chapter 13: acids & bases - welcome to dr. shanzer's flipped...
TRANSCRIPT
Name:
Regents Chemistry: Dr. Shanzer
PracticePacketChapter13:Acids&Bases
176
Chapter 13: Acids, Bases and Salts
Monoprotic acid - acids that contain only 1 hydrogen ion (H+) HNO3
Diprotic acid - acids that contain 2 hydrogen ion (H+) H2SO4
Conjugate acid – the particle formed when a base gains a hydrogen ion
Conjugate base – the particle that remains when an acid has donated a hydrogen ion
Hydronium ion (H3O+) – a water molecule that gains a hydrogen ion and becomes positivity charged
Arrhenius Acid – hydrogen-containing compounds that ionize or yield hydrogen ions (H+)
Arrhenius Base – compounds that ionize hydroxide ions (OH-)
Neutral Solution – an aqueous solution in which [H+] = [OH-]
Acidic Solution – a solution in which [H+] is greater than [OH-]
Basic Solution - a solution in which [H+] is less than [OH-]
pH – negative logarithm of the hydrogen-ion concentration pH = -log [H+]
Strong Acid - completely ionizes (separate into ions) greater [H+]
Strong Base - completely ionizes (separate into ions) greater [OH-]
Neutralization Reaction – a reaction in which an acid reacts with a base to
produce salt and water (neutral)
Equivalence Point – when the number of moles of hydrogen ions equals the number of moles of hydroxide ions
Titration – process of adding known concentration to determine the concentration of another solution
End point – the point in which the indicator changes color in a titration
177
Characteristics
of
Acids & Bases
Chemistry 200 Video Lesson 13.1
Objective:How can we recognize characteristics of acids and bases?
How can we determine the difference between Arrhenius and Bronsted/Lowry acids and bases?
Acids&BasesAcids&Basescanberecognizedbytheirproperties.
• Dilutesolutionshaveasourtastelemons-->citricacid
vinegar-->aceticacidcarbonateddrinks-->carbonicacidtomato-->ascorbicacid
• Haveabittertaste&slippery,soapyfeel.soap,milkofmagnesia
ex:
ex:
Acids
Bases
AcidsTwotheoriesusedtoexplainthebehaviorofacids:
• asubstancethatreleaseshydrogenions[H+]inanaqueoussolutionex:HCl,H2SO4
• notallsubstancesthatcontainhydrogenareacids.CH4(methane)isnotanacidbecausethehydrogenatomsdonotionize.
ArrheniusAcid(SvanteArrhenius)
• A Hydrogen Ion is produced when a hydrogen atom
loses an electron to become a positive ion
• AHydroniumIon[H3O]+isformedwhenahydrogenionreactsw/H2O
TableE
178
ex:HCl(g)-------->H3O+(aq)+Cl-(aq)H2O
Arrheniusacidscanbedividedinto2maincategoriesInorganicAcid• anacidthatstartsw/hydrogen&doesnothavecarbon
ex:HCl,H2SO4
OrganicAcid• anacidthathascarbon&endsw/COOH
(CarboxylGroup)
ex:CH3COOH(ethanoicacid)HCOOH(methanoicacid)
AnArrheniusacidcansometimesyieldmorethanonehydrogenioninaqueoussolution
MonoproticAcid
• acidsthatproduceasinglehydrogenion(ionizesin1step) ex:HCl+H2O-->H3O++Cl-
DiproticAcid
• acidsthatproduce2hydrogenions(ionizesin2steps)
ex:H2SO4-->H++HSO4-(step1)HSO4--->H++SO4-2(step2)Triproticacids:3hydrogenionsex:H3PO4
TheBrønsted-Lowrytheoryfocusessolelyonthehydrogenionasaproton&canbeusedtofurtherexplainacidbehavior:
Brønsted-LowryAcid• anyspecies(anyparticle:atom,moleculeorion)thatcandonateaproton[H+]toanotherspecies(aprotondonor)
• allArrheniusacidsarealsoBrønsted-Lowryacids
ex:HCl(g)+H2O(l)-->H3O+(aq)+Cl-(aq)
• accordingtoBrønsted-Lowry,HClisanacidbecauseitdonatesaproton(H+)
BasesTheoriesusedtoexplainacidbehaviorcanalsobeusedto
explainbasesaswellArrheniusBase
• asubstancethatreleaseshydroxideions[OH-]inanaqueoussolution
ex:NaOH------>Na+(aq)+OH-(aq)H2O
AnAlcoholshouldnotbeconfusedw/anArrheniusBaseasitdoesnotformahydroxideion[OH-]inan(aq)solution,thereforeitisNOTabase!!!
CH3OH
C2H5OH
Brønsted-LowryBase• anyspeciesthatcanacceptaproton[H+]fromanotherspecies-->aprotonacceptor
• accordingtoArrhenius,onlymetallichydroxidesarebases
• Brønsted-LowrydeUinitionsexplainwhyasubstancethathasno[OH-],likeNH3,behaveslikeabaseinH2O
ex:NH3+H2O-->NH4++OH-
179
Amphiprotic(Amphoteric)
• asubstancethatcanactasanacidorabase
ex:H2O
NH3+H2O-->NH4++OH-
HCl+H2O-->H3O++Cl-Base
Acid
Electrolytes
Chemistry 200 Video Lesson 13.2
Objective:How do we determine which acids and bases are strong or weak electrolytes?
Electrolytes
Ex: Acids
Bases
Salts
• are substances that have mobile ions when put into
solution (aq). This allows it to conduct electricity.
AWeakAcidisaweakelectrolytebecauseitpartiallyionizesinwatertoproduceasmallnumberofH+ionsinsolution.
AStrongAcidisastrongelectrolytebecauseitcompletelyionizesinwatertoproducealargenumberofH+ionsinsolution.
