2015 chemhandouts1

Partido State University
SE Enhancement Class 2015
Chemistry-handouts
ALV 2015 Page 1
SI or Metricsystem
 powers/multiplesof 10
 2 typesof SI units
 fundamental
 derived
 7 base units/fundamental units
 meter– length.widthorheight/ distance
 Kelvin –temperature
 second – time
 kilogram– mass
 Ampere – electriccurrent
 mole – amountof substance
 candela– luminousof intensity
precisionvs. accuracy
precision
 valuesof setor seriesof measurementsare closertoeachother
accuracy
 measure of howclose yourmeasuredvalue istothe actual value
Percenterror (% error)
% 𝑒𝑟𝑟𝑜𝑟 =
𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑣𝑎𝑙𝑢𝑒
𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝑥 100
Mass vs. Weight
 Mass - measure of the amountof matterin an object (constant).
 Weight - measure of the force of gravitational attractionbetweenthe objectandthe Earth
(changing)
Significantfigures
1. Nonzerodigitsare ALWAYSsignificant
2. Leadingzerosare NOT significant
3. Captive zerosare ALWAYSsignificant
4. Trailingzerosat the rightendof a numberare NOT significant;but are ALWAYSsignificantafter
a decimal point
Significantfiguresin Measurements
 Multiplication/Division
 numberof significantfiguresinthe answer—isthe same asthe numberwithLEAST
significantfigures
 Addition/Subtraction
 answerwill have the same numberof placesasthe numberwiththe LEAST placestothe
rightof the decimal point
ScientificNotation
 Whenmultiplyingexponential terms, addexponents.
 Whendividingexponential terms, subtractexponents.
 Whenraising exponential termstoapower, multiplyexponents.
Molecules: The Law of Definite Proportions
 Law of definite proportions - a givenpure compoundalwayscontains:the same elementsin
exactlythe same proportionsbymass

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 e.g.: Water will alwaysconsistof 2atoms of hydrogenandone atom of oxygen.
 Molecularformulas
 containa collectionof elementalsymbolswhichasa grouprepresentsone molecule.
Formula Name Proportions
CO Carbonmonoxide
1 atomof carbon
1 atomof oxygen
H2O water
2 atomsof hydrogen
1 atomof oxygen
PropertiesofPure Substances
a. Physical properties
 can be observedwithoutchangingthe compositionof the substance.
 include:color, odor,taste, solubility , density,meltingpoint and boilingpoint
b. Chemical properties
 can be observedwhenasubstance undergoesachange incomposition.
 include:iron rusting, gasoline burninginair,waterundergoingelectrolysis andchlorine
reacts withsodium
Changesof Pure Substances
a. Physical changes
 occur withouta change inthe compositionof the substance (conversionfromone state
of matterto another)
b. Chemical changes
 observedwhenachange inthe compositionof asubstance occurs. New substances are
formedwithdifferentphysical andchemical properties.
Cl2 + Na ---> 2 NaCl
 chlorine gas(poisonous) plussodium(reactive metal) producesanew substance--
sodiumchloride (tablesalt) whichhastotallydifferentphysical andchemical
properties
Energy
 Energy isdefinedasthe abilitytodoworkor to transferheat
 principal typesof energyare:mechanical,heat,electrical,chemical andlight
 Energycan eitherbe
 potential energy- energypossessedbyitspositioninspace oritschemical composition
 kineticenergy- due to motion
 Heat energy- energyistransferredfromone substance toanotherwhenthere isatemperature
difference betweenthe substances.
Measurementof Energy(units)
 Calorie (cal)
 Joule (j)
 1 cal = 4.184 J
A calorie is the quantityof heatrequiredtoraise the temperature of 1 gram of water from14.5
o
C to 15.5 o
C
SpecificHeat
 Specificheat (SpH)
 physical property

