Types of Chemical Reactions & Solution Stoichiometry

Chapter 4Chapter 4

Aqueous Solutions

Water is the dissolving Water is the dissolving medium, or medium, or solventsolvent..

Some Properties of Water

Water is able to dissolve so many substances Water is able to dissolve so many substances because:because:

- Water is “bent” or Water is “bent” or V-shapedV-shaped..- The O-H bonds are The O-H bonds are covalentcovalent..- Water is a Water is a polarpolar molecule. molecule.- HydrationHydration occurs when salts dissolve in occurs when salts dissolve in

water.water.

04_40

H

O2

105

H

Water is a polar molecule because it is a bentmolecule. The hydrogen end is + while the oxygenend is -, Delta () is a partial charge--less than 1.

A Solute

- dissolves in water (or other “solvent”)dissolves in water (or other “solvent”)

- changes phase changes phase (if different from the (if different from the solvent)solvent)

- is present in is present in lesserlesser amount (if the same amount (if the same phase as the solvent)phase as the solvent)

A Solvent

- retains its phase retains its phase (if different from the (if different from the solute)solute)

- is present in is present in greatergreater amount (if the same amount (if the same phase as the solute)phase as the solute)

04_41

+

+

++

+

+

+

+

++

+

OH

H

O HH

+

+

Cation

Anion

Polar water molecules interact with the positiveand negative ions of a salt, assisting in the dissolving process. This process is called hydration.

Solubility

The general rule for solubility is:The general rule for solubility is:

““Like dissolves like.”Like dissolves like.”

Polar water molecules can dissolve other polar Polar water molecules can dissolve other polar molecules such as alcohol and, also, ionic molecules such as alcohol and, also, ionic substances such as NaCl.substances such as NaCl.

Nonpolar molecules can dissolve other Nonpolar molecules can dissolve other nonpolar molecules but not polar or ionic nonpolar molecules but not polar or ionic substances. Gasoline can dissolve grease.substances. Gasoline can dissolve grease.

Miscibility

MiscibleMiscible -- two substances that will mix -- two substances that will mix together in any proportion to make a together in any proportion to make a solution. Alcohol and water are miscible solution. Alcohol and water are miscible because they are both polar and form because they are both polar and form hydrogen bonds.hydrogen bonds.

ImmiscibleImmiscible -- two substances that will not -- two substances that will not dissolve in each other. Oil and vinegar are dissolve in each other. Oil and vinegar are immiscible because oil is nonpolar and immiscible because oil is nonpolar and vinegar is polar.vinegar is polar.

Solubility

How does the rule “Like dissolves like.” How does the rule “Like dissolves like.” apply to cleaning paint brushes used for apply to cleaning paint brushes used for latex paint as opposed to those used with latex paint as opposed to those used with oil-based paint? oil-based paint?

Simple Rules for Solubility1.1. Most nitrate (NOMost nitrate (NO33

) salts ) salts are solubleare soluble..

2.2. Most alkali (group 1A) salts and NHMost alkali (group 1A) salts and NH44++ are solubleare soluble..

3.3. Most ClMost Cl, Br, Br, and I, and I salts salts are soluble are soluble ((NOTNOT AgAg++, Pb, Pb2+2+, , HgHg22

2+2+))

4.4. Most sulfate salts Most sulfate salts are soluble are soluble ((NOTNOT BaSOBaSO44, PbSO, PbSO44, ,

HgSOHgSO44, CaSO, CaSO44))

5.5. Most OHMost OH salts salts are only slightly soluble are only slightly soluble (NaOH, KOH (NaOH, KOH are soluble, Ba(OH)are soluble, Ba(OH)22, Ca(OH), Ca(OH)22 are marginally soluble) are marginally soluble)

6.6. Most SMost S22, CO, CO3322, CrO, CrO44

22, PO, PO4433 salts salts are only slightly are only slightly

solublesoluble. .

Electrolytes & Nonelectrolytes

An electrolyte is a material that dissolves in An electrolyte is a material that dissolves in water to give a solution that conducts an water to give a solution that conducts an electric current.electric current.

A nonelectrolyte is a substance which, when A nonelectrolyte is a substance which, when dissolved in water, gives a nonconducting dissolved in water, gives a nonconducting solution.solution.

