Solutions. – Solution , A homogeneous mixture in which all of the material is in the same state. – S ubstances present in lesser amounts, called solutes , are dispersed uniformly throughout the substance in the greater amount, the solvent
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– Solution, A homogeneous mixture in which all of the material is in the same state.
– Substances present in lesser amounts, called solutes, are dispersed uniformly throughout the substance in the greater amount, the solvent
– Aqueous solution — a solution in which the solvent is water
– Nonaqueous solution — any substance other than water is the solvent
– Many of the chemical reactions that are essential for life depend on the interaction of water molecules with dissolved compounds.
– When strong electrolytes dissolve, constituent ions dissociate completely, producing aqueous solutions that conduct electricity very well.
– When weak electrolytes dissolve, they produce relatively few ions in solution. Aqueous solutions, of weak electrolytes do not conduct electricity as well as solutions of strong electrolytes.
– Nonelectrolytes dissolve in water as neutral molecules and have no effect on conductivity.
CH3CO2H(aq) → CH3CO2-(aq) +H+(aq)
abbreviated as M
Calculating Volume from Mass
Problem: What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution?
0.250 L of water was used to make 0.250 L of solution. Notice the water left over.
(Vs) (Ms) = moles of solute = (Vd) (Md).
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate molarity of the solution and the concentration of each of the ions.
Have 10.0 g of Zn
What volume of 2.50 M HCl is needed to convert the Zn completely?
1. Overall equation — shows all of the substances present in their undissociated form
2. Complete ionic equation — shows all of the substances present in the form in which they actually exist in solution
3. Net ionic equation
– Derived from the complete ionic equation by omitting all spectator ions, ions that occur on both sides of the equation with the same coefficients
– Demonstrate that many different combinations of reactants can give the same net chemical reaction
The molecular formula above can be written as the total ionic formula
Mg(s)+ 2H+(aq)+ 2Cl-(aq)→ H2(g)+ Mg2+(aq)+ 2Cl-(aq)
The two Cl- ions are SPECTATOR IONS — they do not participate. Could have used NO3- for the spectator ion as salts of nitrates are all soluble.
By leaving out the spectator ions out you get the net ionic reaction
Mg(s) + 2 H+(aq) ---> H2(g) + Mg2+(aq)
Net Ionic Equations
The anions exchange places between cations.
Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2KNO3(aq)
Pb2+(aq) + 2 I-(aq) → PbI2(s)
NaOH(aq) + HCl(aq)→ NaCl(aq)+ H2O(l)
OH-(aq) + H+(aq) → H2O(lq)
MgCO3(s)+ 2HCl(aq) → 2Mg(Cl)2(aq)+ H2O(l) + CO2(g)
If one ion from the “Soluble Compound” list is present in a compound, the compound is water soluble.
In addition to understanding solubility. It is equally important to know if a reaction will occur.
KBr(aq) + NaCl(aq) → KCl(aq) + NaBr(aq)
1. Reactions that go to completion
2. Reaction that does not go to completion is a reaction of a weak acid or a weak base with water
A brief history of
A Brönsted-Lowry Acid → H+ in water
Weak Brönsted-Lowry acids are weak electrolytes
CH3CO2H acetic acid (CH3COOH)
H2CO3 carbonic acid
H3PO4 phosphoric acid
HF hydrofluoric acid
– Monoprotic acids are compounds capable of donating a single proton per molecule.
– Polyprotic acids can donate more than one hydrogen ion per molecule.
NaOH(aq) → Na+(aq) + OH-(aq)
Brönsted-LowryBase → OH- in water
NaOH is a strong base
Nonmetal oxides can be acids
CO2(aq) + H2O(l) → H2CO3(aq)
SO3(aq) + H2O(l) → H2SO4(aq)
NO2(aq) + H2O(l) → HNO3(aq)
Acid Rain is an example of nonmetal oxides behaving as acids. This process can result from burning coal and oil.
CaO in water. Phenolphthalein indicator shows a of calcium oxide solution is basic.
pH: a way to express acidity -- the concentration of H+ in solution.
Low pH: high [H+]
High pH: low [H+]
Acidic solution pH < 7
Neutral pH = 7
Basic solution pH > 7
pH = log (1/ [H+]) = - log [H+]
In a neutral solution, [H+] = [OH-] = 1.00 x 10-7 M at 25 oC
pH = - log [H+] =
If the [H+] of soda is 1.6 x 10-3 M, the pH is ____.
If the pH of Coke is 3.12, it is _____.
