~ 'IJe rinceton Rev1ew d! MCAT GENERAL CHEMISTRY Steven A Leduc Kendra Bowman Copyright© 2003 by Princeton Review, Inc All rights reserved Contributors Steven A Leduc National Director of MCAT Research, Production, and Development, The Princeton Review Kendra Bowman, Ph.D Christopher Volpe, Ph.D Matthew B A Patterson, M.D Staff of The Princeton Review Copyright© 2003, 2002, 2001, 2000, 1999, 1998, 1997 by Princeton Review, Inc All rights reserved Version 5.1 MCAT is a registered trademark of the Association of American Medical Colleges (AAMC) The Princeton Review is not affiliated with Princeton University or AAMC This manual is for the exclusive use of Princeton Review course students, and is not legal for resale ~ '11e rinceton Review (]! www Princeton Review com 030930 306 MCAT GENERAL CHEMISTRY TABLE OF CONTENTS ~ l/ '(/ /{,.o j)~ ~~ 11 REFERENCE 311 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Metric Units 311 Density ~ 312 Molecular Formulas 313 Formula and Molecular Weight 313 The Mole 314 Empirical Formulas 315 Percentage Composition by Mass 315 Chemical Equations and Stoichiometric Coefficients 318 Balancing Equations 318 0.9 Stoichiometric Relationships in Balanced Equations 319 0.10 The Limiting Reagent 320 0.11 Some Notations Used in Chemical Equations 321 12 Oxidation States 322 ATOMIC STRUCTURE 325 1.1 Elements and the Periodic Table 325 1.2 Atoms 325 1.3 Isotopes 326 Atomic Weight 327 1.4 Ions 327 1.5 Electron Quantum Numbers 328 1.6 Electron Configurations 329 Blocks in the Periodic Table 332 Some Anomalous Electron Configurations 335 Electron Configurations of Ions 335 Diamagnetic and Paramagnetic Atoms 337 Electron Energy Levels 338 Line Spectra 339 1.8 Nuclear Structure 341 Nucleus Stability and Radioactivity 341 Alpha Decay 341 Beta Decay 342 Gamrna Decay 343 Half-Life 346 Nuclear Binding Energy 348 307 PERIODIC TRENDS AND BONDING • 349 2.1 Groups of the Periodic Table 349 2.2 Periodic Trends 351 2.3 Bonding 354 Lewis Dot Symbols 354 Covalent Bonds • • 355 Formal Charge • 356 Resonance • 358 Polar Covalent Bonds • 359 Coordinate Covalent Bonds 360 Ionic Bonds • 361 2.4 VSEPR Theory .• 362 ·A Note on Hybridization ~ 364 2.5 Solids • 365 2.6 Intermolecular Forces 366 Hydrogen Bonding • • 367 PHASES ························~····· 369 3.1 Physical Changes 369 Phase Transitions 369 3.2 Heats of Phase Changes 371 3.3 Calorimetry 372 3.4 Phase Transition Diagram ~ 374 3.5 Phase Diagrams 376 The Phase Diagram for Water 377 ~ GASES 3~9 4.1 Gases and the Kinetic-Molecular Theory 379 Units of Volume, Temperature, and Pressure 380 4.2 The Ideal-Gas Law ~ · 381 Other P-V-T Gas Laws .: 382 4.3 Dalton's Law of Partial Pressures 385 4.4 Graham's Law of Effusion 386 4.5 Approaching Ideal-Gas Behavior 390 308 SOLUTIONS 393 5.1 Dissolution and Solubility 393 Concentration Measurements 393 Electrolytes 394 Solubility Rules 396 5.2 Colligative Properties 397 Vapor-Pressure Depression 397 Boiling-Point Elevation 399 Freezing-Point Depression 399 Osmotic Pressure 401 KINETICS 403 6.1 Reaction Mechanism: An Analogy 403 Rate-Determining Step 404 6.2 Reaction R·ate 404 Activation Energy 404 6.3 Catalysts 406 6.4 Rate Laws 407 The Rate Constant , 408 EQUILIBRIUM 411 7.1 Equilibrium 411 The Equilibrium Constant 411 7.2 The Reaction Quotient 414 7.3 Le Chatelier's Principle 415 7.4 Solubility Product Constant 419 Solubility Computations 419 7.5 lon Product 420 7.6 The Common-len Effect 421 ACIDS AND BASES 423 8.1 Definitions 423 8.2 Conjugate Acids and Bases 424 8.3 The Strengths of Acids and Bases 425 8.4 The len-Product Constant for Water 429 8.5 pH 429 8.6 Neutralization Reactions 434 8.7 Indicators 435 8.8 Hydrolysis of Salts 437 8.9 Buffer Solutions 438 8.1 Acid-Base Titrations 441 309 9· THERMODYNAMICS ······················"'·························· 445 9.1 System and Surroundings 445 9.2 Laws of Thermodynamics 446 The First Law 446 The Second law ~ 446 Physical Thermodynamics: Heat and Work 446 9.3 Entropy 451 9.4 Enthalpy 452 9.5 Heat of Formation 453 Enthalpy and Heats of Formation 454 Standard State Conditions 454 Enthalpy and Balanced Equations 454 Hess's Law of Heat Summation 454 9.6 Gibbs Free Energy : 456 A.G and