180
AStrongBaseisastrongelectrolytebecauseitcompletelyionizesinwatertoproducealargenumberofOH-ionsinsolution
AWeakBaseisaweakelectrolytebecauseitpartiallyionizesinwatertoproduceasmallnumberofOH-ionsinsolution.
ex:
HCl(aq) + NaOH(aq) --> NaCl(aq) + HOH(l)
Acid/Base Reactions
• Acid react with a base to produce a salt and water
Acid/Metal Reactions
• Acids react with a metal to produce
a salt and hydrogen gas (H2(g))
2HNO3(aq) + Ca(s) --> Ca(NO3)2(aq) + H2(g)
ex:
2HCl(aq) + Cu(s) --> CuCl2(aq) + H2(g)
NO REACTION!!
Conjugate Acid-
Base pairsVideo Lesson 13.3
Objectives• Compare and contrast acids and bases as
defined by the theories of Arrhenius andBronsted-Lowry (alternate acid/basetheory).
• QUESTION: Why is ammonia (NH3) a base?
181
• Explains the behavior of WEAK acids and BASES (Na2CO3& NH3)
• Bronsted-Lowry Acids are PROTON DONORS o The prefer to give away a proton (H+)
Alternate Acid-Base Theory
AKA Bronsted-Lowry Theory• Bronsted-Lowry Bases are PROTON ACCEPTORS
Alternate Acid Base
Theory
• An acid is a H+ donor while a base is a H+ acceptor. • The substance produced when an acid
has donated its proton is called theconjugate base.
• The substance produced when thebase accepts a H+ is called theconjugate acid
Conjugate Acid/Base Pairs Label the acids and bases according to Bronsted-Lowry:
Acid
Conjugate Base
Base Conjugate Acid
Conjugate Acid-Base Pairs
Practice
Label conjugate acid base pairs in the following equation
Acid
Base
c. base
c. acid
Practice
pH & Acid-Base
IndicatorsVideo Lesson 13.4
182
Objectives• Describe how [H+] and [OH-] are related in
an aqueous solution.• Classify a solution as neutral, acidic or basic.• Describe the purpose of an acid-base
indicator.
The pH Scale• pH = direct measurement of H+ ion concentration in
a solution • “power of hydrogen”
H+H+H+H+ H+
OH-OH-OH-OH-
O 7 14
Acidic Neutral Basic
• The pH scale is logarithmic o Based on exponents of the number 10
• The pH of a solution is the negative log of the [H+] or [OH-] ions
Logarithmic
• Each change in a single pH unit signifies a tenfoldchange in [H+] concentration
A Tenfold Change
• Indicator o A substance that changes color as a result of a
pH change • Table M• Example: Phenolphthalein à colorless up until a pH
of 8, light pink from 8 to 9 and pink from pH of 9 up
Acid-Base Indicators
183
pH = 5 pH = 11
Acid-Base
TitrationVideo Lesson 13.5
Objectives• Define products of an acid-base reaction.• Explain how acid-base titration is used to
calculate the concentration of an acid orbase.
• Explain the concept of equivalence inneutralization reactions.
184
Neutralization Reactions• Neutral
o Neither acidic nor basic o Equal concentrations of H+ and OH-
o Occurs when Arrhenius acid and an Arrhenius base react to form water and salt
Acid + Base à Salt + Water
HCl + KOH à KCl + HOH
Double Replacement!
+1 -1 -1+1 +1 -1
Salt
Or
Ionic Substance
Water
Acid Base
+1 -1
Neutralization Reactions
• Stomach acido pH between 2-3
• Heart burn o Happens when pH
level drops in the stomach
• Milk of Magnesia bringsthe pH up
• Mg(OH)2
Magnesium hydroxide
Acid-Base Titration• Titrations are used to calculate the concentration
(Molarity) of an unknown solution. • Lab process in which an unknown solution is
systematically reacted with a solution of known concentration by adding measured volumes of an acid or a base to the unknown until neutralization occurs.
• In all neutralizationreactions there mustbe a 1:1 ratio of molesof H+ ions and moles ofOH- ions.
• Equivalence Pointo [H+] = [OH-]
185
Example #1• What is the concentration of a solution of HI if 0.3 L is
neutralized by 0.6 L of 0.2 M solution of KOH?
Example #2
• You have 50 mL of 1.0 M H2SO4 (aq). What volume of 0.5 M NaOH would be required to neutralize the acid? (Diprotic Acids yield 2 H+ ions in solution!)
186
Sketch Notes
16187
PracticePacket:AcidsandBases
http://drshanzerchemistry.weebly.com
Video13.1:CharacteristicsofAcidsandBases
Acid/Base/SaltCharacteristics:Onthelinetotheleft,writeAifthestatementisapropertyofanacid,writeBifthepropertyisthatofabase,andwriteX,ifitisapropertyofbothacidicandbasicsolutions.
Completethechart.
Answerthefollowingquestions
1. _____Whichspeciescanconductanelectriccurrent?1. H2O(s)2. CH3OH(aq)
3. NaOH(s)4. HCl(aq)
2. _____AccordingtoArrheniustheory,whichspeciesdoesanacidproduceinaqueoussolution?
1. hydroxideions2. sodiumions
3. hydrogenions4. chlorideions
3. _____WhichsubstanceisanArrheniusacid?1. Mg(OH)2(aq)2. LiF(aq)
3. CH3CHO(aq)4. HBr(aq)
4. _____AccordingtotheArrheniustheory,whenabaseisdissolvedinwater,itproducesasolutioncontainingonlyonekindofnegativeion.Whatisthenameofthisnegativeion?
1. hydrogensulfateion2. hydrogencarbonateion
3. hydrideion4. hydroxideion
5. IntermsofH+ions,explainthedifferencebetweenmonoprotic,diproticandtriproticacids.
Substance Acid,BaseorSalt?