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 amountof energy(calories) requiredtoproduce agivenchange intemperature ( o
C) in
relationtothe mass of a substance
 inequation:
𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 =
𝑐𝑎𝑙𝑜𝑟𝑖𝑒𝑠
𝑚𝑎𝑠𝑠 (∆𝑇)
Example:
Calculate the SpHof a metal witha massof 25 grams and ittakes250 caloriesof heatenergyto
raise the temperature of the metal from20 o
C to 25 o
C.
𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 =
250 𝑐𝑎𝑙𝑜𝑟𝑖𝑒𝑠
25𝑔 (25℃ − 20℃)
= 2.0 𝑐𝑎𝑙/𝑔℃
Law of Conservation of Mass and Conservation of Energy
 Law of Conservation of Mass:
Mass can neither be created nor destroyed. The total mass of any system always
remains constant.
 Law of Conservation of Energy:
Energycan neitherbe creatednordestroyed.Energymaybe transformedfromone type
to another.
Division of the Elements
 metals
 nonmetals
 metalloids
Metals have certain physical properties:
 high luster; conduct electricity; malleable; ductile; most have high densities; many have high
m.p.; most are hard
Metals have certain chemical properties:
 do not readily combine with other metals
 do combine with nonmetals
 few are found in the free state (example Au, Ag, Cu, and Pt)
Nonmetals have certain physical properties:
 Notlusterous;poorconductors;notmalleable orductile;brittle;mosthave low densities; many
have low melting points (mp); most are soft
Nonmetals have certain chemical properties:
 combine with metals or other nonmetals
 few exist in nature in the free state
Dalton's Atomic Theory
 elements are composed of atoms
 atoms are indestructible
 atoms of the same element are identical
 atoms combine in whole number ratios
 atoms of different elements can unite in different ratio
The Structure of the Atom
 The atom consists of three principle subatomic particles:
Particle Symbol Charge Relative mass
Electron E Positive (+) Approx. zero (0)
Proton P Negative (-) 1

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Neutron N Neutral (o) 1
Mass Number
 sum of the number of protons and neutrons.
Mass Number = number of protons + number of neutrons
Atomic number
 refer to the number of protons or electrons of an atom
Ions
 charged atoms
 could be:
 cation (+)
 anion (- )
The Mole
 amountof substance
 1 doz= 12
 1 mole = 6.02 x 10 23 atomsmolecules,particles
 equation:
𝑛 =
𝑀
𝑀𝑊
where:
n =amountof substance,moles
M = mass of substance,kg
MW = molarmass/molecularmass
Convertingmolesto grams
How manygrams of lithiumare in3.50 molesof lithium?
𝑀 = 𝑛 𝑥 𝑀𝑊
𝑀 = 3.50 𝑚𝑜𝑙𝑒𝑠 𝑥
7 𝑔 𝐿𝑖
1 𝑚𝑜𝑙𝑒 𝐿𝑖
= 24.5 𝑔 𝐿𝑖
Convertinggrams to moles
How manymolesof lithiumare in18.2 grams of lithium?
𝑛 =
𝑀
𝑀𝑊
𝑛 =
18.2𝑔𝐿𝑖
7𝑔 𝐿𝑖
= 2.6 𝑚𝑜𝑙𝑒 𝐿𝑖
UsingAvogadro’s Number
How many atoms of lithiumare in3.50 molesof lithium?
𝑁𝐴 = 𝑛 𝑥 𝑁
𝑁𝐴 = 3.50 𝑚𝑜𝑙𝑒𝑠 𝐿𝑖 𝑥 6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠
𝑁𝐴 = 2.07 𝑥 1024 atoms
Formulas
 Empirical formula:the lowestwholenumberratioof atomsina compound.
 Molecularformula: the true numberof atomsof eachelementinthe formulaof acompound.