Temperature 456 9.7 Reaction Energy Diagrams 458 Kinetics vs Thermodynamics 458 Reversibility 459 The Thermodynamics of Catalysts 459 9.8 Thermodynamics and Equilibrium 460 10 ELECTROCHEMISTRY 461 10.1 Oxidation-Reduction Reactions 461 10.2 Galvanic Cells 462 t0.3 Reduction Potentials 464 Oxidizing and Reducing Agents - 466 10.4 Electrolytic Cells 469 10.5 Faraday's Law of Electrolysis 470 10.6 Concentration Cells 471 GLOSSARY , :··· 473 MCAT G-CHEM FORMULA SHEET 497 310 MCAT GENERAL CHEMISTRY- CHAPTER 0: REFERENCE 311 REFERENCE §0.1 METRIC UNITS Before we begin our study of chemistry, we'll briefly go over metric units Scientists use the Systeme International d'Unites (the International System of Units), abbreviated SI, to express the measurements of physical quantities Six of the seven base units of the SI are given below: SI hil5~ :unit abhr~yiatiQn mea5:ures meter m length kilogram kg mass second s time mole mol amount of substance kelvin K temperature ampere A electric current [The seventh SI base unit, the candela (cd), measures luminous intensity, but we won't need to worry about this one.] The units of any physical quantity can be written in terms of the SI base units For example, the SI unit of speed is meters per second (m/s), the SI unit of energy (the joule) is kilograms times meters2 per second2 (kg·m2 /s2), and so forth Multiples of the base units that are powers of ten are often abbreviated and precede the symbol for the unit For example, m is the symbol for milli-, which means 10-3 (one thousandth) So, one thousandth of a second,l millisecond, would be Written as ms The letter M is the symbol for mega-, which means 106 (one million); a distance of one million meters, megameteriwould be abbreviated as Mm Some of the most common power-of-ten prefixes are given in the list below: mulnpl~ prefix symbol nano- n 10-9 micro- 10-6 milli- Jl m centi- c 10-2 kilo- k 103 mega- M 106 10-3 Two other units, ones that are common in chemistry, are the liter and the angstrom The liter (abbreviated L) is a unit of volume equal to 1/1000 of a cubic meter: lOOt> L ::;: m 3,_ ~l ~ ,/''\ /(,::' 1l =1000 cm3 ~ /' The standard SI unit of volume, the cubic meter, is inconveniently large for most laboratory, work The liter is a smaller unit Furthermore, the most common way of expressing solution concentrations, molarity (M), uses the liter in its definition: M =moles of solute per liter of solution 312 MCAT PHYSICAL SCIENCES REVIEW In addition, you'll see the milliliter (mL) as often as you'll see the liter A simple consequence of the definition of a liter is the fact that one milliliter is the same volume as one cubic centimeter: mL = cm3 = cc While the volume of any substance can, strictly speaking, be expressed in liters, you rarely hear of a milliliter of gold, for example Ordinarily, the liter is used to express the volumes of liquids and gases, but not of solids The angstrom, abbreviated A, is a unit of length equal to 10-10 m The angstrom is convenient because atomic radii and bond lengths are typically around to A ., Example 0-1: By how many orders of magnitude is a centimeter longer than an angstrom? Solution An order of magnitude is a factor of ten Since em = 10-2 m and A = 10-10 m, a centimeter is factors of ten, or orders of magnitude, greater than an angstrom §0.2 DENSITY The density of a substance is its mass per volume: mass _ m density: P = volume - V In SI units, density is expressed in kilograms per cubic meter (kg/m3) However, in chemistry, densities are more often expressed in grams per cubic centimeter (g/ cm3) This unit of density is convenient because most liquids and solids have a density of around to 20 g/ cm3 • Here's the conversion between these two sets of density units: _L cm3 multiply by 1000 kg m3 divide by 1000 For example, water has a density of g/ cm3 (it varies slightly with temperature, but this is the value the MCATwill expect you to use) To write this density in kg/m3, we'd multiply by 1000: The density of water is 1000 kg/m3 • As another example, the density of copper is about 9000 kg/m3, so to express this density in g/cm3, we'd divide by 1000: The density of copper is g/cm3• , Example 0-2: Diamond