H+(aq) + CaCO3(s) → H2CO3(aq) + Ca2+(aq) i.e. calcium carbonate dissolves in acid. driving force: further decomposition of H2CO3 to CO2(g)

H2CO3(aq) → CO2(g) + H2O(l) escapes (origin of bubbles)

C. Oxidation-Reduction (Redox) Reactions. (introduction) (Section 4.5)

1. Many reactions involve transfer of electrons between species, called oxidation-reduction reactions, abbreviated redox reactions.

2. To recognize these, we first need to introduce a concept called the oxidation number

or oxidation state. 3. “Oxidation” originally referred to combining with Oxygen, resulting in increase in

oxidation number. D. Oxidation Numbers. (keeping track of electrons)

1. Oxidation number (ox#), or oxidation state, of an element:

a. In an ionic compound, oxidation number of an element = number of e- gained or

lost by an atom of that element. Example: NaCl compound consists of Na+ and Cl- ion pair.

ox# of Na = +1 Na+ said to be in “+1 oxidation state” ox# of Cl = -1 CaCO3 ox# of Ca = +2

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b. In a molecular compound, ox# is assigned according to rules that aid in balancing redox equations. Meaning is different.

Example: CO2 ox# of O is -2 ox# of C is +4 (+4)(-2) CO2 Example: CO ox# of O is -2 ox# of C is +2 (+2)(-2) CO C is said to be less fully oxidized to CO than in CO2.

2. Rules for assigning ox#’s:

a. Ox#’s always assigned on a per atom basis. b. Ox#’s of any free uncombined element is zero. Examples: O2, S8, Cu(s)

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c. Ox#’s of an element in a simple monatomic ion equals the charge of the ion. Na+, Na is +1 d. Sum of ox#’s in a compound = 0. e. Sum of ox#’s in polyatomic ion = charge of ion.

CO3

-2 C is +4; O is -2 sum = +4 + 3(-2) = -2

f. H is always +1 except when combined with a metal in a metal hydride: NaH,

CaH2.

(+1)(-1) NaH

g. Group IA are always +1. h. Group IIA are always +2. i. Group IIIA are always +3, except in exotic compounds. j. Group VA are -3 in binary compound with a metal or with hydrogen. (+1)(-3) Na3N -sodium nitride (-3)(+1) NH3 -ammonium Otherwise, most often +3 or +5.

k. Oxygen is -2 except in OF2 (+2) and in peroxides. (-1). (+1)(-1) H2O2 -example of exception (+1)(-1) Na2O2 -sodium peroxide l. Halogens are -1 in binary compounds with metals or H.

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3. Examples of assigning ox# in some compounds. (Hint: 1st do elements having hard and fast rules, then others by elimination.) a. Sodium hypochlorite (Clorox bleach): NaOCl (+1) -do Group IA first always; Na (-2) -do O next; O -Cl has to be what to sum to zero? +1 (+1) (-2)(+1) Na O Cl

After bleach does its thing, Cl ends up as NaCl. Cl is -1 there. b. Potassium chromate: K2CrO4 (+1) -do Group IA first always; K (-2) -do O; O -Cr has to be what to sum to zero? +6 (+1) (+6)(-2) K2 Cr O4 c. Sulfuric acid: H2SO4 H is +1, except in hydrides. O is -2, except in peroxides and OF2 S is what to make sum = 0? +6 (+1)(+6)(-2) H2 S O4 d. Sulfurous acid: H2SO3 (+1)(+4)(-2) H2 S O3 Note: S is (+4) here and (+6) in sulfuric acid.

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4. “Oxidation” originally referred to combining with Oxygen, resulting in increase in

ox#.

Example - Combustion of methane is an oxidation of the Carbon:

(-4)(+1) (0) (+4)(-2) CH4 + 2O2 → CO2 + 2H2O

(-4) (+4) {C} → {C} oxidation of C 5. Oxidation = algebraic increase in ox# with corresp. loss of e-. 6. Reduction = algebraic decrease in ox# with corresp. gain of e-. 7. Oxidation and reduction always occur simultaneously.

(0) (0) (+1)(-2) 2H2 + O2 → 2 H2O

Oxygen (0) → (-2) reduced

Hydrogen (0) → (+1) oxidized 8. Oxidizing agent:

a. Oxidizes other species.

b. Is itself reduced. 9. Example: Immerse copper wire in silver nitrate solution. Copper is dissolved as the

Cupric ion and silver is deposited as metal. AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + Ag(s)

Net ionic equation: Ag+(aq) + Cu(s) → Cu2+(aq) + Ag(s) What species is oxidized? Cu (0) → (+2) What species is reduced? Ag (+1) → (0) What is the oxidizing agent? AgNO3 or Ag+(aq)

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E. Many Oxidation-Reduction Reactions also fit under other categories, such as:

1. Combination reactions. 2. Decomposition reactions. 3. Displacement reactions. 4. Combustion reactions.

F. Combination Reactions.

1. Combination reaction = reaction in which 2 or more substances combine to form a third substance.

2. Element + Element Compound

a. 2 Mg(s) + O2 2 MgO (metal + nonmetal ionic cmpd) b. P4(s) + 6 Cl2(g) 4 PCl3(s) (nonmetal+nonmetal molec cmpd)

3. Compound + Element Compound

PCl3(s) + Cl2(g) PCl5(s) 4. Compound + Compound Compound

CaO(s) + CO2(g) CaCO3(s) (this one is NOT redox)

G. Decomposition Reactions.

1. Example: 2 KClO3 2 KCl + 3O2

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H. Displacement Reactions. 1. Reaction in which one element displaces another from a compound.