ElectrolytesStrongStrong - conduct current efficiently and are - conduct current efficiently and are

soluble salts, strong acids, and strong bases.soluble salts, strong acids, and strong bases.

NaCl, KNONaCl, KNO33, HNO, HNO33, NaOH, NaOH

Weak Weak - conduct only a small current and are - conduct only a small current and are

weak acids and weak bases.weak acids and weak bases.

HCHC22HH33OO22, aq. NH, aq. NH33, tap H, tap H22OO

NonNon - no current flows and are molecular - no current flows and are molecular

substancessubstances

pure Hpure H22O, sugar solution, glycerolO, sugar solution, glycerol

04_43 Power Source

(a) (b) (c)

+

+

+

+

+

+

Electrical conductivity of aqueous solutions. a) strong electrolyte b) weak electrolyte c) nonelectrolyte in solution.Svante Arrhenius first identified these electrical properties.

04_1529

= Ba2+

= Cl

BaCl2(s)dissolves

When BaCl2 dissolves, the Ba2+ and Cl- ions are randomlydispersed in the water. BaCl2 is a strong electrolyte.

Acids

Strong acids Strong acids -- dissociate completely (~100 %) to dissociate completely (~100 %) to produce Hproduce H++ in solution in solution

HCl, HHCl, H22SOSO44, HNO, HNO33, HBr, HI, & HClO, HBr, HI, & HClO44

Weak acids Weak acids - dissociate to a slight extent (~ 1 %) - dissociate to a slight extent (~ 1 %) to give Hto give H++ in solution in solution

HCHC22HH33OO22, HCOOH, HNO, HCOOH, HNO22, & H, & H22SOSO33

04_1530

= H+

+

+

+ +

+

+

+

+

+ +

= Cl

+

HCl is completely ionized and is a strong electrolyte.

BasesStrong bases Strong bases - react completely with water to - react completely with water to give OHgive OH ions. ions. sodium hydroxidesodium hydroxide

NaOHNaOH(s)(s) ---> Na ---> Na++(aq)(aq) + OH + OH--

(aq)(aq)

Weak bases Weak bases - react only slightly with water to - react only slightly with water to give OHgive OH ions. ions. ammoniaammonia

NHNH3(aq)3(aq) + HOH + HOH(l)(l) <---> NH <---> NH44++

(aq)(aq) + + OHOH--(aq)(aq)

04_1531

+

+

+

+

+

++

+

+

+

+

- = OH

= Na+

An aqueous solution of sodium hydroxide which isa strong bases dissociating almost 100 %.

04_1532

Acetic acid(CH3COOH) exists in water mostly as undissociatedmolecules. Only a small percent of the molecules are ionized.

Molarity

Molarity (Molarity (MM) = moles of solute per volume of ) = moles of solute per volume of solution in liters:solution in liters:

M

M

molaritymoles of soluteliters of solution

HClmoles of HCl

liters of solution3

62

Molarity Calculations

Calculate the molarity of a solution prepared by Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.water to make 1.50 L of solution.

(11.5g NaOH/1.50L)(1 mol NaOH/40.00g (11.5g NaOH/1.50L)(1 mol NaOH/40.00g NaOH)NaOH)

= 0.192M NaOH= 0.192M NaOH

Molarity Calculations

Calculate the molarity of a solution prepared by Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.water to make 26.8 mL of solution.

(1.56g HCl/26.8mL)(1 mol HCl/36.46g HCl) (1.56g HCl/26.8mL)(1 mol HCl/36.46g HCl) (1000mL/1L) = 1.60M HCl(1000mL/1L) = 1.60M HCl

Molarity Calculations

How many moles of nitrate ions are present in How many moles of nitrate ions are present in 25.00 mL of a 0.75 M Co(NO25.00 mL of a 0.75 M Co(NO33))22 solution? solution?

(25.00mL)(1L/1000mL)(0.75mol Co(NO(25.00mL)(1L/1000mL)(0.75mol Co(NO33))22/1L)/1L)

(2 mol NO(2 mol NO33--/1 mol Co(NO/1 mol Co(NO33))22) = 3.8 x 10) = 3.8 x 10-2 -2 mol NOmol NO33

--

Molarity Calculations

Typical blood serum is about 0.14M NaCl. What Typical blood serum is about 0.14M NaCl. What volume of blood contains 1.0 mg of NaCl?volume of blood contains 1.0 mg of NaCl?