has a density of 3500 kg/m3 • What is the volume, in cm3, of a t-carat diamond (where, by definition, carat = 0.2 g)? Solution If we divide mass by density, we get volume, so, converting 3500 kg/m3 into 3.5 g/ cm3, we find that m 1.75(0.2 g)_ 0.35 o =0.1 cm3 V=-= p g 3.5em3 - g 3.5 cm3 MCAT GENERAL CHEMISTRY ,;,;_ CHAPTER 0: REFERENCE 313 §0.3 MOLECULAR FORMULAS When two or more atoms form a covalent bond they create a molecule For example, when two atoms of hydrogen (H) bond to one atom of oxygen (0), the resulting molecule is H 20, water A compound's molecular formula gives the identities and numbers of the atoms in the molecule For example, the formula C4H 4N2 tells us that this molecule contains carbon atoms, hydrogen atoms, and nitrogen atoms Example 0-3: What is the molecular formula of para-nitrotoluene? A C6H 5N02 CH ! HC/)H B C7~N02 C C7H 8N02 D C 7H 9N02 H~v Solution There are a total of seven C's, seven H's, one N, and two 0's, so choice B is the correct answer H ~~ §0.4 FORMULA AND MOLECULAR WEIGHT If we know the chemical formula, we can figure out the formula weight, which is the sum of the atomic weights of all the atoms in the molecule The unit for atomic weight is the atomfc mass unit, abbreviated amu (or simply u) (Note: Although weight is the popular term, it should really be mass.) One atomic mass unit is, by definition, equal to exactly 1/12 the mass of an atom of carbon12, the most abundant naturally-occurring form of carbon The periodic table lists the mass of each element; it is actually a weighted average of the atomic masses of all its naturally-occurring forms (isotopes) For example, the atomic mass of hydrogen is listed as 1.0 (amu), and that of nitrogen as 14.0 (amu) Therefore, the formula weight for C4H 4N is 4{12) + 4(1) + 2(14) = 80 (The unit amu is often not explicitly included.) When a compound exists as discrete molecules, the term molecular weight (MW) is usually used instead of formula weight For example, the molecular weight of water, H 20, is 2{1) + 16 = 18 The term formula weight is usually used for ionic compounds, such as NaCl The formula weight of NaCl is 23 + 35.5 = 58.5 Example 0-4: What is the molecular weight of calcium phosphate, Ca3(P04) 2? A B C D 310 g/mol 350 g/mol 405g/mol 450 g/mol Solution The masses of the elements areCa= 40 g/mol, P = 31 g/mol, and = 16 g/mol Therefore, the molecular weight of calcium phosphate is 3(40 g/mol) + 2(31 g/mol) + 8{16 g/mol) = 310 g/mol Choice A is the answer 314 MOAT PHYSICAL SCIENCES REVIEW §0.5 THE MOLE A mole is simply a particular number of things, like a dozen is any group of 12 things One mole of anything contains 6.02 x 1023 entities A mole of atoms is a collection of 6.02x 1023 atoms; a mole of molecules contains 6.02 x 1023 molecules, and so on This number, 6.02 x 1023,is called Avogadro's number, denoted by NA (or N0) What's so special about 6.02 x 1023? The answer is based on the atomic mass unit, which is defined so that the mass of a carbon-12 atom is exactly 12 u: The number of carbon-12 atoms in a sample of mass of 12 grams is 6.02 x 1023 • Avogadro's number is the link between atomic mass units and grams For example, the periodic table lists the mass of sodium (Na, atomic number 11) as 23.0 This means that atom of sodium has a mass of 23 atomic mass units, or that mole of sodium atoms has a mass of 23 grams Since mole of a substance has a mass in grams equal to the mass in amu' s of formula unit of the substance, we have the following formula: mass in grams # moles- molecular weight (MW) _ Example 0-5: (a) Which has the greater molecular weight: potassium dichromate (~Cr20 7) or lead azide (PbN6)? (b) Which contains more molecules: a 1-mole sample of potassium dichromate or a 1-molesample of le~d azide? Solution (a) The molecular weight of potassium dichromate is 2(39.1) + 2(52) + 7(16) =294.2 and the molecular weight of lead azide is 207.2 + 6(14) =291.2 Therefore, potassium dichromate has the greater molecular weight (b) Trick question Both samples contain the same number of molecules, namely 1.Il).ole of them (Which weighs more: a pound of rocks or a pound of feathers?) _ Example 0-6: How many molecules of hydrazine, N 2H 