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 molecularformula = (empirical formula)n [n =integer]
 molecularformula = C6H6 = (CH)6
 empirical formula = CH
STOICHIOMETRY
Chemical Equations
C2H5OH + 3O2 2CO2 + 3H2O
reactants products
Quantitative significance:
1 mole of ethanol reactswith 3 molesof oxygentoproduce 2 molesof carbon dioxide
and 3 molesof water
StoichiometryProblem1
6.50 grams of aluminumreactswithan excessof oxygen.How manygramsof aluminumoxide
are formed?
4 Al + 3 O2  2Al2O3
6.50 𝑔 𝐴𝑙 𝑥
1 𝑚𝑜𝑙𝑒 𝐴𝑙
27 𝑔 𝐴𝑙
𝑥
2 𝑚𝑜𝑙 𝐴𝑙2 𝑂3
4 𝑚𝑜𝑙 𝐴𝑙
𝑥
101.96 𝑔 𝑚𝑜𝑙 𝐴𝑙2 𝑂3
1 𝑚𝑜𝑙 𝑚𝑜𝑙 𝐴𝑙2 𝑂3
= 12.3 𝑔 𝑚𝑜𝑙 𝐴𝑙2 𝑂3
StoichiometryProblem2
How many grams of hydrogen will be produced if 10.0 grams of calcium is added to an excess of
hydrochloric acid?
2 HCl + Ca  CaCl2 + H2
10𝑔 𝐶𝑎 𝑥
1 𝑚𝑜𝑙𝑒 𝐶𝑎
40𝑔 𝐶𝑎
𝑥
1 𝑚𝑜𝑙𝑒 𝐻2
1 𝑚𝑜𝑙𝑒 𝐶𝑎
𝑥
2𝑔 𝐻2
1 𝑚𝑜𝑙𝑒 𝐻2
= 0.5 𝑔 𝐻2
GASES
Characteristics of Gases
 Expansion - gases expand indefinitely and uniformly to fill all the space in which they are placed.
 Indefinite Shape and Volume - gas has no definite shape or volume, but will fit the vessel in which it is
placed.
 Compressibility - gases can be highly compressed.
 Low density - density of gases is very low and, therefore, measured in grams/liter (g/l)
 Diffusion - two or more different gases will normally mix completely and uniformly when in contact with
each other.
Pressure of Gases
 Pressure is defined as force per unit area (force/area).
 STP (standard temperature and pressure) - the conditions are 0 oC and 1 atm.
 Units of pressure
 14.7 psi = 1 atm
 76.0 cm Hg = 1 atm
 760 mm Hg = 1 atm
 760 torr = 1 atm
 1.013 x 105 Pa = 1 atm
Gas Laws
A. Boyle's Law
 Boyles Law - at constant temperature, the volume is inversely proportional to the pressure.

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 As the pressure increases, the volume decreases; as the pressure decreases, the volume increases
Example:
A gas has a volume of 2 liters at a pressure of 6 atm. Calculate the new volume if the pressure was
decreased to 3 atm and the temperature remained the same.
B. Charles's Law
 Charles Law - at constant pressures the volume is directly proportional to the temperature.
 As the temperature increases, the volume increases, as the temperature decreases, the volume decreases
Example:
A gas has a volume of 2 liters at a temperature of 600 K. Calculate the new volume if the temperature
was decreased to 300 K and the pressure remains the same.
C. Gay-Lussac's Law
 Gay-Lussac Law - at constant volume, the pressure is directly proportional to the temperature.
 As the temperature increases, the pressure increases; as the temperature decreases, the pressure
decreases
Example:
A gas has a pressure of 700 torr at a temperature of 300 K. Calculate the new pressure if the temperature
was increased to 600 K and the volume remains the same.
In Gay-Lussac's Law, the pressure is proportional to the temperature. Therefore, if the temperature is doubles, the
pressure would be expected to double.
D. General Gas Law
 Boyles and Charles Law can be combined into a single equation.
Example: solve for pressure
2 liters of a gas has an initial pressure of 800 torr at a temperature of 300 K. What would be the new
pressure if the temperature was increased to 600 K and the volume remained the same?
Solution:
Using the general gas law formula, solve for P2

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Example: solve for temperature
A gas has an initial pressure of 4 atm at a temperature of 300 K with an initial volume of 2 liters. What
would be the new temperature if the pressure was increased to 8 atm and the volume doubled?
Solution:
Using the general gas law formula, solve for T2
Example: solve for volume
A gas has an initial pressure of 30 psi at a temperature of 600 K with an initial volume of 2 liters. What
would be the new volume if the pressure was decreased to 15 psi and the temperature remained the
same?
Solution:
Using the general gas law formula, solve for V2
E. Ideal Gas Law
Example: solve for moles
A gas has a pressure of 4 atm. at a temperature of 300 K with an initial volume of 2000 ml. Calculate the
moles of the gas.
Solution:
Using the ideal gas law formula, solve for moles.
2. Convert all units to either atm., liters, or kelvin.
Example: solve for pressure
2 moles of a gas has a volume of 0.75 liters at a temperature of 27 oC. Calculate the pressure of the gas.
Solution:
Using the ideal gas law formula, solve for pressure.