(1.0mg NaCl)(1g/1000mg)(1mol/58.45g)(1L/0.14mol) (1.0mg NaCl)(1g/1000mg)(1mol/58.45g)(1L/0.14mol) = 1.2 x 10= 1.2 x 10-4-4 L blood L blood

serumserum

Standard SolutionA standard solution is a solution whose concentration is A standard solution is a solution whose concentration is

accurately known. accurately known.

Standard solutions are made using a volumetric flask as Standard solutions are made using a volumetric flask as follows:follows:

• mass the solute accurately and add it to the mass the solute accurately and add it to the volumetric flaskvolumetric flask

• add a small quantity of distilled HOHadd a small quantity of distilled HOH• dissolve the solute by gently swirling the flaskdissolve the solute by gently swirling the flask• add more distilled HOH until the level of the solution add more distilled HOH until the level of the solution

reaches the mark on the neckreaches the mark on the neck• invert the capped volumetric 25X to thoroughly mix invert the capped volumetric 25X to thoroughly mix

the solution.the solution.

04_44

Volume marker(calibration mark)

Weighedamountof solute

Wash Bottle

(a) (b) (c) (d)

Steps involved in making a standard solution.

Common Terms of Solution Concentration

StockStock - routinely used solutions prepared in - routinely used solutions prepared in concentrated form.concentrated form.

ConcentratedConcentrated - - relativelyrelatively large ratio of solute large ratio of solute to solvent. (to solvent. (5.0 5.0 MM NaCl NaCl))

DiluteDilute - - relativelyrelatively small ratio of solute to small ratio of solute to solvent. (solvent. (0.01 0.01 MM NaCl NaCl))

Dilution of Stock Solutions

When diluting stock solutions, the moles of When diluting stock solutions, the moles of solute after dilution must equal the moles solute after dilution must equal the moles of solute before dilution.of solute before dilution.

Stock solutions are diluted using either a Stock solutions are diluted using either a measuring or a delivery pipet and a measuring or a delivery pipet and a volumetric flask.volumetric flask.

04_46

(a) (b) (c)

Rubber bulb

500 mL

Steps to dilute a stock solution.

Dilution Calculations

What volume of 16 M sulfuric acid must be used to What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M Hprepare 1.5 L of a 0.10 M H22SOSO44 solution? solution?

(0.10mol H(0.10mol H22SOSO44/1L)(1.5L)(1L/16mol)(1000mL/1L)/1L)(1.5L)(1L/16mol)(1000mL/1L)

= 9.4 mL conc = 9.4 mL conc HH22SOSO44

Types of Solution Reactions

- Precipitation reactionsPrecipitation reactionsAgNOAgNO33((aqaq) + NaCl() + NaCl(aqaq) ) AgCl( AgCl(ss) + NaNO) + NaNO33((aqaq))

- Acid-base reactionsAcid-base reactionsNaOH(NaOH(aqaq) + HCl() + HCl(aqaq) ) NaCl( NaCl(aqaq) + H) + H22O(O(ll))

- Oxidation-reduction reactionsOxidation-reduction reactionsFeFe22OO33((ss) + 2Al() + 2Al(ss) ) 2Fe( 2Fe(ll) + Al) + Al22OO33((ss))

Describing Reactions in Solution

1.1. Molecular equation Molecular equation (reactants and products as (reactants and products as compoundscompounds))

AgNOAgNO33((aqaq) + NaCl() + NaCl(aqaq) ) AgCl( AgCl(ss) + NaNO) + NaNO33((aqaq))

2.2. Complete ionic equation Complete ionic equation (all strong (all strong electrolytes shown as electrolytes shown as ionsions))

AgAg++((aqaq) + NO) + NO33((aqaq) + Na) + Na++((aqaq) + Cl) + Cl((aqaq) )

AgCl(AgCl(ss) + Na) + Na++((aqaq) + NO) + NO33((aqaq))

Describing Reactions in Solution (continued)

3.3. Net ionic equation Net ionic equation (show only (show only components that actually react)components that actually react)

AgAg++((aqaq) + Cl) + Cl((aqaq) ) AgCl( AgCl(ss))

NaNa++ and NO and NO33 are spectator ions. are spectator ions.

Simple Rules for Solubility1.1. Most nitrate (NOMost nitrate (NO33

) salts ) salts are solubleare soluble..