4, are in a sample whose mass is 96 grams? Solution The molecular weight of N 2H is 2(14) + 4(1) = 32 This means that mole of N 2H has a mass of 32 grams Therefore, a sample that has a mass of 96 grams contains moles of molecules, because the formula above tells us that '' = 96 g , 32 g/ mol = moles 540 MCAT PHYSICAL SCIENCES REVIEW It's important to notice that the actual numerical value of the proportionality constant was irrelevant in the statements made above For example, the fact that 1t is the proportionality constant in the equation C =1td did not affect the conClusions made above If C and d were some other quantities and C happened to always be equal to {17,000)d, we'd still say C oc: d, and all the conclusions made in Example 4-4 above would still be correct Graphically, proportions are easy to spot If the horizontal and vertical axes are labelled linearly (as they usually are), then the graph af a proportion is a straight line through the origin Be careful not to make the common mistake of thinking that any 'ole straight line is the graph of a proportion If the line doesn't go through the origin, then it's not the graph of a proportion y y y x This is the ~raph of a proportion -+ ~X Not a proportion: line does not go through 1he origin , .- X Not a proportion: Graph is not a straight line The examples we;ve seen so far have been the equations C = 1td, PE = wh, and F =qvB Notice that in all of these equations, all the variables are present to the first power But what about an equation like this: K = ? This equation gives the kinetic energy of an object of mass m moving with speed v So, if m is constant, K is proportional to v2• Now, what if v were multiplied by, say, a factor of 3what would happen to K? Because K oc: v2, if v increases by a factor of 3, then K will increase by a factor of 32, which is (By the way, this does not mean that if we graph K versus v, we'll get a straight line through the origin K is not proportional to v; it's proportional to v2• If we were to graph K vs v2, then we'd get a straight line through the origin.) Here's another example using the same proportion, K oc: v2: If v were decreased by a factor of 2, then K would decrease by a factor of 22 = !mv Here are a few more examples: IIJo- IIJo- fxr rs 3• Example 4-7: The volume of a sphere of radius r is given by the equation V = Therefore, the volume is proportional to the radius cubed: v oc: So, for example, if r were doubled, then V would increase by a factor of 23 = If r were decreased by a factor of 3, then V would decrease by a factor of 33 = 27 Example 4-8: Consider anobject starting from rest and accelerating at a constant rate of a in a straight line Let the distance it travels be x and its final velocity be v Then it's known that x = (1/2a)v2, which means xis proportional to v squared: x oc: v2• If v were quadrupled, then x would increase by a factor of 42 = 16 If v were reduced by a factor of 5, then x would be reduced by a factor of 52 = 25 Now, how about this: What if x were increased by a factor of 9-what would happen to v? Since x is proportional to v2, if x increases by a factor of 9, then v2 also increases by a factor of 9;this means that v increases by a factor of MCAT MATH- CHAPTER 4: PROPORTIONS 541 ,_ Example 4-9: The intensity of energy radiated by an object of absolute temperature T is given by the formula I= crT', where cr is a constant Therefore, I o: T' If T were increased by a factor of 3, then I would increase by a factor of 34 = 81 And, in order to reduce I to 1/16 its original value, the temperature T would have to be reduced by a factor of 2, since 1;; =( • f) §4.2 INVERSE PROPORTIONS If one quantity is always equal to a nonzero constant divided by another quantity (that is, if A =k! B, where k is some constant), we say that the two quantities are inversely proportional Here are two equivalent ways of saying this: (i) If the product of two quantities is a constant (AB = k), then the quantities are inversely proportional (ii) If A is proportional to 1/B [that is, if A= k(1/B)], then A and Bare inversely proportional In fact, we'll use this final description to symbolize an inverse proportion That is, if A is inversely