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Example: solve for the gas constant
1 mole of a gas at STP (273 K and 1 atm) has a volume of 22.4 liters. Calculate the gas constant R.
Solution:
Using the ideal gas law formula, solve for the gas constant R.
Example: solve for temperature
Nitrogen gas (N2 - 28 amu) has a volume of 5 liters, a pressure of 15 atm and a mass of 56
grams. Calculate the temperature of the gas.
Solution:
Using the ideal gas law formula, solve for temperature.
Example: solve for volume
A gas has a temperature of 300 K, a pressure of 22 psi and 2 moles. Calculate the volume of the gas.
Solution:
Using the ideal gas law formula, solve for volume.

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Example: solve for mass
Chlorine gas (Cl2 - 71 amu) has a volume of 2 liters, a pressure of 4 atm and a temperature of 300
kelvin. Calculate the mass of chlorine.
Solution:
Using the ideal gas law formula, solve for moles.
grams = moles x mol wt
grams = 0.32 moles x 71 amu = 22.7 grams
SOLUTIONS
Solution
 homogeneous mixture of two or more substances.
 generally composed of two substances
 solute ---- substance being dissolved and is in smaller quantity.
 solvent --- substance that dissolves the solute and is in larger quantity
Types of solutions
 liquid solution
 e.g. salt solution (salt & water)
 gaseous solution
 e.g. air solution (various gases like oxygen, nitrogen, carbon dioxide, water and
other gases)
 solid solutions
 e.g. metal alloys like brass (copper and zinc)
Factors that Affect Solubility
 nature of solute and solvent
 “LIKE DISSOLVES LIKE". Water dissolves salts, but not oil or gasoline.
 Temperature
 the higher the temperature, more solute will dissolve in a given solvent.
 Pressure
 pressure has little or no effect on the solubility of a solid or liquid, but the
solubility of gases is greatly affected by pressure (Henry's Law).
Henry's Law
 Henry's Law - the solubility of a gas in a liquid is directly proportional to the applied
pressure. ---the higher the pressure, the more gas dissolves in a liquid.

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 e.g. Carbonated soft drinks (carbon dioxide gas dissolved in a liquid (water) at
high pressure)
Factors that affect the Rate of Dissolution
 Particle size
 the smaller the particle size, the faster the rate of dissolution. Powders dissolve
faster than large lumps.
 Temperature
 the higher the temperature, the faster is the rate of dissolution. It is easier to
dissolve sugar in a glass of hot tea than it is in a glass of iced tea.
 Rate of stirring
 stirring or agitation of the solute in a solvent causes it to dissolve faster.
 Concentration
 if a solution has some solute dissolved in it, the rate of dissolving additional
solute will be slower.
Relative Terms for Expressing Solute Concentration
Solubility
 the amount of solute that can be dissolved in a given amount of solvent.
Non-quantitative terms:
 Concentrated solution: the solution contains more solute than a dilute solution.
 Dilute solution: the solution contains less solute than concentrated solution.
 Unsaturated solution: the solution can dissolve more solute.
 Saturated solution: the solution contains the maximum amount of solute the solvent
can dissolve.
 Supersaturated solution: the solution contains more solute than a saturated solution.
This is very unstable condition and slight disturbance causes the excess solute to settle
out.
Note: All of these are temperature dependent. A solution that is saturated at 25o
C may be
unsaturated at 45o
C.
Quantitative Terms for Expressing Solute Concentration
A. Percent Concentration of Solute
 expressed as the percentage of the solute based on the entire solution.
a. The solution is the total of the solute and the solvent.
 The concentration can be expressed as percent :
a. the percent by mass
b. percent by volume (for two liquids)

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c. combination of the mass of the solute and the volume of the solution.
Example: (% by mass)
20 grams of NaCl is is mixed with 180 ml of water, what would be concentration as
%NaCl by mass in this solution?
 mass water = 180 g (remember that the density of water = 1.0 g/ml)
Example (% by volume):
200 ml of antifreeze is mixed with 0.8 liters of water, what would be % concentration of
antifreeze by volume in this solution?
 volume water = 800 ml (remember 0.8 liters = 800 ml)
Example (combination of the mass of the solute and the volume of the solution)
150 ml of ethanol is mixed with 1.5 liters of water, what would be % concentration of
ethanol by mass/volume in this solution?
 volume water = 1500 ml (remember 1.5 liters = 1500 ml)