2.2. Most alkali (group 1A) salts and NHMost alkali (group 1A) salts and NH44++ are solubleare soluble..

3.3. Most ClMost Cl, Br, Br, and I, and I salts salts are soluble are soluble ((NOTNOT AgAg++, Pb, Pb2+2+, , HgHg22

2+2+))

4.4. Most sulfate salts Most sulfate salts are soluble are soluble ((NOTNOT BaSOBaSO44, PbSO, PbSO44, ,

HgSOHgSO44, CaSO, CaSO44))

5.5. Most OHMost OH salts salts are only slightly soluble are only slightly soluble (NaOH, KOH (NaOH, KOH are soluble, Ba(OH)are soluble, Ba(OH)22, Ca(OH), Ca(OH)22 are marginally soluble) are marginally soluble)

6.6. Most SMost S22, CO, CO3322, CrO, CrO44

22, PO, PO4433 salts salts are only slightly are only slightly

solublesoluble. .

Solubility

Using the solubility rules, predict what will happen Using the solubility rules, predict what will happen when the following pairs of solutions are mixed.when the following pairs of solutions are mixed.

a) KNOa) KNO3(aq)3(aq) & BaCl & BaCl2(aq)2(aq)

b) Nab) Na22SOSO4(aq)4(aq) & Pb(NO & Pb(NO33))2(aq)2(aq)

c) KOHc) KOH(aq)(aq) & Fe(NO & Fe(NO33))3(aq)3(aq)

No precipitate forms

PbSO4(s) forms

Fe(OH)3(s) forms

04_STOICHIOMETRY FOR REACTIONS IN SOLUTION

STEP 1Identify the species present in the combined solution, and determinewhat reaction occurs.

STEP 2Write the balanced net ionic equation for the reaction.

STEP 3Calculate the moles of reactants.

STEP 4Determine which reactant is limiting.

STEP 5Calculate the moles of product or products, as required.

STEP 6Convert to grams or other units, as required.

Precipitation CalculationsWhen aqueous solutions of NaWhen aqueous solutions of Na22SOSO44 & &

Pb(NOPb(NO33))22 are mixed, PbSO are mixed, PbSO4 4 precipitates. precipitates.

Calculate the mass of PbSOCalculate the mass of PbSO44 formed when formed when

1.25 L of 0.0500 M Pb(NO1.25 L of 0.0500 M Pb(NO33))22 & 2.00 L of & 2.00 L of

0.0250 M Na0.0250 M Na22SOSO44 are mixed. are mixed.

1. Species present are Na1. Species present are Na++, SO, SO442-2-, Pb, Pb2+2+, & , &

NONO33--..

2. Pb2. Pb2+2+(aq)(aq) + SO + SO44

2-2-(aq)(aq) ----> PbSO ----> PbSO4(s)4(s)

Precipitation CalculationsContinued

3. (1.25L)(0.500mol Pb3. (1.25L)(0.500mol Pb2+2+/L) = 0.0625 mol Pb/L) = 0.0625 mol Pb2+2+

(2.00L)(0.0250mol SO(2.00L)(0.0250mol SO442-2-/L) = 0.0500 mol SO/L) = 0.0500 mol SO44

2-2-

4.4. (0.0625mol Pb(0.0625mol Pb2+2+)(1mol SO)(1mol SO442-2-/1mol Pb/1mol Pb2+2+) =) =

0.0625 mol SO0.0625 mol SO442-2-

SOSO442- 2- is the limiting reactant.is the limiting reactant.

5.5. (0.0500mol SO(0.0500mol SO442-2-)(1mol PbSO)(1mol PbSO44/1mol SO/1mol SO44

2-2-))

(303.3g/1mol) = 15.2 g PbSO(303.3g/1mol) = 15.2 g PbSO44

Performing Calculations for Acid-Base Reactions

1.1. List initial species and predict reaction.List initial species and predict reaction.

2.2. Write balanced net ionic reaction.Write balanced net ionic reaction.

3.3. Calculate moles of reactants.Calculate moles of reactants.

4.4. Determine limiting reactant.Determine limiting reactant.

5.5. Calculate moles of required reactant/product.Calculate moles of required reactant/product.

6.6. Convert to grams or volume, as required.Convert to grams or volume, as required.

Acid-Base Calculations

What volume of a 0.100M HCl solution is needed What volume of a 0.100M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH?to neutralize 25.0 mL of 0.350 M NaOH?