Howdoyouknow?
1. NaOH2. HCl3. NaCl4. HF5. K2SO46. Fe(OH)37. H3PO4
188
Continue to Question #6 on page 190
$UUKHQLXV�$FLGV�DQG�%DVHV 13.1 Continued
$�%�&�'�
�� :KLFK�FRPSRXQGV�DUH�FODVVLILHG�DV�$UUKHQLXV�DFLGV"
$� WKH�EHKDYLRU�RI�PDQ\�DFLGV�DQG�EDVHV%� WKH�HIIHFW�RI�VWUHVV�RQ�D�SKDVH�HTXLOLEULXP&� WKH�RSHUDWLRQ�RI�DQ�HOHFWURFKHPLFDO�FHOO'� WKH�VSRQWDQHRXV�GHFD\�RI�VRPH�QXFOHL
�� :KDW�FDQ�EH�H[SODLQHG�E\�WKH�$UUKHQLXV�WKHRU\"
$� 2+±�LRQV %� 2�±�LRQV&� .��LRQV '� +��LRQV
�� 3RWDVVLXP�K\GUR[LGH�LV�FODVVLILHG�DV�DQ�$UUKHQLXVEDVH�EHFDXVH�.2+�FRQWDLQV
$� D�VDOW %� D�K\GURFDUERQ&� DQ�$UUKHQLXV�DFLG '� DQ�$UUKHQLXV�EDVH
�� :KHQ�RQH�FRPSRXQG�GLVVROYHV�LQ�ZDWHU��WKH�RQO\SRVLWLYH�LRQ�SURGXFHG�LQ�WKH�VROXWLRQ�LV+�2���DT���7KLV�FRPSRXQG�LV�FODVVLILHG�DV
$� K\GUR[LGH %� K\GURQLXP&� K\SRFKORULWH '� SHUFKORUDWH
�� *LYHQ�WKH�HTXDWLRQ�
+&O�J����+�2� ��� �;�DT����&O±�DT�:KLFK�LRQ�LV�UHSUHVHQWHG�E\�;"
$� DPPRQLXP�LRQ %� K\GURQLXP�LRQ&� K\GUR[LGH�LRQ '� VXOIDWH�LRQ
�� 7KH�RQO\�SRVLWLYH�LRQ�IRXQG�LQ�+�62��DT��LV�WKH
$� %D�2+�� %� &+�&22&+�
&� +�32� '� 1D&O
�� :KLFK�VXEVWDQFH�LV�DQ�$UUKHQLXV�DFLG"
$� &+�&22+�DQG�&+�&+�2+%� +&�+�2���DQG�+�32�
&� .+&2���DQG�.+62�
'� 1D6&1�DQG�1D�6�2
�� :KLFK�WZR�IRUPXODV�UHSUHVHQW�$UUKHQLXV�DFLGV"
$� +&�+�2��DT����1D2+�DT�� �1D&�+�2��DT���+�2� �
%� &�+��J������2��J�� ���&2��J������+�2� �&� =Q�V������+&O�DT�� �=Q&O��DT����+��J�'� %D&O��DT����1D�62��DT� �%D62��V�����
1D&O�DT�
�� :KLFK�FKHPLFDO�HTXDWLRQ�UHSUHVHQWV�WKH�UHDFWLRQ�RI�DQ$UUKHQLXV�DFLG�DQG�DQ�$UUKHQLXV�EDVH"
$� &2��±�DV�WKH�RQO\�QHJDWLYH�LRQ�LQ�VROXWLRQ%� 2+±�DV�WKH�RQO\�QHJDWLYH�LRQ�LQ�VROXWLRQ&� 1+���DV�WKH�RQO\�SRVLWLYH�LRQ�LQ�VROXWLRQ'� +��DV�WKH�RQO\�SRVLWLYH�LRQ�LQ�VROXWLRQ
��� $FFRUGLQJ�WR�WKH�$UUKHQLXV�WKHRU\��ZKHQ�D�EDVHGLVVROYHV�LQ�ZDWHU�LW�SURGXFHV
$� 0J�2+�� %� &�+��2+��&� 0J&O� '� &+�&O
��� :KLFK�VXEVWDQFH�\LHOGV�K\GUR[LGH�LRQ�DV�WKH�RQO\QHJDWLYH�LRQ�LQ�DTXHRXV�VROXWLRQ"
$� .2+��&D�2+����DQG�&+�2+%� .2+��1D2+��DQG�/L2+&� /L2+��&D�2+����DQG�&�+��2+��'� 1D2+��&D�2+����DQG�&+�&22+
��� $FFRUGLQJ�WR�WKH�$UUKHQLXV�WKHRU\��ZKLFK�OLVW�RIFRPSRXQGV�LQFOXGHV�RQO\�EDVHV"
189
PracticePacket:AcidsandBases
http://mintzchemistry.weebly.com
6. Identifyeachofthefollowingacidsasmonoprotic,diproticortriproticandstatehowmanyH+ionsarereleasedinsolution.
1. H2CO32. H3PO43. HNO34. CH3COOH
5. H2SO46. HCl7. C2H5COOH8. H2S
Video13.2:Electrolytes
Answerthequestionsbelowbasedontheinformationaboveandonyourknowledgeofchemistry.1. Usingthediagrambelow,drawadiagramofthemoleculeformedwhenonehydrogen
(proton)ispulledoffonewatermoleculeandattachedtoanother.
2. Whataretheformulasoftheionsformed?
3. Writeanequationshowingtheformationoftheionsfromtwomoleculesofwater.
4. Thehydroniumion(H3O+)andthehydroxideion(OH-)areformedbythereactionbetweenwatermolecules.Whatwouldformfromthereactionbetweenhydroniumandhydroxideions?Writeanequationshowingthereaction.
