4 - 1 Reactions in Solution Some Important Definitions Electrolytes Reactions in Solution Ionic Equations Single Replacement Reactions Concentration Solution.
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4 - 1
Reactions in SolutionReactions in Solution
Some Important DefinitionsSome Important Definitions
ElectrolytesElectrolytes
Reactions in SolutionReactions in Solution
Ionic EquationsIonic Equations
Single Replacement ReactionsSingle Replacement Reactions
ConcentrationConcentration
Solution StoichiometrySolution Stoichiometry
TitrationsTitrations
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Sugar in waterOxygen in water
AirDental fillings
Saline
Sugar in waterOxygen in water
AirDental fillings
Saline
SolutionsSolutions
A solutionA solution A homogeneous mixture of twoor more components.
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A solutionA solution
In a solution•The solute can’t be filtered out.•The solute always stays mixed.•Particles are always in motion.•Volumes may not be additive.•A solution will have different
properties than the solvent
A solution consists of two component types.
solventsolvent - component in the greater extent
solutesolute - component in the lesser extent(You may have more than one.)
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Physical states of solutionsPhysical states of solutions
Solutions can be made that exist in any of the three states.
Solid solutionsSolid solutionsdental fillings, 14K gold, sterling silver
Liquid solutionsLiquid solutionssaline, vodka, vinegar, sugar water
Gas solutionsGas solutionsthe atmosphere, anesthesia gases
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SolubilitySolubility
A measure of how much of a solute can be dissolved in a solvent.
Common unit- grams / 100 ml
Factors affecting solubilityFactors affecting solubility•Temperature•Pressure•Polarity
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Solubility of some substancesSolubility of some substances
Temperature SolubilitySubstance oC g/100 ml water
NaCl (s) 100 39.12
PbCl2 (s) 100 3.34
AgCl (s) 100 0.0021
CH3CH2OH (l) 0 - 100 infinity
CH3CH2OCH2CH3 (l) 15 8.43
O2 (g) 60 0.0023
CO2 (g) 40 0.097
SO2 (g) 40 5.41
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SaturationSaturation
When a solution contains as much solute as it can at a given temperature.
UnsaturatedUnsaturated Can still dissolve more.
SaturatedSaturated Have dissolved all you can.
SupersaturatedSupersaturated Temporarily have dissolved
too much.
PrecipitatePrecipitate Excess solute that falls out of solution.
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Properties ofProperties ofaqueous solutionsaqueous solutions
There are two general classes of solutes.
ElectrolyticElectrolytic•ionic compounds in polar solvents•dissociate in solution to make ions•conduct electricity•may be strong (100% dissociation) or
weak (less than 100%)
NonelectrolyticNonelectrolytic•do not conduct electricity•solute is dispersed but does not
dissociate
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Dissolving ionic compoundsDissolving ionic compounds
When an ionic solid dissolves in water, the solvent removes ions from the crystal.
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Dissolving covalent Dissolving covalent compoundscompounds
Covalent compounds do not dissociate.
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Acids, bases and saltsAcids, bases and salts
Three types of compounds are electrolytes:
• AcidAcid - a compound that increases the concentration of hydrogen ions in water.
HCl H+ + Cl-
• BaseBase - a compound that increases the concentration of hydroxide ion in water.
NaOH Na+ + OH-
• SaltSalt - the ions that remain after an acid and base react with each other - neutralization.
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
water
water
water
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Some acids, bases and their saltsSome acids, bases and their salts
AcidAcid Sodium salt Sodium saltNameName FormulaFormula NameName FormulaFormulaAcetic acid HC2H3O2 Sodium acetate NaC2H3O2
Hydrogen chloride HCl Sodium chloride NaClNitric acid HNO3 Sodium nitrate
NaNO3
Phosphoric acid H3PO4 Sodium phosphate Na3PO4
Sulfuric acid H2SO4 Sodium sulfate Na2SO4
BaseBase Chloride salt Chloride saltNameName FormulaFormula NameName FormulaFormulaSodium hydroxide NaOH Sodium chloride NaClBarium oxide BaO Barium chloride BaCl2Sodium oxide Na2O Sodium chloride NaClAmmonia NH3 Ammonium chloride NH4Cl
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Reactions between ions in solutionReactions between ions in solution
Neutralization is an example of a reaction between ions in solution.
When ions react, we might observe the formation of a precipitate or a gas.AgNO3 (aq) + NaCl (aq) AgCl AgCl (s)(s) + NaNO3 (aq)
Na2CO3 (aq) + 2HNO3 (aq) 2NaNO3 (aq) +H2O (l) CO CO2 2 (g)(g)
However, not all ions will react in solution.
KNO3 (aq) + NaCl (aq) No reaction
Solubility rules can help predict reactions.
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Some simple solubility rulesSome simple solubility rules
• All acids are soluble.
• All Na+, K+ and NH4+ salts are soluble.
• All nitrate and acetate salts are soluble.