1. H1. H++, Cl, Cl--, Na, Na++, & OH, & OH--

2. H2. H++(aq)(aq) + OH + OH--

(aq)(aq) ----> HOH ----> HOH(l)(l)

(25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol (25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol NaOH)(1L/0.100mol) = 87.5 mL HCl solutionNaOH)(1L/0.100mol) = 87.5 mL HCl solution

Key Titration Terms

TitrantTitrant - solution of known concentration used - solution of known concentration used in titrationin titration

AnalyteAnalyte - substance being analyzed - substance being analyzed

Equivalence point Equivalence point - enough titrant added to - enough titrant added to react exactly with the analytereact exactly with the analyte

Endpoint Endpoint - the indicator changes color so you - the indicator changes color so you can tell the equivalence point has been reached.can tell the equivalence point has been reached.

Titration CalculationsIf 41.20 mL of NaOH is required to neutralize 1.3009 g If 41.20 mL of NaOH is required to neutralize 1.3009 g

of KHP (KHCof KHP (KHC88HH44OO66), what is the concentration of ), what is the concentration of

the NaOH solution? Assume there is one acidic the NaOH solution? Assume there is one acidic hydrogen in KHP.hydrogen in KHP.

HCHC88HH44OO66--(aq) (aq) + OH+ OH--

(aq)(aq) ----> HOH ----> HOH(l)(l) + C + C88HH44OO662-2-

(aq)(aq)

(1.3009g KHP)(1mol/204.22g)(1mol NaOH/1mol KHP)(1.3009g KHP)(1mol/204.22g)(1mol NaOH/1mol KHP)(1/41.20mL)(1000mL/1L) = 0.1546 M NaOH(1/41.20mL)(1000mL/1L) = 0.1546 M NaOH

Redox Reactions

Redox reactions are reactions in which Redox reactions are reactions in which electrons are transferred. electrons are transferred.

Decomposition and synthesis reactions Decomposition and synthesis reactions maymay be be redox.redox.

Single replacement reactions are Single replacement reactions are alwaysalways redox. redox.

Double replacement reactions are Double replacement reactions are nevernever redox. redox.

Combustion reactions are Combustion reactions are alwaysalways redox. redox.

OIL RIG

Oxidation Is Loss.

Reduction Is Gain.

Redox

Oxidizing agent is the electron acceptor--Oxidizing agent is the electron acceptor--usually a nonmetal.usually a nonmetal.

Reducing agent is the electron donor--usually Reducing agent is the electron donor--usually a metal. a metal.

CHCH4(g)4(g) + 2O + 2O2(g)2(g) ----> CO ----> CO2(g)2(g) + 2HOH + 2HOH(g)(g)

Carbon is oxidized.Carbon is oxidized.

Oxygen is reduced.Oxygen is reduced.

CHCH4 4 is the reducing agent.is the reducing agent.

OO2 2 is the oxidizing agent.is the oxidizing agent.

Rules for Assigning Oxidation States

1.1. Oxidation state of an atom in an element = 0Oxidation state of an atom in an element = 0

2.2. Oxidation state of monatomic element = chargeOxidation state of monatomic element = charge

3.3. Oxygen = Oxygen = 2 in covalent compounds (except in 2 in covalent compounds (except in peroxides where it = peroxides where it = 1)1)

4.4. H = +1 in covalent compoundsH = +1 in covalent compounds

5.5. Fluorine = Fluorine = 1 in compounds1 in compounds

6.6. Sum of oxidation states = 0 in compounds Sum Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ionof oxidation states = charge of the ion

Determining Oxidation States

SFSF66 [NO[NO33]]--

+6+6 -6 = 0-6 = 0 +5 -6 = -1 +5 -6 = -1

(-1 for each F) (-1 for each F) (-2 for each (-2 for each O)O)

Balancing by Half-Reaction Method

1.1. Write separate reduction, oxidation Write separate reduction, oxidation reactions.reactions.

2.2. For each half-reaction:For each half-reaction:

-- Balance elements (except H, O)Balance elements (except H, O)

-- Balance O using HBalance O using H22OO

-- Balance H using HBalance H using H++

-- Balance charge using electronsBalance charge using electrons

Balancing by Half-Reaction Method (continued)

3.3. If necessary, multiply by integer to If necessary, multiply by integer to equalize electron count.equalize electron count.

