chapter 4 solution stoiciometry. solutions = homogeneous mixtures solute – thing being dissolved...

Chapter 4 Solution Stoiciometry

Solutions = Homogeneous Mixtures

Solute – thing being dissolved (lesser part of Homogeneous mixture)

Solvent – medium that there is more of (dissolves solute)

Solubility

Ability of substance to be dissolved in solvent

(greater than .01 mol/L = soluble)

Molarity = moles solute/liters solvent

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

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

Calculations:

Calculate the molarity of each ion in solution of 2M CrCl3

CrCl3 Cr3+ + 3 Cl-

2M 2M 6M

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

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

Solubility Rules

Memorize table 4.1 on page 118 and handout

Used to predict a precipitate (insoluble product)

Test on Friday.

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.

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

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)

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)

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

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

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+

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

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)

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)

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)

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

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

Sum of all oxidation numbers in a neutral compound is zero.

In a charged compound, it equals the charge

Oxidation numbers

Redox Practice

Determine the oxidation number on the following green atoms:

H2O H2SO4 P2O5 NO3- LiH BaO2

-2 +6 +5 +5 -1 -1

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