proportional to B, then we'll write A o:: 1/B (There's no commonly.:.accepted single symbol for inversely proportional to.) Of course, if A= k/B, then B = k/ A, so we could also say that B o:: 1/A Here are a few examples: ,_Example 4-10: The electric potential,tfr, at a distance r from a point charge q is given by the equation tfr = kq/r, where k is a constant Therefore, tfr is inversely proportional to r: tfr o: 1/ r • ,_ Example 4-11: The pressure P and volume V of a sample containing n moles of an ideal gas at a fixed temperature Tis given by the equation PV = nRT, where R is a constant Therefore, the pressure is inversely proportional to the volume: P o: 1/ V ,_Example 4-12: For electromagnetic waves traveling through space, the wavelength A and frequency! are related by the equation A.f = c, where cis the speed of light (a universal constant) Therefore, wavelength is inversely proportional to frequency: A o: 1/f The mos! important fact about inverse proportions is this: If A o: 1/B, and B is multiplied by afactor ofb, then A will be multiplied by a factor ofl/b After all, if A= k/B, then (1/b)A = k!(bB) Intuitively, if one quantity is increased by a factor of b, the other quantity will decrease by the same factor, and vice versa ,_Example 4-13: Since the electric potential is inversely proportional to the distance from the source charge, tfr o: 1/r, then, if the distance is doubled, the potential is cut in half If the distance is cut in half, the potential is doubled If the distance is tripled, the potential is multiplied by 1/3 542 MCAT PHYSICAL SCIENCES REVIEW E'(ample 4-14: Since the pressure of an ideal gas at constant temperature is inversely proportional to the volume, P oc 1/V, then if the volume is doubled, the pressure is reduced by a factor of If the volume is quadrupled, the pressure is reduced by a factor of If the vohtme is divided by (which is the same as saying it's multiplied by 1/3), then the pressure will increase by a factor of Example 4-15: Because for electromagnetic waves traveling through space, the ·wavelength is inversely proportional to frequency, A oc If, iff is increased by a factor of 10, Awill decrease by a factor of 10 If the frequency is decreased by a factor of 2, the wavelength will increase by a factor of The graph of an inverse proportion is a hyperbola In the graph below, xy = k, sox andy are inversely proportional to each other y I - -, x This is the graph of an inverse proportion The examples we've see~ so far have been where one quantity is inversely proportional to the first power of another quantity But what about an equation like this: F=GMm r2 This equation gives the gravitational force between two objects of masses M and m separated by a distance r (G is the universal gravitational constant.) So, if M and m are constant, F is inversely proportional to r 2• Now, what if r were increased by, say, a factor of 3-what would happen to F? Because F cc 1/r, if r increases by a factor of 3, then F will decrease by a factor of 32, which is Here's another example using the same proportion, F oc 1/r: If r were decreased by a factor of 2, then F would increase by a factor of 22 =4 Example 4·16: The electrostatic··force between two charges Q and q separated by a distance r is given by the equationF = kQq/r, where k is a constant Therefore, the electrostatic force is inveJ:'sely proportional to the distance squared: F oc 1/r2• S(), for example, if r were doubled, then F would decrease by a factor of 22 =4 If r were decreased by a factor of 3, then F would increase by a factor of 32 = MCAT MATH- CHAPTER 4: PROPORTIONS ., Example 4·17: The frequency of a simple harmonic oscillator consisting of a block of mass m attached to a spring of force constant k is given by the formula f = "i;;~k/m Therefore, the frequency is inversely proportional to the square root of the block's mass: f o:: 1/ [;;; So, if m were multiplied by 4, fwould be multiplied by 1/2, since /4· = 1/2 If m were increased by a factor of 36, thenfwould decrease by a factor of If we wanted f to increase by a factor of 3, we'd have to decrease m by a factor of9 1/ ., Example 4-18: An object starting from rest travels a distance given