5. Howdoesthereactionbetweenwatermoleculescomparetothereactionbetweenhydroniumandhydroxideions?
6. Purewaterisactuallyamixtureofwatermolecules,hydroniumions,andhydroxideions:
a. Howdoestheconcentrationofhydroxideionscomparetotheconcentrationofhydroniumionsinpurewater?Explain
b. Howdoestheconcentrationofionscomparetotheconcentrationofmoleculesinpurewater?
190
(OHFWURO\WHV
$�%�&�'�
�� :KLFK�WZR�FRPSRXQGV�DUH�HOHFWURO\WHV"
$� VDWXUDWHG�K\GURFDUERQV %� DOFRKROV&� DON\QHV '� RUJDQLF�DFLGV
�� :KLFK�FRPSRXQGV�FDQ�EH�FODVVLILHG�DV�HOHFWURO\WHV"
$� HOHFWULFDO�FRQGXFWLYLW\�RI�.&O�DT�%� S+�RI�.&O�V�&� HOHFWULFDO�FRQGXFWLYLW\�RI�.&O�V�'� S+�RI�.&O�DT�
�� :KLFK�ODERUDWRU\�WHVW�UHVXOW�FDQ�EH�XVHG�WR�GHWHUPLQH�LI�.&O�V��LVDQ�HOHFWURO\WH"
$� ����/�RI�����0�+&,�DT�%� ����/�RI�����0�+&,�DT�&� ����/�RI������0�+&,�DT�'� ����/�RI������0�+&,�DT�
�� :KLFK�VDPSOH�RI�+&,�DT��FRQWDLQV�WKH�JUHDWHVW�QXPEHU�RI�PROHVRI�VROXWH�SDUWLFOHV"
$� LQFUHDVHV�EHFDXVH�WKH�FRQFHQWUDWLRQ�RI�LRQV�GHFUHDVHV%� GHFUHDVHV��EXW�WKH�FRQFHQWUDWLRQ�RI�LRQV�UHPDLQV�WKH�VDPH&� LQFUHDVHV��EXW�WKH�FRQFHQWUDWLRQ�RI�LRQV�UHPDLQV�WKH�VDPH'� GHFUHDVHV�EHFDXVH�WKH�FRQFHQWUDWLRQ�RI�LRQV�GHFUHDVHV
�� $V�ZDWHU�LV�DGGHG�WR�D������0�1D&O�DTXHRXV�VROXWLRQ��WKHFRQGXFWLYLW\�RI�WKH�UHVXOWLQJ�VROXWLRQ
$� &�+�2+ %� &&��&� &+�&22+ '� &+�
�� :KLFK�FRPSRXQG�GLVVROYHV�LQ�ZDWHU�WR�IRUP�DQ�DTXHRXV�VROXWLRQWKDW�FDQ�FRQGXFW�DQ�HOHFWULF�FXUUHQW"
$� /L2+ %� /L%U &� &+�2+ '� +%U
�� $�VWXGHQW�WHVWHG�D�����0�DTXHRXV�VROXWLRQ�DQG�PDGH�WKHIROORZLQJ�REVHUYDWLRQV�� FRQGXFWV�HOHFWULFLW\� WXUQV�EOXH�OLWPXV�WR�UHG� UHDFWV�ZLWK�=Q�V��WR�SURGXFH�JDV�EXEEOHV
:KLFK�FRPSRXQG�FRXOG�EH�WKH�VROXWH�LQ�WKLV�VROXWLRQ"
$� ����0�1D2+ %� �����0�1D2+&� �����0�&+�2+ '� ����0�&+�2+
�� :KLFK�RI�WKH�IROORZLQJ�DTXHRXV�VROXWLRQV�LV�WKH�EHVW�FRQGXFWRU�RIHOHFWULFLW\"
$� EDULXP�VXOIDWH %� VRGLXP�QLWUDWH&� VLOYHU�FKORULGH '� PDJQHVLXP�FDUERQDWH
�� %DVHG�RQ�5HIHUHQFH�7DEOH�)��ZKLFK�RI�WKHVH�VDOWV�LV�WKH�EHVWHOHFWURO\WH"
$� .�62��DT� %� 1D&O�DT�&� +&O�DT� '� &�+��2��DT�
��� $Q�H[DPSOH�RI�D�QRQHOHFWURO\WH�LV
$� &��+��2���DT� %� /L2+�DT�&� &�+O�2��DT� '� &+�2+�DT�
��� 7KH�GLDJUDP�EHORZ�VKRZV�DQ�DSSDUDWXV�XVHG�WR�WHVW�WKHFRQGXFWLYLW\�RI�YDULRXV�PDWHULDOV�
:KLFK�DTXHRXV�VROXWLRQ�ZLOO�FDXVH�WKH�EXOE�WR�OLJKW"
$� $�DQG�' %� &�DQG�'&� %�DQG�& '� $�DQG�%
��� %HDNHUV�$��%��&��DQG�'�VKRZQ�EHORZ�HDFK�FRQWDLQ�D�GLIIHUHQWVROXWLRQ�
7KH�EXOE�ZLOO�JORZ�ZKHQ�WKH�FRQGXFWLYLW\�DSSDUDWXV�LV�SODFHGLQWR�ZKLFK�EHDNHUV"
191
PracticePacket:AcidsandBases
http://drshanzerchemistry.weebly.com
Completethefollowingreactions,balanceandindicatephaseoftheproducts
1. ____HNO3(aq)+____Mg(s)à
2. ____H2SO4(aq)+____Li(s)à
3. ____HBr(aq)+____Zn(s)à
4. Aluminummetalreactsw/phosphoricacid.
5. Calciummetalreactsw/hydrochloricacid
UsingTableJanswerthefollowingquestions.