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 mass ethanol = 118.5 g (density = 0.79 g/ml and mass = density x vol)
B. The concentration can be expressed as parts by mass of solute per million parts of solution
(ppm) (ppm)
Example:
A drinking water sample has 15 mg of Pb per 500 ml of solution. Calculate the ppm of
Pb in the water.
 mass solution = 500 ml = 500 g = 500,000 mg
C. Molarity (M)
 moles of solute per liter of solution
Example:
120 grams of NaOH is diluted to 750 ml. What would be the molarity of this solution.
First, write the molarity formulas
 0.75 liters (750 ml = 0.75 l)
 determine the formula mass of NaOH
 calculate the moles of NaOH
 calculate the molarity of NaOH

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Example:
500 ml of a 2.5 M solution of Ca(OH)2 needs to be made. What would be the mass of
Ca(OH)2 needed to dilute to 800 ml.
 write the molarity formulas
 convert ml to liters (500 ml / 1000 = liters)
0.5 liters (500 ml = 0.5 l)
 determine the formula mass of Ca(OH)2
Ca(OH)2 = 74.1 amu
 calculate the moles of Ca(OH)2
 calculate the grams of Ca(OH)2
 To make the 2.5 M Ca(OH)2 solution, weigh out 92.6 grams of Ca(OH)2 and dilute with
water to 800 ml.
D. Normality (N)
 used when more precise measurements of the concentration of solutions are needed
and the solutions are being use in reactions of acids and bases.
Note that equivalents can be defined in several ways. Here are just a few.
 The number of moles of H+ or OH- ions replaced in a chemical reaction.
 The number of replaceable H+ or OH- ions in a compound.
 The number of moles of electrons transferred in a chemical reaction (red-ox).

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Example:
296.4 grams of Ca(OH)2 is diluted to 600 ml. What would be the normality of this
solution.
 write the normality formulas
 convert 600 ml to liters (600ml / 1000 = liters)
0.6 liters (600 ml = 0.6 L)
Ca(OH)2 = 74.1 amu
 calculate the number of replaceable OH- in Ca(OH)2
Ca(OH)2 has 2 replaceable hydroxides
 calculate the equivalent mass of Ca(OH)2
 calculate the equivalents of Ca(OH)2
 calculate the normality of Ca(OH)2
Example:
600 ml of a 2.5 M solution of Ca(OH)2 needs to be made. What would be the mass of
Ca(OH)2 needed to dilute to 600 ml.
 write the normality formulas

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 convert 600 ml to liters (600 ml / 1000 = liters)
0.6 liters (600 ml = 0.6 L)
Ca(OH)2 = 74.1 amu
 calculate the number of replaceable OH- in Ca(OH)2
Ca(OH)2 has 2 replaceable hydroxides
 calculate the equivalent mass of Ca(OH)2
 calculate the equivalents of Ca(OH)2
 calculate the normality of Ca(OH)2
 To make the 2.5 M Ca(OH)2 solution, weigh out 55.6 grams of Ca(OH)2 and dilute with
water to 600 ml.
E. Dilution Formula
 used to dilute a concentrated stock solution (these stock solutions are concentrated so
as to save room during storage and costs in shipping) to a desired concentration.
C1 = the initial concentration
C2 = the final concentration
V1 = the initial volume
V2 = the final volume
Example:
A stock solution of hydrochloric acid is has a concentration of 12M (C1). 100 ml (V2) of a
6M HCl (C2) solution needs to be made from the stock solution. How many ml (V1) of
the stock solution needs to be diluted to 100 ml (V2) to make the 6M (C2) solution?