by d = at2!2 in a time t if its undergoes a constant acceleration a (a) If t is doubled, then what happens to d? (b) If tis tripled, what happens to d? Solution (a) Since d cc t 2, if t increases by a factor of 2, then d increases by a factor of 22 = (b) Since d cc t 2, if t increases by a factor of 3, then d increases by a factor of ., Example 4-19: The kinetic energy of an object of mass m traveling with speed v is given by the formula K = mv2 /2 (a) If v is increased by a factor of 6, what happens to K? (b) In order to increase K by a factor of 6, what must happen to v? Solution (a) Since K v2, if v increases by a factor of 6, then K increases by a factor of 62 = 36 (b) Since K cc v2, it follows that fi cc v So, if K is to increase by a factor of 6, then v must be increased by a factor of J6 cc ., Example 4-20: The area of a circle of radius r is given by A = nr2 (a) If r is halved, then what happens to A? (b) If r increases by 50%, by what percentage will A increase? (c) If the Solution (a) Since A cc diameter of the circle is halved, what happens to A? r 2, if r decreases by a factor of 2, then A will decrease by a factor of22 =4 (b) If r increases by r/2 (which is 50% of r) to 3r/2, then r increases by a factor of 3/2 Therefore, A increases to (3/2) =9/4 =225% times its original value This represents an increase of 225% - 100% = 125% (c) Since d = 2r, we know that d o:: r Therefore, if dis halved, then so is r, and A decreases by a factor of [just as in part (a)] Alternatively, we could first write the formula for the area of a circle in terms ofd, its diameter, as.follows: A= 7tr =n(td) = }ni' Therefore, A cc dl, so if dis decreased by a factor of 2, then A will be decreased by a factor of 22 = 543 544 MCAT PHYSICAL SCIENCES REVIEW , Example 4-21: A metal rod of length L will be stretched by a distance ilL= mgL/EA when a weight w = mg is suspended from it In this formula, Eisa constant (depending on the material of which the rod is made) and A= nr2 is the rod's cross-sectional area (a) How will ilL change if the mass m is doubled? (b) If identical weights are suspended from two metal rods of the same initial length and made of the same material, but with the radius of Rod #1 being twice that of Rod #2, which rod will be stretched more, and 'by what factor? Solution (a) Since ilL oc t L ~ ~ m, if m increases by a factor of 2, then so does ilL (b) Because A oc r2, it follows that ilL oc 1/r2 • Since the radius of Rod #2 is half the radius of Rod #1, the stretch of Rod #2 will be 1/(!)2 = times that of Rod #1 Example 4-22: Newton's Law of Gravitation states that the force between two objects is given by F = GMm/r2 , where M and mare the masses of the objects, r is the distance between them, and G is the universal gravitational constant (a) If both masses are doubled, and the distance between them is also doubled, what happens to F? (b) If r is decreased by 50%, what happens to F? Solution (a) If both masses are doubled, then the product Mm increases by a factor of (2)(2) = If the distance r is doubled, then r2 increases by a factor of Since F oc Mm/r and both Mm and r2 increase by the same factor (in this case, 4), F will remain unchanged: F' =G(2M)(2m) =G4Mm =G Mm =F (2r) 4r r (b) If r decreases by 50%, then r changes tor- (r /2) = r12; that is, r decreases by a factor of Since F oc 1/ r 2, if r decreases by a factor of 2, F will increase by a factor of 22 = Example 4-23: Consider the equation xy = z (a) If z is a constant, how are x andy related? (b) If xis constant, how are y and z related? (c) If xis a positive constant, and we increase y by adding 2, by what factor does z increase? Solution (a) If z is a constant, then x andy are inversely proportional: x oc 1/y (b) If xis constant, then y and z are (directly) proportional: z oc y (c) If xis constant, then z oc y If we add toy, all we can say is that z will increase, but we cannot say by what factor If we multiplied y by 2, then we could say that z will increase by a factor of But adding toy does not allow us to say by what factor z will increase Predictions can be made with propo~tions only when we're told the factor by which some quantity changes (that is, what the