1. Whydoesgoldoccuruncombinedwhereaszincdoesnot?_________________________________________________________________________________________________
2. Whywassilverusedtomakecoinsinthepast?_________________________________________________________________________________________________
3. Whyiscopperusedtomakeelectricalwiresandcables?_________________________________________________________________________________________________
4. WhydoweknowlittleaboutthelifestylesofthepeopleoftheIronAge?_________________________________________________________________________________________________
Video13.3:ConjugateAcid-BasePairsOperationalDefinition:Acidsandbasesarechemicalspeciesthatexhibitdistinctivesetsofobservableproperties.Acidstastesour,basesarebitter.
ConceptualDefinitions:Acidsandbasesaredefinedconceptuallytohelpaccountforwhatishappeningonthemicroscopiclevel.
ArrheniusConcept:Anacidisasubstancethatwhendissolvedinwaterformshydrogenions(H+).Abaseisasubstancethatwhendissolvedinwaterproduceshydroxideions(OH-).ThisconceptislimitedbecauseotherchemicalshaveoperationalacidorbasepropertiesbutdonotformH+orOH-.Anexamplewouldbeammonia(NH3)hasbasicpropertiesbutdoesn’tcontainhydroxideions.
AlternateAcidBaseConcept(Bronsted-Lowry):Anacidisaproton(H+)donorandabaseisaproton(H+)acceptor.
192
PracticePacket:AcidsandBases
http://drshanzerchemistry.weebly.com
HF+H2O⇔H3O++F- AcidBaseC.AcidC.Base
Fillinthefollowingtable.Identifytheacid,base,conjugateacidandconjugatebaseineachoftheequations.
6. Writetheequationthatshowsammonia,NH3reactingwithhydrobromicacid,HBr.Labeltheacid,thebase,theconjugateacidandtheconjugatebase.
7. Writetheequationthatshowsthereactionofhydrogensulfide,HS-withhydroxideion,OH-.Labeltheacid,thebase,theconjugateacidandtheconjugatebase.
Video13.4:pH&Acid-BaseIndicatorsIndicators:GiventhepHofthefollowingsubstances,usetableMofyourreferencetodeterminewhatcolortheindicatorwillturnwhenplacedineachsubstance.
Equation Acid Base ConjugateAcid
ConjugateBase
1. HCl+NH3àNH4+Cl-
2. PO43-+HNO3àNO3-+HPO42-
3. HCO3-+OH-àH2O+CO32-
4. NH4++H2OàNH3+H3O+
5. HPO42-+H2OàOH-H2PO4-
Solution pHRange
MethylOrange
Bromothymolblue
Phenolphthalein Litmus Bromocresolgreen
Thymolblue
Acid/Base
Vinegar 1.3Soap 8.4Cola 3.2Ammonia 12Coffee 5.2
193
$OWHUQDWH�$FLG�%DVH�7KHRU\
�� DFFHSWV�DQ� LRQ�� GRQDWHV�DQ� �LRQ�� DFFHSWV�DQ� �LRQ�� GRQDWHV�DQ� �LRQ
�� $FFRUGLQJ�WR�RQH�DFLG�EDVH�WKHRU\��1+��DFWV�DV�DEDVH�ZKHQ�DQ�1+��PROHFXOH
�� ���� ��
�� $FFRUGLQJ�WR�RQH�DFLG�EDVH�WKHRU\��D�EDVH�LV�DQ
�� DQ�DOFRKRO �� DQ�DFLG�� D�EDVH �� D�VDOW
�� $�VXEVWDQFH�WKDW�GLVVROYHV�LQ�ZDWHU�DQG�SURGXFHVK\GURQLXP�LRQV�DV�WKH�RQO\�SRVLWLYH�LRQV�LQ�WKHVROXWLRQ�LV�FODVVLILHG�DV
�� D�EDVH�EHFDXVH�LW�DFFHSWV�DQ�+�
�� D�EDVH�EHFDXVH�LW�GRQDWHV�DQ�+�
�� DQ�DFLG�EHFDXVH�LW�DFFHSWV�DQ�+�
�� DQ�DFLG�EHFDXVH�LW�GRQDWHV�DQ�+�
�� *LYHQ�WKH�GLDJUDP�UHSUHVHQWLQJ�D�UHDFWLRQ�
$FFRUGLQJ�WR�RQH�DFLG�EDVH�WKHRU\��WKH�ZDWHU�DFWVDV
�� D�EDVH�EHFDXVH�WKH\�DFFHSW�+��LRQV �� D�EDVH�EHFDXVH�WKH\�GRQDWH�+��LRQV�� DQ�DFLG�EHFDXVH�WKH\�DFFHSW�+��LRQV �� DQ�DFLG�EHFDXVH�WKH\�GRQDWH�+��LRQV
�� *LYHQ�WKH�EDODQFHG�HTXDWLRQ�UHSUHVHQWLQJ�D�UHDFWLRQ�
$FFRUGLQJ�WR�RQH�DFLG�EDVH�WKHRU\��WKH�+�2� ��PROHFXOHV�DFW�DV
�� +��GRQRU �� +��DFFHSWRU�� +�GRQRU �� +�DFFHSWRU
�� 2QH�DFLG�EDVH�WKHRU\�GHILQHV�D�EDVH�DV�DQ
�� +��GRQRU �� +��DFFHSWRU�� 2+��GRQRU �� 2+��DFFHSWRU
�� 2QH�DOWHUQDWH�DFLG�EDVH�WKHRU\�VWDWHV�WKDW�DQ�DFLG�LVDQ
194
�� DQ�DFLG�EHFDXVH�WKH\�DFFHSW�+��LRQV�� DQ�DFLG�EHFDXVH�WKH\�GRQDWH�+��LRQV�� D�EDVH�EHFDXVH�WKH\�DFFHSW�+��LRQV�� D�EDVH�EHFDXVH�WKH\�GRQDWH�+��LRQV
�� *LYHQ�WKH�EDODQFHG�HTXDWLRQ�UHSUHVHQWLQJ�DUHDFWLRQ�
1+��J����+�2� �� 1+����DT����2+±�DT�$FFRUGLQJ�WR�RQH�DFLG�EDVH�WKHRU\��WKH�1+��J�PROHFXOHV�DFW�DV
�� +�2� � �� 1+��J��� 1+���DT� �� 2+±�DT�
�� *LYHQ�WKH�HTXDWLRQ�UHSUHVHQWLQJ�D�UHDFWLRQ�DWHTXLOLEULXP�
1+��J���+�2� �� 1+����DT����2+±�DT�
7KH�+��DFFHSWRU�IRU�WKH�IRUZDUG�UHDFWLRQ�LV
�� EDVH �� DFLG�� SURWRQ�DFFHSWRU �� HOHFWURQ�GRQRU
��� *LYHQ�WKH�UHDFWLRQ������������
1+����+�2�l�1+�����2+±
7KH�ZDWHU�DFWV�DV�WKH
195
S+��$FLG�%DVH�,QGLFDWRUV
�� GHFUHDVHG�E\�D�IDFWRU�RI���� GHFUHDVHG�E\�D�IDFWRU�RI����� LQFUHDVHG�E\�D�IDFWRU�RI���� LQFUHDVHG�E\�D�IDFWRU�RI���
�� :KHQ�WKH�S+�RI�DQ�DTXHRXV�VROXWLRQ�LV�FKDQJHGIURP���WR����WKH�FRQFHQWUDWLRQ�RI�K\GURQLXP�LRQVLQ�WKH�VROXWLRQ�LV
�� GHFUHDVHV�E\�D�IDFWRU�RI����� LQFUHDVHV�E\�D�IDFWRU�RI����� GHFUHDVHV�E\�D�IDFWRU�RI������ LQFUHDVHV�E\�D�IDFWRU�RI����
�� :KHQ�WKH�S+�RI�D�VROXWLRQ�LV�FKDQJHG�IURP���WR���WKH�K\GURQLXP�LRQ�FRQFHQWUDWLRQ�RI�WKH�VROXWLRQ
�� GHFUHDVHV���S+�XQLW�� GHFUHDVHV����S+�XQLWV�� LQFUHDVHV���S+�XQLW�� LQFUHDVHV����S+�XQLWV
�� :KHQ�WKH�K\GURQLXP�LRQ�FRQFHQWUDWLRQ�RI�DVROXWLRQ�LV�LQFUHDVHG�E\�D�IDFWRU�RI�����WKH�S+YDOXH�RI�WKH�VROXWLRQ
�� GHFUHDVHV���S+�XQLW�� GHFUHDVHV����S+�XQLWV�� LQFUHDVHV���S+�XQLW�� LQFUHDVHV����S+�XQLWV
�� :KHQ�WKH�K\GURQLXP�LRQ�FRQFHQWUDWLRQ�RI�DVROXWLRQ�LV�LQFUHDVHG�E\�D�IDFWRU�RI�����WKH�S+YDOXH�RI�WKH�VROXWLRQ
�� GHFUHDVHV�E\�D�IDFWRU�RI���� LQFUHDVHV�E\�D�IDFWRU�RI���� GHFUHDVHV�E\�D�IDFWRU�RI����� LQFUHDVHV�E\�D�IDFWRU�RI���
�� :KHQ�WKH�S+�YDOXH�RI�D�VROXWLRQ�LV�FKDQJHG�IURP��WR����WKH�FRQFHQWUDWLRQ�RI�K\GURQLXP�LRQV
�� S+��WR�S+�� �� S+���WR�S+���� S+���WR�S+�� �� S+���WR�S+��
�� :KLFK�FKDQJH�LQ�S+�UHSUHVHQWV�D�KXQGUHGIROGLQFUHDVH�LQ�WKH�FRQFHQWUDWLRQ�RI�K\GURQLXP�LRQV�LQD�VROXWLRQ"
�� � �� � �� � �� �
�� :KDW�LV�WKH�S+�RI�D�VROXWLRQ�WKDW�KDV�D�K\GURQLXPLRQ�FRQFHQWUDWLRQ�����WLPHV�JUHDWHU�WKDQ�D�VROXWLRQZLWK�D�S+�RI��"
�� LQFUHDVHV�E\�D�IDFWRU�RI���� LQFUHDVHV�E\�D�IDFWRU�RI������� GHFUHDVHV�E\�D�IDFWRU�RI���� GHFUHDVHV�E\�D�IDFWRU�RI�����
�� $V�WKH�S+�RI�D�VROXWLRQ�LV�FKDQJHG�IURP���WR����WKHFRQFHQWUDWLRQ�RI�K\GURQLXP�LRQV
�� � �� � �� � �� ��
�� :KLFK�S+�LQGLFDWHV�D�EDVLF�VROXWLRQ"
�� � �� � �� �� �� ��
��� :KLFK�RI�WKHVH�S+�QXPEHUV�LQGLFDWHV�WKH�KLJKHVWOHYHO�RI�DFLGLW\"
�� � �� � �� � �� ��
��� 3KHQROSKWKDOHLQ�LV�SLQN�LQ�DQ�DTXHRXV�VROXWLRQKDYLQJ�D�S+�RI
�� UHG �� EOXH�� SLQN �� \HOORZ
��� :KDW�LV�WKH�FRORU�RI�WKH�LQGLFDWRU�WK\PRO�EOXH�LQ�DVROXWLRQ�WKDW�KDV�D�S+�RI���"
�� ��� �� ��� �� ��� �� ���
��� 7KUHH�VDPSOHV�RI�WKH�VDPH�VROXWLRQ�DUH�WHVWHG�HDFK�ZLWK�D�GLIIHUHQW�LQGLFDWRU��$OO�WKUHHLQGLFDWRUV��EURPWK\PRO�EOXH��EURPFUHVRO�JUHHQDQG�WK\PRO�EOXH��DSSHDU�EOXH�LI�WKH�S+�RI�WKHVROXWLRQ�LV
�� D�VROXWLRQ�LQ�ZKLFK�EURPWK\PRO�EOXH�LV�EOXH�� D�VROXWLRQ�LQ�ZKLFK�EURPFUHVRO�JUHHQ�LV�EOXH�� D�VROXWLRQ�LQ�ZKLFK�SKHQROSKWKDOHLQ�LV�SLQN�� D�VROXWLRQ�LQ�ZKLFK�PHWK\O�RUDQJH�LV�UHG
��� %DVHG�RQ�WKH�UHVXOWV�RI�WHVWLQJ�FRORUOHVV�VROXWLRQVZLWK�LQGLFDWRUV��ZKLFK�VROXWLRQ�LV�PRVW�DFLGLF"
�� EURPWK\PRO�EOXH �� EURPFUHVRO�JUHHQ�� OLWPXV �� WK\PRO�EOXH
��� :KLFK�LQGLFDWRU�ZRXOG�EHVW�GLVWLQJXLVK�EHWZHHQ�DVROXWLRQ�ZLWK�D�S+�RI�����DQG�D�VROXWLRQ�ZLWK�DS+�RI����
196
Modeling Neutralization Reactions Engage:
● Predict what will happen to the pH when equal volumes of acid and base are combined?● Test pH of solutions then mix together. Test pH of resulting solution after mixing. Make
observations● Test conductivity before and after
Activity: 1. Write the balanced neutralization reaction for the reaction of HCl and NaOH similar to
the one you saw in the “Upset Tummy” phenomena. 2. Using the key, draw particle models to represent the reaction before and after mixing.
Draw five dissociated acid particles in beaker 1 and five dissociated base particles in beaker 2.
3. Match each hydrogen ion in beaker one with one hydroxide ion in beaker two. Thendraw the number of water molecules that form in beaker three.
Key: H+ =
Cl-1 =
Na+ =
OH-1 =