You should know the strong acids & bases
– Metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule.
– Oxygen atoms acquire a negative charge and form oxide ions (O2-).
– Metals lose electrons to oxygen and have been oxidized—oxidation is the loss of electrons.
– Oxygen atoms have gained electrons and have been reduced—reduction is the gain of electrons.
2 Ag(s) + Cu2+(aq) → Al+(aq) + Cu(s)
Hydrogen Fuel Cell
2 H2(g) + O2(g) → 2H2O(l)
Fe2O3(s) + 2Al(s) → 2 Fe(s) + Al2O3(s)
Will be covered
in Chem 102
– A method in which a measured volume of a solution of known concentration, called the titrant, is added to a measured volume of a solution containing a compound whose concentration is to be determined (the unknown)
– Reaction must be fast, complete, and specific (only the compound of interest should react with the titrant)
– Equivalence point — point at which exactly enough reactant has been added for the reaction to go to completion (computed mathematically)
1. Add titrant solution from the buret.
2. Reagent (base) reacts with compound (acid) in solution in the flask.
3. Indicator shows when exact stoichiometric reaction has occurred.
4. Net ionic equation
H++ OH- → H2O
5. At equivalence point
moles H+= moles OH-
H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)
Carry out this reaction using a TITRATION.
PROBLEM: Standardize a solution of NaOH — i.e., accurately determine its concentration. 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?
Quantitative Analysis: Titrations
PROBLEM : Use standardized NaOH to determine the amount of an acid in an unknown. Apples contain malic acid, C4H6O5. 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?
Quantitative Analysis: Titrations
HOOCCH2COHCOOH(aq) + 2NaOH(aq) → Na2C4H4O5(aq) + 2H2O(l)
2H2(g) + O2(g) → 2H2O(g) + heat and light
This can be set up to provide
ELECTRIC ENERGY in a fuel cell.
2 H2 → 4 H+ + 4 e-
4 e- + O2 + 2 H2O → 4 OH-
H2/O2 Fuel Cell Energy
ENERGYis the capacity to do work (w) or transfer heat (q).
HEATis the thermal energy that can be transferred from an object at one temperature to an object at another temperature
– Net transfer of thermal energy stops when the two objects reach the same temperature.
Other forms of energy —
and radio waves
atomic and molecular motion – Temperature of an
object is a measure of the thermal energy content
of atoms in a chemical compound; radiant and thermal
energy produced in this reaction due to energy released
during the breaking and reforming of chemical bonds
particles in the nucleus of the atoms are rearranged
All of thermodynamics depends on the law of
CONSERVATION OF ENERGY.
Energy Change in Chemical Processes
Potential Energy of system dropped. Kinetic energy increased. Therefore, you often feel a Temperature increase.
Heat transfers until thermal equilibrium is established. ∆T measures energy transferred.
∆H along one path = ∆H along another path
Most chemical reactions occur at constant P, so
Heat transferred at constant P = qp
qp = ∆H where H = enthalpy
and so ∆E = ∆H + w (and w is usually small)
∆H = heat transferred at constant P ≈ ∆E
∆H = change in heat content of the system
∆H = Hfinal - Hinitial
∆H = Hfinal - Hinitial
If Hfinal > Hinitial then ∆H is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then ∆H is negative
Process is EXOTHERMIC
Consider the formation of water
H2(g) + 1/2 O2(g) → H2O(g) + 241.8 kJ
Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ
Example of HESS’S LAW— Making liquid H2O from H2 + O2 involves twoexothermic steps.
H2 + O2 gas
Making liquid H2O from H2 involves two steps.
H2(g) + 1/2 O2(g) → H2O(g) + 242 kJ
H2O(g) → H2O(l) + 44 kJ
H2(g) + 1/2 O2(g) → H2O(l) + 286 kJ
If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.
H2(g) + 1/2 O2(g) → H2O(g) ∆H˚ = -242 kJ
2H2(g) + O2(g) → 2H2O(g) ∆H˚ = -484 kJ
H2O(g) → H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ
H2(g) + 1/2 O2(g) → H2O(l) ∆H˚ = -286 kJ
Depend on how the reaction is written and on phases of reactants and products
P in standard state is P4
Phosphoric acid in standard state is H3PO4(s)
H2(g) + ½ O2(g) → H2O(g) ∆Hf˚ (H2O, g)= -241.8 kJ/mol
C(s) + ½ O2(g) → CO(g)∆Hf˚ of CO = - 111 kJ/mol
By definition, ∆Hfo= 0 for elements in their standard states.