quantity is multiplied by) MCAT MATH- CHAPTER 5: LOGARITHMS 545 LOGARITHMS §5.1 THE DEFINITION OF A LOGARITHM A logarithm (or just log, for short) is an exponent For example, in the equation 23 = 8, is the exponent, so is the logarithm More precisely, since is the exponent that gives when the base is 2, exponent (logarithm) + 23 = base/ we say that the base-2log of is 3, symbolized by the equation log2 = Here's another example: Since 102 = 100, the base-10 log of 100 is 2; that is, log 10 100 = The logarithm of a number to a given base is the exponent the base needs to be raised to give the number What's the log, base 3, of 81? It's the exponent we'd have to raise to in order to give 81 Since 34 =81, the base-3 log of 81 is 4, which we write as log3 81 =4 The exponent equation 23 = is equivalent to the log equation log = 3; the exponent equation 102 =100 is equivalent to the log equation log10 100 =2; and the exponent equation 34 =81 is equivalent to the log equation log3 81 = For every exponent equation, bi = y, there's a corresponding log equation: 1ogb y =x, and vice versa To help make the conversion, use the following mnemonic, which I call the two arrows method: , log = 2~ 23 = lr=Y , logy= x b~ You should read the log equations with the two arrows like this: equals , log = 2~ 2~3~8 b -Joo-X -Joo-y 23 = tr=y to tne equals , logb~x to tne Always remember: The log is the exponent to the equals 546 MCAT PHYSICAL SCIENCES REVIEW §5.2 LAWS OF LOGARITHMS There are only a few rules for dealing with logs that you'll need to know, and they follow directly from the rules for exponents (given earlier, in Chapter 1) After all, logs are exponents In stating these rules, we will assume that in an equation like Iogb y = x, the base b is a positive number that's different from 1, and that y is positive {Why these restrictions? Well, if b is negative, then not every number has a log For example, log-3 is 2, but what is log_3 27? If b were 0, then only would have a log; and if b were 1, then every number x could equallog1 y if y =1,: and no number x could equallog1 y if y :t-1 And why musty be positive? Because if b is a positive number, then lr (which is y) is always positive, no matter what real value we use for x Therefore, only positive numbers have logs.] · LAWS OF LOGARITHMS Law The log of a product is the sum of the logs: Iogb (yz) = Iogb y + Iogb z Law The log of a quotient is the difference of the logs: Iogb (y I z) = Iogb y -Iogb z Law The log of (a number to a power) is that power times the log of the number: Iogb (yz) =z Iogb y · We could also add to this list that the log of is 0, but this fact just follows from the definition of a log: Since b0 =1 for any allowed base b, we'll always have log,; =0 ·For the MCAT, the two mostimportantbases are b = 10 and b logs, and the "10" is often not written at all: =e Base-10 logs are called common log y means log10 y The base-10 log is useful because we use a decimal number system, which is based (pun intended) on the number 10 For example, the number 273.15 means (2 · 102} + (7 · 101) + (3 · 10°) + (1 · 10-1) + (5 ·10-2) In physics, the formula for the decibel level of a sound uses the base-10 log In chemistry, the base-10 log has many uses, such as finding values of the pH, pOH, pK., and pKb Base-e logs are known as natural logs Here, e is a particular constant, approximately equal to 2.7 This may seem like a strange number to choose as a base, but it makes calculus run smoothly-which is why it's called the natural logarithm-because (and you don't need to know this for the MCAT) the only numerical value of b for which the function j(x) =lr is its own derivative is b =e = 2.71828 Base-e logs are often used in the mathematical description of physical processes in which the rate of change of some quantity is proportional to the quantity itself; radioactive decay is a typical example The notation "In" (the abbreviation-in reverse-for natural logarithm) is often used to mean log11: In y means loge y The relationship between the base-10 log and the base-e log of a given number can be expressed as In y = 2.3log y For example, if y =1000 = 103, then In 1000 = 2.3log 1000 =2:3 · =6.9 You may also