4. What was the pH of the resulting solution?
5. In a second neutralization reaction H2SO4 and NaOH are mixed. Using the key below,draw three dissociated acid particles in beaker one and three dissociated base particlesin beaker two.
197
Modeling Neutralization Reactions Engage:
● Predict what will happen to the pH when equal volumes of acid and base are combined?● Test pH of solutions then mix together. Test pH of resulting solution after mixing. Make
observations● Test conductivity before and after
Activity: 1. Write the balanced neutralization reaction for the reaction of HCl and NaOH similar to
the one you saw in the “Upset Tummy” phenomena. 2. Using the key, draw particle models to represent the reaction before and after mixing.
Draw five dissociated acid particles in beaker 1 and five dissociated base particles in beaker 2.
3. Match each hydrogen ion in beaker one with one hydroxide ion in beaker two. Thendraw the number of water molecules that form in beaker three.
Key: H+ =
Cl-1 =
Na+ =
OH-1 =
4. What was the pH of the resulting solution?
5. In a second neutralization reaction H2SO4 and NaOH are mixed. Using the key below,draw three dissociated acid particles in beaker one and three dissociated base particlesin beaker two.
198
6. Match each hydrogen ion in beaker one with one hydroxide ion in beaker two. Thendraw the number of water molecules that form in beaker three. Draw the remaining ions in breaker three.
Key: H+ =
SO4-2 =
Na+ =
OH-1 =
7. What additional ion(s) would be needed to neutralize the solution? How many more wouldyou need?
8.Draw a model below to justify that neutralization has occurred.
8. Write the balanced equation for the complete neutralization of H2SO4 and NaOH .
199
6. Match each hydrogen ion in beaker one with one hydroxide ion in beaker two. Thendraw the number of water molecules that form in beaker three. Draw the remaining ions in breaker three.
Key: H+ =
SO4-2 =
Na+ =
OH-1 =
7. What additional ion(s) would be needed to neutralize the solution? How many more wouldyou need?
8.Draw a model below to justify that neutralization has occurred.
8. Write the balanced equation for the complete neutralization of H2SO4 and NaOH .
200
Titration and Neutralization Calculations
Do Now: Using your knowledge of chemistry, fill in each blank to complete the statements.
1. The chemicals HCl, HBr, HNO3, H2SO4 are all categorized as acids because when dissolved in water they
produce a _________________ ion.
2. The chemicals NaOH, LiOH, and Ca(OH)2 are all categorized as bases because when dissolved in water
they produce a ______________ ion.
3. When acids react with bases the reaction is known as a _______________________ reaction because
the properties of the original acid and base are lost to form a ______________ and water.
4. Calculate the Molarity of an acid created by dissolving 73.0g of HCl in 200.0mL if solution.
5. Calculate the moles of NaOH dissolved in a 2.0M basic solution with a total volume of 500.0mL.
READ ME! Titrations are procedures used to determine the concentration or Molarity (M) of an
unknown acid or a base. The acid and base solutions are combined together in a ratio of 1 mole of Acid for
every 1 mole of Base, thus neutralizing the reaction. Observe the titration set up below. A student placed
20.00mL of an unknown acid and two drops of phenolphthalein indicator in the flask. Then the student added
the known base dropwise until the solution was neutralized (indicated by the color change of the
phenolphthalein). Answer questions 1 through 4.
Titration setup Read the volume of the base that was released from the burette
If the base concentration was known to be 2.00M, calculate the number of moles of basic ion are dissolved in the solution.