• All chlorides except AgCl and Hg2Cl2 are soluble. PbCl2 is slightly soluble.
• All sulfates are soluble except PbSO4, Hg2SO4, SrSO4 and BaSO4. Ag2SO4 and CaSO4 are slightly soluble.
• All sulfides are insoluble except those of the Group IA (1), IIA (2) and ammonium sulfide.
• All hydroxides are insoluble except those of the group IA(1) and Ba(OH)2. Sr(OH)2 and Ca(OH)2 are slightly soluble.
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Ionic equationsIonic equations
When ionic substances dissolve in water, they dissociate into ions.
AgNO3 Ag+ + NO3
-
KCl K+
+ Cl-
When a reaction occurs, only some of the ions are actually involved in the reaction.
AgAg++ + NO3
- + K
+ + ClCl- AgClAgCl(s(s) + K+ +
NO3-
H2O
H2O
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Ionic equationsIonic equations
To help make the reaction easier to see, we commonly list only the species actually involved in the reaction.
Full ionic equationFull ionic equation
AgAg++ + NO3
- + K
+ + ClCl- AgCAgCll(s)(s) + K+ + NO3
-
Net ionic equationNet ionic equationAgAg++
+ ClCl-- AgClAgCl(s)(s)
NO3- and K
+ are referred to as spectatorspectator ions.
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Single replacement reactionSingle replacement reaction
Where one element displaces another in a chemical compound.
H2 + CuO Cu + H2O
• In this example, hydrogen replaces copper.
• This type of reaction always involves oxidation and reduction (REDOX).
• Since one species is replacing another, there are no spectator ions.
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Single replacement reactionsSingle replacement reactions
If various metals are in water, we observe that some are more reactive than others.
2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g) (fast)
Ca (s) + 2H2O (l) Ca(OH)2 (s) + H2 (g) (slow)
Mg (s) + H2O (l) no reaction
This indicates that the order of reactivity of these metals towards water is
Na > Ca > MgNa > Ca > Mg
We can show the reactivity of metals towards water and acids using an activity series.activity series.
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Activity series of metalsActivity series of metalspotassium
sodiumpotassium
sodium
calciumcalcium
magnesiumaluminum
zincchromium
magnesiumaluminum
zincchromium
ironnickel
tinlead
ironnickel
tinlead
coppersilver
platinumgold
coppersilver
platinumgold
incr
easi
ng r
eact
ivit
y
Reacts violently with cold water
Reacts slowly with cold water
Reacts very slowly with steambut quite reactive in acid
Reacts moderately with acid
Unreactive in acid
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Activity series of metals -Activity series of metals -various metals in HClvarious metals in HCl
Iron Zinc Magnesium
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Reactivity of nonmetalsReactivity of nonmetals
I
BrSe
ClS
FO
P
NC
Increased reactivity
Incr
ease
d r
eact
ivit
y
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Concentration of solutionsConcentration of solutions
We need a way to tell how much solute is in a solution - concentrationconcentration.
There are many systems - we will cover four.
•Weight / volume percent•Volume / volume percent•Weight / weight percent•Molarity
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Weight/Volume %Weight/Volume %
Weight/Volume % = Mass solute Total Volume
x 100
If 5 grams of NaCl is dissolved in water to make 200 ml of solution, what is the concentration?
5 g / 200 ml * 100 = 2.5 wt/v%
Saline is a 0.9 wt/v% solution of NaCl in water.
use g and mluse g and ml
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Volume/Volume %Volume/Volume %
Volume/Volume % = Volume SoluteTotal Volume
x 100
If 10 ml of alcohol is dissolved in water to make 200 ml of solution, what is the concentration?
10 ml / 200 ml * 100 = 5 V/V%
Alcohol in wine is measured as a V/V%.
Use the same units for bothUse the same units for both
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Weight/Weight %Weight/Weight %
Weight/Weight % = Mass SoluteTotal Mass
x 100
If a ham contained 5 grams of fat in 200 gof ham, what is the % wt/wt?
5 g / 200g * 100 = 2.5 wt/wt%
On the label, it would say 97.5 % fat free.
Use the same units for bothUse the same units for both
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Very low concentrationsVery low concentrations
Pollutants in air and water are typically found at very low concentrations. Two common units are used to express these trace amounts.
Parts per million - ppmParts per million - ppmParts per billion - ppbParts per billion - ppb
Both are modifications of the % system which could be viewed as parts per hundred - pph.
Both mass and volume % systems are used.
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Low concentrations in airLow concentrations in air
Trace amounts in are are expressed as volume/volume ratios.
ppm = x 106
ppb = x 109
Example.Example. One cm3 of SO2 in one m3 of air would be expressed as 1 ppm or 1000 ppb.
volume solutevolume solution
volume solutevolume solution
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Low concentrations in waterLow concentrations in water
Mass percentages are used for water pollutants.
ppm = x 106
ppb = x 109
Example.Example. One ppm of a toxin in water is the same as 1 mg / liter since one liter of water has a mass of approximately 106 mg.
mass solutemass solution
mass solutemass solution
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MolarityMolarity
M = moles solute molliters of solution L
=
MolarityMolarity• Recognizes that compounds have different
formula weights.