find it useful to know the following approximate values: log 2"' 0.3 In 2"' 0.7 log =0.5 In =1.1 log = 0.7 InS"' 1.6 MCAT MATH - CHAPTER 5: LOGARITHMS , Example 5-1: (a) What is log3 9? (b) Find logs (1/25) (c) Find log4 (d) What is the value of log16 4? (e) Given that log =0.7, what's log 500? (f) Given that log =0.3, find log (2 x 10-6) (g) Given that log 2"' 0.3 and log 3"' 0.5, find log (6 x 1023) Solution (a) log3 =xis the same as 3x = 9, from which we see that x = So,log3 = logs (1/25) =xis the same as 5x = 1/25 = 1/52 =5-2, sox = -2 Therefore, log5 (1/25) = -2 (c) log4 = x is the same as 4x = Since 4x = (22)x = 22x and = 23, the equation 4x = is the same as 22x = 23, so 2x = 3, which gives x = 3/2 Therefore,log4 =3/2 (d) log16 = x is the same as 16x = To find x, you might notice that the square root of 16 is 4, so 16112 = 4, which means log16 = 1/2 Alternatively, we can write 16x as (42)x = 42x and as 41 • Therefore, the equation 16x = is the same as 42x = 41, so 2x = 1, which gives x = 1/2 (e) log 500 = log (5 · 100) = log + log 100, where we used Law in the last step Since log 100 = log 102 = 2, we find that log 500 "" 0.7 + = 2.7 (f) log (2 x 10-6) =log +log 10-6, by Law Since log 10-6 = -6, we find that log (2 x 10-6) =0.3 + (-6) = -5.7 (g) log (6 x 1023) = log +log + log 1023, by Law Since log 1023 23, we find that log (6 x 1023) =0.3 + 0.5 + 23 = 23.8 (b) = , Example 5-2: In each case, find y (a) log2 y = (d) logy= 7.5 (e) logy= -2.5 (b) log y = -3 (c) logy=4 (f) lny = Solution (a) log2 y =5 is the same as 2s = y, soy = 32 log y = :3 is the same as 2-a = y, which gives y =1/23 = 1/8 (c) logy= is the same as 104 = y, soy= 10,000 (d) logy= 7.5 is the same as 107·5 =,lb We'll rewrite 7.5 as + 0.5, soy =107+ = 107 x 10°·s Because 10°·s = 10112 = flO, which is approximately 3, we find that y"" 107 X = X 107• (e) logy= -2.5 is the same as 10-2·s = y We'll rewrite -2.5 as -3 + 0.5, soy= tQ-3+(0.S) = 10-3 x 10°·5 Because 10°·s = 10112 = /10, which is app~oximately 3; we have that y =10-a x = 0.003 (f) ln y = means loge y = 3; this is the same as y e3 (which is about 20) (b) = 547 548 MCAT PHYSICAL SCIENCES REVIEW _Example 5-3: If a sound wave has intensity I (measured in W /m2), then its loudness level, fJ (measured in decibels), is found from the formula I /J=10log- Io where I is a constant equal to 10-12 W /m2 What is the loudness level of a sound wave whose intensity is 10-5 W /m2? Solution Using the given formula, we find that 10-5 fJ =10log10-12 =10log (10 )= 10· = 70 decibels _ Example 5-4: The definition of the pH of an aqueous solution is pH= , log [H30+J (or, simply, -log [H+]) where [H30+] is the hydronium ion concentration (in M) Part 1: Find the pH of each of the following solutions: coffee, with [H30+] = x 10-6M (b) seawater, with [H30+] = x 10-9 M (c) vinegar, with [H30+] 1.3 x 10-3M (a) = Part II: Find [H30+] for each of the following pH values: pH= (e) pH= 11.5 (f) pH= 4.7 (d) Solution (a) pH =-log (8 x 10-6) =-[log + log (10-6)] =-log + We can now make a quick approximation by simply noticing that log is a little less than log 10; that is, log is a little less than Let's say it's 0.9 Then pH= -Q.9 + = 5.1 (b) pH= -log (3 x 10-9) =-[log +log (10-9)] =-log + We now make a quick approximation by simply noticing that log is about 0.5 (after all, 9°5 is 3, so 10°·5 is close to 3) This gives pH =-Q.5 + = 8.5 (c) pH = -log (1.3 x 10-3) = -[log 1.3 + log (l0-3)] = -log 1.3 + We can now make a quick approximation by simply noticing that log 1.3 is just a little more than log 1; that is, log 1.3 is a little more than Let's say it's 0~1 This gives pH"" -o.1 + ;.· 2.9 *Note 1: We can generalize these three calcula.tions as follows: If [H30+] = m x 10-nM, where :S m < 10 and n is an integer, then the pH is between (n -1) and n; it's closer to (n- 1) if m > and it's closer ton if m < [We use as the cutoff since log =0.5.] (d) If pH = 7, then -log [H3Q+] = 7, so log [H30+] = -7, which means [H30+] =10-7 M (e) If pH= 11.5, then -log [H30+] =11.5, so log [H30+] = -11.5, which means [H30+] = 10-11.5 = 10(0.5)-12 = 100.5 x 10-12., X 10-12 M (f) If pH= 4.7, then -log [HsO+] = 4.7, so log [H 30+] = -4.7, which means[H 30+] = 10