4.4. Add half-reactions.Add half-reactions.

5.5. Check that both Check that both elementselements and and chargescharges are are balanced.balanced.

Balancing By Half-Reaction Acidic Solution

HH++(aq)(aq) + Cr + Cr22OO77

2-2-(aq)(aq) + C + C22HH55OHOH(l)(l) ---> Cr ---> Cr3+3+

(aq)(aq) + CO + CO2(g)2(g) + HOH + HOH(l)(l)

Red CrRed Cr22OO772-2-

(aq)(aq) ---> Cr ---> Cr3+3+(aq) (aq)

OxOx CC22HH55OHOH(l)(l) ---> CO---> CO2(g)2(g)

Red 2(6eRed 2(6e-- + 14H + 14H++(aq) (aq) + Cr+ Cr22OO77

2-2-(aq)(aq) ---> 2Cr ---> 2Cr3+3+

(aq) (aq) + 7HOH+ 7HOH(l)(l)))

OxOx C C22HH55OHOH(l)(l) + 3HOH + 3HOH(l) (l) ---> 2CO---> 2CO2(g)2(g) + 12H + 12H++(aq)(aq) + 12e + 12e--

16H16H++(aq)(aq) + 2Cr + 2Cr22OO77

2-2-(aq)(aq) + C + C22HH55OHOH(l)(l) ---> 4Cr ---> 4Cr3+3+

(aq)(aq) + 11HOH + 11HOH(l)(l) + +

2CO2CO2(g)2(g)

12+ = 12+12+ = 12+

Half-Reaction Method - Balancing in Base

1.1. Balance as in acid.Balance as in acid.

2.2. Add OHAdd OH that equals H that equals H++ ions (both sides!) ions (both sides!)

3.3. Form water by combining HForm water by combining H++, OH, OH..

4.4. Check elements and charges for balance.Check elements and charges for balance.

Balancing By Half-Reaction Basic Solution

AgAg(s)(s) + CN + CN--(aq)(aq) + O + O2(g)2(g) ---> Ag(CN) ---> Ag(CN)22

--(aq)(aq)(Basic)(Basic)

Ox CNOx CN--(aq) (aq) + Ag+ Ag(s)(s) ---> Ag(CN) ---> Ag(CN)22

--(aq)(aq)

RedRed O O2(g)2(g) ---> --->

Ox 4(2CNOx 4(2CN--(aq) (aq) + Ag+ Ag(s)(s) ---> Ag(CN) ---> Ag(CN)22

--(aq) (aq) + e+ e--))

RedRed O O2(g)2(g) + 4H + 4H++(aq) (aq) + 4e+ 4e-- ---> 2HOH ---> 2HOH(l)(l)

8CN8CN--(aq)(aq) + 4Ag + 4Ag(s) (s) + O+ O2(g)2(g) + 4H + 4H++

(aq) (aq) ---> 4Ag(CN)---> 4Ag(CN)22--(aq) (aq) + +

2HOH2HOH(l)(l)

Balancing By Half-Reaction Basic Solution(Continued)

8CN8CN--(aq)(aq) + 4Ag + 4Ag(s) (s) + O+ O2(g)2(g) + 4H + 4H++

(aq)(aq) + 4OH + 4OH--(aq) (aq) ---> --->

4Ag(CN) 4Ag(CN)22--(aq) (aq) + 2HOH+ 2HOH(l) (l) + 4OH+ 4OH--

(aq)(aq)

8CN8CN--(aq)(aq) + 4Ag + 4Ag(s) (s) + O+ O2(g)2(g) + 4HOH + 4HOH(l) (l) ---> 4Ag(CN)---> 4Ag(CN)22

--(aq) (aq) + +

2HOH 2HOH(l) (l) + 4OH+ 4OH--(aq)(aq)

8CN8CN--(aq)(aq) + 4Ag + 4Ag(s) (s) + O+ O2(g)2(g) + 2HOH + 2HOH(l) (l) ---> 4Ag(CN)---> 4Ag(CN)22

--(aq) (aq) + +

4OH4OH--(aq)(aq)

88-- = 8 = 8--