M + AX → MX + A 2. The more readily a metal M displaces another element and forms a positive ion, the

more active that metal is. 3. The metal doing the displacing is being oxidized. 4. Table 4.6 is Activity Series:

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5. [More Active Metal + → [Less Active Metal + Salt of Less Active Metal] Salt of More Active Metal] Example: Zinc metal is placed in copper(II) sulfate solution. Copper metal falls to

bottom. Resulting solution contains zinc sulfate, ZnSO4. Formula unit equation:

CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq) Net ionic equation:

Cu2+ (aq) + Zn(s) → Cu(s) + Zn2+(aq) Displacement reaction: the more active metal, zinc, displaces ions of less active metal,

copper, from aqueous solution. 6. [Active Metal + Nonoxidizing Acid] → [Hydrogen + Salt of Acid]

a. Example:

Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) strong acid soluble salt Net ionic equation:

Zn(s) + H+(aq) + HSO4-(aq) → Zn2+(aq) + SO4

2-(aq) + H2(g) b. HNO3 is the common oxidizing acid. It reacts with active metals to produce

oxides of nitrogen, but not hydrogen, H2. c. Very active metals can even displace hydrogen from water. Example:

2K(s) + 2H2O(l) → 2[K+(aq) + OH-(aq)] + H2(g) 7. [Active Nonmetal + → [Less Active Nonmetal + Salt of Less Active Nonmetal] Salt of More Active Nonmetal]

a. Example:

Cl2(g) + 2[Na+(aq) + Br-(aq)] → 2[Na+(aq) + Cl-(aq)] + Br2(l) chlorine sodium bromide sodium chloride bromine

Chapter 4 Page 18

b. Each halogen will displace less active (heavier) halogens from their binary salts;

i.e., the order of increasing activities is:

(least active) I2 < Br2 < Cl2 < F2 (most active) I. Combustion Reactions.

1. Combustion reaction = a reaction in which a substance reacts with oxygen, sometimes

with rapid release of heat to product a flame. 2. Oxygen goes from 0 to -2 oxidation state. 3. Of the combusted material, all Carbon ends up as CO or CO2, all Hydrogen ends up in

H2O. 4. Example: Combustion of propane: ___ C3H8(g) + ___ O2(g)

J. Balancing Simple Redox Reactions. 1. The following redox reaction looks balanced, but what is wrong? Zn(s) + Ag+(aq) Zn2+(aq) + Ag(s) The charge is not balanced. 2. Balance using the Half-Reaction method: (separate into two hypothetical half-reactions, one an oxidation, the other a reduction,

by specifically putting electrons in the equations. Zn(s) Zn2+(aq) + 2 e- oxidation 2[Ag+(aq) + 1 e- Ag(s) ] reduction Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s) 3. More complicated cases are handled in Chapter 19, in connection with

electrochemical cells.

Chapter 4 Page 19

Part Three: Working with Solutions A. Molar Concentrations of Solution. (Section 4.7)

1. Many reactions carried out in solution for convenience. 2. Solution = homogeneous mixture of 2 or more substances. 3. Solvent = the dispersing medium. (substance usually present in greatest abundance) 4. Solute = substance dissolved in the solvent. (usually present in least amount) 5. Aqueous solutions most common. (water is solvent) 6. Concentrations expressed in various ways:

a. Mass Percent of solute:

% solute =

€

mass of solutemass of solution

× 100%

b. Molarity (M) or molar concentration:

molarity =

€

number of moles of solutenumber of liters of solution

B. Mass Percent of Solute.

1. Problem involving percent solute: A saline solution is 5.00% NaCl. Calculate the mass of NaCl present in 400 g of

solution.

5.00% means →

€

5.00 g NaCl100 g solution

€

5.00 g NaCl100 g solution

× 400g solution = 20.0g NaCl present

€

% solute( )100

× mass solution = mass solute

Chapter 4 Page 20

2. Variation involving density: A saline solution is 5.00% NaCl. Calculate the mass of NaCl present in 400 mL of

solution. The density of the solution is 1.10 g/mL. 400 mL solution x 1.10 g/mL = 440 g solution Then work like previous example:

€

5.00 g NaCl100 g solution

× 440g solution = 22.0g NaCl present

3. Another variation: What volume of a 30% glucose solution contains 50 grams of

glucose? The density of solution is 1.15 g/mL. 50 g glucose ---> vol of solution required

€

50g glu × 100g soln30g glu

×1mL soln

1.15g soln=145mL soln

€

100g soln30g glu is the % solute by mass

€

1mL soln1.15g soln is the density of solution

C. Molarity, or Molar Concentration (M).

1. Definition:

molarity =

€

moles of soluteLiters of solution

2. Preparation of solutions of desired concentration:

Chapter 4 Page 21

What mass of NaOH would you weigh out to prepare 2.00 L of 0.100 M NaOH? 0.100 M reads 0.100 molar meaning 0.100 moles NaOH per Liter of solution.