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 substitute the actual values for the variables in the formula.
Electrolytesvs. Nonelectrolytes
electrolytes
 substance whose aqueoussolutionconductsanelectriccurrent
nonelectrolytes
 substance whose aqueoussolutiondoesnotconductan electriccurrent
ACIDS AND BASES
Acid/Base Definitions
 Arrhenius Model
 Acids produce hydrogen ions in aqueous solutions
 Bases produce hydroxide ions in aqueous solutions
 Bronsted-Lowry Model
 Acids are proton donors
 Bases are proton acceptors
 Lewis Acid Model
 Acids are electron pair acceptors
 Bases are electron pair donors
Properties of Acids
 Acids are proton (hydrogen ion, H+
) donors
 Acids have a pH lower than 7
 Acids taste sour
 Acids effect indicators
 Blue litmus turns red
 Methyl orange turns red
 Acids react with active metals, producing salts and hydrogen gas(H2)
 Acids react with carbonates
 Acids neutralize bases
Strong Acids
 assumed to be 100% ionized in solution (good H+
donors).
 include:
 hydrochloric acid ----- HCl
 Sulfuric acid ------ H2SO4
 Nitric acid ------ HNO3
 Hydriodic acid ----HI
 Perchloric acid -----HClO4
 Hydrobromic acid ----HBr
 Hydronium ion -----H3O+
Weak acids

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 include:
 Iodic acid ----HIO3
 Oxalic acid ---- H2C2O4
 sulfurous acid ----H2SO3
 phosphoric acid ----H3PO4
 nitrous acid ----HNO2
 Hydroflouric acid –HF
 formic acid ---- HCOOH
 benzoic acid ---C6H5COOH
 acetic acid -----CH3COOH
 crbonic acid ---H2CO3
 organic acids and their sources
 Citric acid ---H3C6H5O7 – citrus fruit
 Malic acid – apples
 Butyric acid – rancid butter
 Amino acids – protein
 Nucleic acids – DNA and RNA
 Ascorbic acid – Vitamin C
Properties of Bases
 Bases are proton (hydrogen ion, H+
) acceptors
 Bases have a pH greater than 7
 Bases taste bitter
 Bases effect indicators
 Red litmus turns blue
 Phenolphthalein turns purple
 Solutions of bases feel slippery
 Bases neutralize acids
Examples of Bases
 Sodium hydroxide (lye)---- NaOH
 Potassium hydroxide ----KOH
 Magnesium hydroxide ----- Mg(OH)2
 Calcium hydroxide (lime) ----- Ca(OH)2
Weak bases
 ammonia ----NH3
 methylamine -----CH3NH2
 ethylamine -----C2H5NH2
 Diethylamine ---- (C2H5)2NH
 Hdroxylamine --- HONH2
 Hydrazine ------ H2NNH2
 Aniline ---- C6H5NH2
 Pyridine ----C5H5N
Self-Ionization of Water
 At 25, [H3O+
] = [OH-
] = 1 x 10-7
Kw is a constant at 25 C
 Kw = [H3O+
][OH-
]

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 Kw = (1 x 10-7
)(1 x 10-7
) = 1 x 10-14
Calculating pH, pOH
 pH = -log10(H3O+
)
 pOH = -log10(OH-
)
Relationship between pH and pOH
 pH + pOH = 14
Finding [H3O+
], [OH-
] from pH, pOH
 [H3O+
] = 10-pH
 [OH-
] = 10-pOH
Example:
What isthe pH of a 0.50 M solutionof aceticacid,HC2H3O2,Ka = 1.8 x 10-5
? Solve forpH and pOH
of the solution
 write the equation
HC2H3O2  C2H3O2
- + H+
0.50 – x x x
[H+
] = 3.0 x 10-3
M
Reaction of Weak Bases with Water
B + H2O  BH+
+ OH-
Example
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5
?
 Write the equation for the reaction
NH3 + H2O  NH4
+
+ OH-
 Set up the law of mass action
NH3 + H2O  NH4
+
+ OH-
50-x x x
)50.0()50.0(
))((
108.1
2
5 x
x
xx
x 


)50.0(
108.1
2
5 x
x 
52.4)100.3log( 5
 
xpH
[ ][ ]
[ ]
b
BH OH
K
B
 

)50.0()50.0(
))((
108.1
2
5 x
x
xx
x 


)50.0(
108.1
2
5 x
x 

Chemistry-handouts
ALV 2015 Page 19
[OH-
] = 3.0 x 10-3
M
52.4)100.3log( 5
 
xpOH
48.900.14  pOHpH

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