Use ∆H˚’s to calculate enthalpy change for
H2O(g) + C(graphite) → H2(g) + CO(g)
Formation and Reaction
– The enthalpy that occurs when a reaction is carried out with all reactants and products in their standard states
– General reaction: aA + bB cC + dD where A, B, C, and D are chemical substances and a, b, c, and d are their stoichiometric coefficients
– Magnitude of ΔH°rxn is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient , minus the sum of the standard enthalpies of formation of the reactants, multiplied by their coefficients
ΔH°rxn= [cΔH°f(C) + dΔH°f(D)] [aΔH°f(A) + bΔH°f(B)]
ΔH°rxn= mΔH°f(product) nΔH°f(reactants)
where means “sum of” and m and n are the stoichiometric coefficients of each of the products and reactants, respectively
Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.
–Themagnitude of ΔH for a reaction depends on the physical states of the reactants and products, the pressure of any gases present, and the temperature at which the reaction is carried out.
– A specific set of conditions under which enthalpy changes are measured are used to ensure uniformity of reaction conditions and data.
– Standard conditions: a pressure of 1 atmosphere (atm) for gases and a concentration of 1 M for species in solution (1 mol/L), each pure substance must be in its standard state, its most stable form at a pressure of 1 atm at a specified temperature.
– Enthalpies of formation measured under standard conditions are called standard enthalpies of formation (ΔH°f).
– Standard enthalpy of formation of any element in its standard state is zero.
Hess’s law allows the calculation of the enthalpy change for any conceivable chemical reaction by using a relatively small set of tabulated data
calculate Ho for the reaction below.
The heat required to raise an object’s T by 1 ˚C.
Which has the larger heat capacity?
Thermal energy cannot be measured easily.
Temperature change caused by the flow of thermal energy between objects or substances can be measured.
Calorimetry is the set of techniques employed to measure enthalpy changes in chemical processes using calorimeters.
q = nCpΔT where n = number of moles of substance
q = mCsΔT where m = mass of substance in grams
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?
q transferred = (sp. ht.)(mass)(∆T)
Note that T is constant as ice melts or water boils
CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)
50.0mL of 0.400 M CuSO4 at 23.35 oC Tfinal 25.23oC
50.0mL of 0.600 M NaOH at 23.35 oC
Density final solution = 1.02 g/mL CH2O = 4.184 J/goC
Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.
Charles’s and Gay-Lussac’s findings can be stated as: At constant pressure, the volume of a fixed amount of a gas is directly proportional to its absolute temperature
This relationship is referred to as Charles’s law and is stated mathematically as
V = (constant) [T (in K)] or V T (in K, at constant P).
Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.
before reaching 0K
absolute zero = -273.15 0C
11.2 L of a gas at STP = 0.500 mole
44.8 L of a gas at STP = ? Moles
The ideal gas law assumes that all gases behave identically. If V, T, and n for each gas in a mixture are known, the pressure of each gas, its partial pressure, can be calculated. Holding V and T constant, P is directly proportional to n:
P = n(RT/V) = n(constant)
Generally, for a mixture oficomponents, the total pressure is given by
Pt = (n1 + n2 + n3 + - - - +ni) (RT/V).
The above equation makes it clear that, at constant T and V, P depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of gaseous species. Nothing in the equation depends on the nature of the gas, only on the quantity.
The total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law is known as Dalton’s law of partial pressures and can be written mathematically as
Pt= P1+ P2+ P3- - - + Pi
where Pt is the total pressure and the other terms are the partial pressures of the individual gases.
John Dalton 1766-1844
2 mol KClO3 yields 2 mol KCl and 3 mol O2
2(122.6g) yields 2 (74.6g) and 3 (32.0g)
Those 3 moles of O2 can also be thought of as:
67.2 L at STP
J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910.
The van der Waals’ equation
(P + an2/V2) (V – nb) = nRT
accounts for the behavior of real gases at low temperatures and high pressures.
At large volumes a and b are relatively small and the van der Waal’s equation reduces to the ideal gas law at high temperatures and low pressures.
PV = nRT
P = nRT/V n = 84.0g * 1mol/17 g T = 200 + 273
P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K)
P = 38.3 atm
n = 84.0g * 1mol/17 g T = 200 + 273
P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K) (4.94 mol)2*4.17 atm L2 mol-2
5 L – (4.94 mol*3.71E-2 L mol-1) (5 L)2
P = 39.81 atm – 4.07 atm = 35.74
P = 38.3 atm