Chapter 4 Solution Stoiciometry. Solutions = Homogeneous Mixtures Solute – thing being dissolved...
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Transcript of Chapter 4 Solution Stoiciometry. Solutions = Homogeneous Mixtures Solute – thing being dissolved...
![Page 1: Chapter 4 Solution Stoiciometry. Solutions = Homogeneous Mixtures Solute – thing being dissolved (lesser part of Homogeneous mixture) Solvent – medium.](https://reader035.fdocuments.us/reader035/viewer/2022062301/56649ec15503460f94bcc7dd/html5/thumbnails/1.jpg)
Chapter 4 Solution Stoiciometry
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Solutions = Homogeneous Mixtures
Solute – thing being dissolved (lesser part of Homogeneous mixture)
Solvent – medium that there is more of (dissolves solute)
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Solubility
Ability of substance to be dissolved in solvent
(greater than .01 mol/L = soluble)
Molarity = moles solute/liters solvent
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Calculations:
How many grams of sodium hydroxide is necessary to make 2L of 3.5M solution?
M=mol solute/L solvent
3.5M = x mol/2L
7mol = x
convert x to grams
7mol x 40g/ 1mol = 280 g
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Calculations:
What volume in mL of water is needed to make 3.6M solution of salt (NaCl) given 4.8g of NaCl?
convert to mol
4.8g x 1mol/58.5g = .08mol NaCl
3.6M = .08mol NaCl / X Liters
X Liters = .08mol/3.6 mol/L
= .022L 22mL
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Calculations:
Calculate the molarity of each ion in solution of 2M CrCl3
CrCl3 Cr3+ + 3 Cl-
2M 2M 6M
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Calculations:
What is the molarity of Fe3+ ions and SO42- ion
in a solution made with 48.05g of Fe2(SO4)3 in Water to make 800 mL of solution?
M = moles solute /moles solution
Find moles Iron (III) sulfate
Find moles Fe3+ and SO42-
Find Molarity
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Dilution
Adding water to get desired molarity
M1V1 = M2V2
How much 18M HCl must be used to make 100 mL of 2M solution?
X =11mL
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Solubility Rules
Memorize table 4.1 on page 118 and handout
Used to predict a precipitate (insoluble product)
Test on Friday.
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Attack the problem
1. Write the products and reactants and balance the equation.
2. Convert to moles if grams or molarity is given
3. Determine the product’s solubility, convert to grams if necessary.
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ExampleCalculate the mass of precipitate produced when 125 mL of .2M AgNO3 is added to an
excess of Na2S to produce silver sulfide and sodium nitrate.
1. Balance: 2AgNO3 + Na2S Ag2S + 2NaNO3
2. Molarity .2M = x mol/.125L
3. Find mol AgNO3 and then convert to Mass
.025 mol AgNO3 x 1molAgS/2mol AgNO3 x 247g AgS/1molAgS
=2.9g
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Net Ionic EquationsWrite balanced molecular equation
Write out ions from strong electrolytes in water (aq)
Make sure to balance your charges
Look for unchanged ions (Spectator ions) on each side (aq of same charge) and cancel
Write the remaining ions in a net ionic equation (charges must balance)
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Net Ionic Equations• Strong Electrolytes solutes dissociate
completely. Strong Acids - amount of H+ dissociated
determines strength (90-100%) Strong Bases - amount of OH- dissociated
determines strength (90-100%) Salts – cation from base, anion from acid
Ex: NaCl NaOH + HCl H2O + NaCl
Memorize the Strong acids and Bases!!! (p122)
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Practice Net Ionic EquationsNaOH + HCl NaCl + H2O (molecular eq)
Na+(aq)
+ OH-(aq) + H+
(aq) + Cl-(aq) Na+(aq)
+ Cl-(aq) + H2O(l) (ion eq)
Na+ + OH- + H+ + Cl- Na+ + Cl- + H2O Spectator ions, no change in oxidation or
state
OH-(aq) + H+
(aq) H2O(l) net ionic equation
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Acids
•Donate H+
Monoprotic = 1H+ Diprotic = 2H+
•Strong acids vs. weak acidsAny ionic = strong (anything not in list is weak)Strong acids dissociate completelyWeak acids don’t dissociate completely and can become proton acceptors (reaction goes both ways) called a conjugate base
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Bases
•Donate OH-
•Strong bases vs. weak basesAny ionic = strong (anything not in list is weak)Strong bases dissociate completelyWeak bases = NH3
Forms conjugate acids, can donate H+
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SaltsReaction of acids and bases
(neutralization) produces metathesis (double replacement) resulting in water
and salt
Salts – cation from base, anion from acidEx: NaOH + HCl H2O + NaCl
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Practice Net Ionic EquationsWrite the net ionic equation for the reaction of
silver nitrate and potassium phosphate
Write the balanced molecular equation:
3AgNO3 + K3PO4 Ag3PO4 + 3KNO3
Determine products solubility
Ag3PO4(s)
Write Ionic equation
3Ag+ (aq) + 3NO3
- (aq) + 3K+
(aq) + PO43-
(aq) Ag3PO4(s) + 3NO3
- (aq) + 3K+
(aq)
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Cancel spectator ions
3Ag+ (aq) + 3NO3
- (aq) + 3K+
(aq) + PO43-
(aq) Ag3PO4(s) + 3NO3
- (aq) + 3K+
(aq)
Net ionic equation:
3Ag+ (aq) + PO4
3-(aq) Ag3PO4(s)
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Oxidation numbers (states)
Reaction in which electrons are transferred between reactants.
Loss of electrons = oxidation
Gain of electrons = reduction
Oxidation numbers = imaginary set of numbers based on rules (provides a way to keep track of electrons)
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Oxidation numbersWhole numbers written as +1, +2, -1 in order
to distinguish them from charges (3+, 2+, 2-)
Elemental form of atom, oxidation number is 0
Includes the diatomic gases (ex Cl2 oxidation number is 0)
Any monatomic ion, oxidation is the charge of the ion
Ex: Na+ oxidation number = +1, Cl- = -1
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Oxidation numbers
Nonmetals usually have negative ox numbers
Oxygen = -2 (except in peroxides O22- where
Oxygen has an oxidation # of –1)
Hydrogen = -1 if bonded to metal and +1 if bonded to nonmetal
Fluorine is –1 for all compounds, other halogens are –1 for binary compounds, but if bound to
oxygen (oxyanions) they have positive oxidation states
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Sum of all oxidation numbers in a neutral compound is zero.
In a charged compound, it equals the charge
Oxidation numbers
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Redox Practice
Determine the oxidation number on the following green atoms:
H2O H2SO4 P2O5 NO3- LiH BaO2
-2 +6 +5 +5 -1 -1
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Redox Practice
Determine which reactant has been oxidized, and which has been reduced. Give their initial and final oxidation numbersCu(s) + 4HNO3 Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)
Cu (02) N (55) N (54)
Oxidized no change Reduced