• A 1 M solution of glucose contains thesame number of molecules as 1 M ethanol.
• [ ] - special symbol which means molar ( mol/L )
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MolarityMolarity
Calculate the molarity of a 2.0 L solution that contains 10 moles of NaOH.
MNaOH = 10 molNaOH / 2.0 L
= 5.0 M
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MolarityMolarity
What’s the molarity of a solution that has 18.23 g HCl in 2.0 liters?
First, you need the FM of HCl.First, you need the FM of HCl.FMHCl = 1.008 x 1 H + 35.45 x 1 Cl
= 36.46 g/mol
Next, find the number of moles.Next, find the number of moles.molesHCl = 18.23 gHCl / 36.46 g/mol
= 0.50 mol
Finally, divide by the volume.Finally, divide by the volume.MHCl = 0.50 mol / 2.0 L
= 0.25 M
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Solution preparationSolution preparation
Solutions are typically prepared by:
Dissolving the proper amount of solute and diluting to volume.
Dilution of a concentrated solution.
Lets look at an example of the calculations required to prepare known molar solutions using both approaches.
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Making a solutionMaking a solution
You are assigned the task of preparing 100.0 ml of a 0.5000 M solution of sodium chloride.
What do you do?
First, you need to know how many moles of NaCl are in 100.0 ml of a 0.5 M solution.
mol = M x V (in liters)
= 0.5000 M x 0.1000 liters
= 0.05000 moles NaCl
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Making a solutionMaking a solution
Next, we need to know how many grams of NaCl to weigh out.
gNaCl = mol x FMNaCl
= 0.05000 mol x 58.44 g/mol
= 2.922 grams
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Making a solutionMaking a solution
Finally, you’re ready to make the solution.
Weigh out exactly 2.922 grams of dry, pure NaCl and transfer it to a volumetric flask.
Fill the flask about 1/3 of the way with pure water and gently swirl until the salt dissolves.
Now, dilute exactly to the mark, cap and mix.
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DilutionDilution
Once you have a solution, it can be diluted by adding more solvent. This is also important for materials only available as solutions
M1V1 = M2V2
1 = initial 2 = final
Any volume or concentration unit can be used as long as you use the same units on both sides of the equation.
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DilutionDilution
What is the concentration of a solution produced by diluting 100.0 ml of 1.5 MNaOH to 2.000 liters?
M1V1 = M2V2
M1 = 1.5 M M2 = ???V1 = 100.0 ml V2 = 2000 ml
M2 = M1V1 / V2
M2 = (1.5 M) (100.0 ml) = 0.075 M (2000. ml)
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Solution stoichiometrySolution stoichiometry
Extension of earlier stoichiometry problems.Extension of earlier stoichiometry problems.
First step is to determine the number of moles based on solution concentration and volume.
Final step is to convert back to volume or concentration as required by the problem.
You still need a balanced equation and must use the coefficients for working the problem.
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Solution stoichiometry Solution stoichiometry exampleexample
Determine the volume of 0.100 M HCl that must be added to completely react with 250 ml of 2.50 M NaOH
Balanced chemical equationBalanced chemical equation
HCl(aq) + NaOH(aq) NaCl(aq) + H2O (l)
The first step is to determine how many moles of NaOH we have.
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Solution stoichiometry Solution stoichiometry exampleexample
We have 250 ml of a 2.50 M solution.
molNaOH = 0.250 L x 2.50 mol/L
= 0.625 molNaOH
From the balanced chemical equation, we know that we need one mole of HCl for each mole of NaOH.
That means we need 0.625 molHCl.
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Solution stoichiometry exampleSolution stoichiometry example
Now we can determine what volume of our 0.100 M HCl solution is required.
L = molHCl / MHCl
= 0.625 mol
= 6.26 L
1 L0.100 mol( )
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TitrationTitration
Method based on measurement of volume.
•You must have a solution of known concentration - standard solution.standard solution.
•It is added to an unknown solution while the volume is measured.
•The process is continued until the end end pointpoint is reached - a change that we can measure.
•Acids and bases are commonly measured using titrations.
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NeutralizationNeutralization
The reaction of an acid with a base to produce a salt and water.
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
If we prepare a standard solution of NaOH, we can then use it to determine the concentration of HCl in a sample.
This is an example of Analytical Chemistry.
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TitrationsTitrations
BuretBuret - volumetric glassware used for titrations.
It allows you to add a known amount of your titrant to the solution you are testing.
An indicator will give you the endpoint.
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TitrationsTitrations
Note the color change which indicates that the ‘endpoint’ has been reached.
Start End
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Indicator examplesIndicator examples
Acid-base indicators are weak acids that undergo a color change at a known pH.
phenolphthalein
pH
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