€

0.100 moles NaOH1 Liter of solution

× 2.00 L solution = 0.200 moles NaOH

0.200 moles NaOH

€

×40.0 g NaOHmol NaOH

= 8.0 g NaOH

Figure 4.18

3. Important relations:

€

M =nV

molarity = moles soluteVolume of solution

so: n = MV or: (moles of solute present) = (conc in molarity) x (Volume in Liters) 4. Problem: What mass of HCl is present in a 2.5 L jug of concentrated HCl solution

(11.6 M)?

a. First compute moles HCl present: moles = conc x volume = 11.6 mol/L x 2.5 L = 29.0 mol

Chapter 4 Page 22

b. Now convert to grams HCl: 29.0 mol x 36.45 g HCl/mol = 1,057 g HCl

5. Problem: Calculate the molarity of a solution in which 10.0 g NaOH are dissolved in 4.000 L solution.

€

molarity = moles NaOHVol solution (L)

=

10.0 g40.0 g/mol

4.000 L= 0.0625 mol/L= 0.0625 M

6. Commercial phosphoric acid is 85% H3PO4 by mass, and its specific gravity is 1.69.

Calculate its molarity.

€

molarity = moles H3PO4

Vol solution (L)

85% H3PO4 =85 g H3PO4

100g soln × 1.69 g soln

1 mL soln×

103mLL

×1 mol H3PO4 97.9 g H3PO4

=14.7 mol/L

D. Dilution of Solutions. (Section 4.8)

1. Dilution is simply the addition of solvent to an existing solution. Moles of solute

remains fixed. 2. Therefore:

moles solute before = moles solute after

n1 = n2

M1 V1= M2 V2

molarity x volume = molarity x volume before before after after

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3. Since V is on both sides, can use V in any units as long as consistent. 4. Dilution problem: If 10.0 mL of 3.0 M NaCl is diluted to 250.0 mL, what is the new molarity of NaCl

solution?

V1M1 = V2M2

(10.0 mL) x (3.0 M) = (250.0 mL) x M2

M2 =

€

10.0 mL × 3.0 M250 mL

= 0.12 M

E. Volumetric Analysis (Section 4.10)

(Solution Stoichiometry) 1. Carrying out reactions in solution: Problem: What volume of 0.100 M HCl(aq) would exactly neutralize 400 mL of

0.200 M NaOH(aq)? moles HCl = moles NaOH (at neutralization) nHCl = nNaOH MHCl VHCl = MNaOH VNaOH 0.100M VHCl = 0.200M × 0.400 L

VHCl =

€

0.200 M × 0.400 L0.100 M

= 0.800 L = 800 mL

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F. Method of Titration: 1. Titration = process in which a solution of one reactant (the titrant) is carefully added

to a solution of another reactant. Volume of titrant required for complete reaction is measured.

2. Complete reaction is determined by a color change in the indicator.

a. Example: the indicator phenolphthalein is colorless in acidic solution and pink in

basic solution. b. The indicator is placed in acidic reactant solution. The base is then added until

the first drop produces a pink color that persists. c. When stoichiometric amounts have reacted, we call this the end point or

equivalence point of the titration. d. Example: reacting H2SO4 with NaOH, the equivalence point is when:

moles of NaOH added = 2 x (moles of H2SO4 present) e. Example: reacting HCl with NaOH, equivalence point is when: moles of HCl = moles of NaOH

Chapter 4 Page 25

3. Problem: 20.0 mL of HCl solution are required to neutralize 10.0 g of NaOH. What is

the molarity of the HCl solution?

4. In the neutralization of phosphoric acid with sodium hydroxide, what volume of 0.200

M NaOH is required to neutralize 400 mL of 0.50 M H3PO4? 1 mole H3PO4 reacts with 3 mol NaOH. 0.400 L H3PO4 x 0.50 mol/L H3PO4 = 0.200 mol H3PO4 Requires 3 x 0.200 mol = 0.600 mol NaOH 0.600 mol = MV = 0.200 M x V

V =

€

0.600 mol0.200 M

= 3.00 L

Chapter 4 Page 26

5. Problem: 50.0 mL of H2SO4 of unknown concentration is titrated with a 0.100 M NaOH titrant solution. The end point is reached after 40.0 mL of titrant are added. Calculate molarity of H2SO4 solution.

H2SO4 + 2 NaOH → Na2SO4 + H2O end point when:

(moles of NaOH) = 2 x (moles of H2SO4)

VbMb = 2 x (VaMa)

40.0 mL x 0.100 M = 2 x (50.0 mL x Ma)

Ma = 0.0400 M H2SO4 6. In the neutralization of 2.00 g of MgO with HCl, 25.0 mL of an HCl solution are

required. What is the molarity of the HCl solution?

2 HCl + MgO MgCl2 + H2O