If the solution in the flask is now neutral, how many moles of acid ion are in the flask?
Calculate the molarity of the unknown acid.
201
Using the equation on Reference Table T, you can solve for either the molarity/concentration (M) or a volume added (V) using the titration formula:
MA= molarity of H+ VA = volume of acid
MB = molarity of OH– VB = volume of base
5. What unit of measurement is obtained when the Molarity of a solution is multiplied by the volume (inliters) of the solution?
6. Explain why the Titration formula is mathematically accurate for finding the concentration of solutionwhen the solution is neutralized.
Notice that the Molarity formula requires the volume to be calculated in Liters. However, the titration formula has no such requirement, as long as the volume unit of measurement is the same for both VA and VB. Since the titration formula has two volumes, the volume unit of measurement will cancel out. Therefore, converting the units will also cancel out. Perform the calculations in question 7 to prove this rule.
7. A 100.00mL solution of 2.00M HCl is neutralized by 50.00mL of NaOH. Calculate the Molarity of theunknown base.
a. Use the titration formula and calculate the Molarity with the volume values in mL.
b. Use the titration formula and calculate the Molarity with the volume values in mL.
c. Compare your answers to questions 7a and 7b.
8. A 25.0-milliliter sample of HNO3 (aq) is neutralized by 32.1 milliliters of 0.150 M KOH (aq). Calculate theconcentration of the acid.
9. The titration formula can also be used to determine the volume needed to neutralize two knownsolutions. Calculate the volume of 0.200 M NaOH needed to neutralize 100. mL of 0.100 M HCl.
buret containingBASE
MA VA = MB
VB
202
PracticePacket:AcidsandBases
http://drshanzerchemistry.weebly.com
Foreachquestion,thetwopHvaluesarebeingcompared.In terms of hydronium, howmanytimesstrongerorweakeristhepHofthesolution?
1. pH5àpH3 ___________________________________
2. pH8àpH4 ___________________________________
3. pH10àpH7 ___________________________________
4. pH14àpH7 ___________________________________
5. pH3àpH6 ___________________________________
Video13.5:Neutralization&Titrations
Completeandbalanceeachoftheacidbaseneutralizationreactionsbelow.
1. _____H2SO4+_____Mg(OH)2à
2. _____HNO3+_____Al(OH)3à
3. _____H3PO4+_____Ca(OH)2à
4. _____HI+_____KOHà
5. _____HBr+_____Ba(OH)2à
6. _____HCl+_____KOHà
7. _____H3PO4+_____LiOHà
8. _____HF+_____Ca(OH)2à
9. InatitrationofHClO4withNaOH,100.0mLofthebasewasrequiredtoneutralize20.0mLof5.0MHClO4.WhatisthemolarityoftheNaOH?(Besuretowritetheneutralizationreaction.)
10. InatitrationofHNO3withNaOH,60.0mLof0.020MNaOHwasneededtoneutralize15.0mLofHNO3.Whatisthemolarityoftheacid?(Writetheneutralizationreaction.)
203
11. In a titration, 20.0 milliliters of 0.15 M HCl(aq) is exactly neutralized by 18.0 milliliters of KOH(aq).(a) Complete the equation below for the neutralization reaction by writing the formula of each
product. KOH(aq) + HCl(aq) → _________ + _________
(b) Compare the number of moles of H+(aq) ions to the number of moles of OH– (aq) ions in the titration mixture when the HCl(aq) is exactly neutralized by the KOH(aq).
(c) Determine the concentration of the KOH(aq).
(d) What is the new pH of the solution?
12. In a laboratory activity, 0.500 mole of NaOH(s) is completely dissolved in distilled water to form 400.milliliters of NaOH(aq). This solution is then used to titrate a solution of HNO3(aq).(a) Identify the negative ion produced when the NaOH(s) is dissolved in distilled water.
(b) Calculate the molarity of the NaOH(aq). Your response must include both a correct numerical setup and the calculated result.
(c) If 26.4 milliliters of the NaOH solution is needed to exactly neutralize 44.0 milliliters of the HNO3 solution, what is the molarity of the HNO3 solution?
(d) Complete the equation below representing this titration reaction by writing the formulas of the products. NaOH(aq) + HNO3(aq) � _____________________ + ______________________
204
PracticePacket:AcidsandBases
http://drshanzerchemistry.weebly.com
13. If10.0mLof0.300MKOHarerequiredtoneutralize30.0mLofstomachacid(HCl),whatisthemolarityofthestomachacid?(Writetheneutralizationreaction.)
14. Ifittakes50mLof0.5MKOHsolutiontocompletelyneutralize125mLofsulfuricacidsolution(H2SO4),whatistheconcentrationoftheH2SO4solution?
TitrationPractice:AtitrationwassetupandusedtodeterminetheunknownmolarconcentrationofasolutionofNaOH.A1.2MHClsolutionwasusedasthetitrationstandard.Thefollowingdatawerecollected.
Trial1 Trial2 Trial3 Trial4Volumeof1.2M
HCl10.0mL 10.0mL 10.0mL 10.0mL
IntialburetreadingofNaOH
0.0mL 12.2mL 23.2mL 35.2mL
FinalburetreadingofNaOH
12.2mL 23.2mL 35.2mL 47.7mL
VolumeofNaOHused(mL)MolarityofNaOH(M)1. CalculatethevolumeofNaOHusedtoneutralizetheacidforeachtrial.Recordinthe
datatableabove.Showonesamplecalculationbelow.
2. UsingtheMAVA=MBVBformula,calculatethemolarityofthebaseforeachtrial.Recordinthedatatableabove.Showonesamplecalculationbelow.
3. CalculatetheaveragemolarityoftheNaOHusingyourresultsfromquestion2.Youranswermustincludethecorrectnumberofsignificantfiguresandthecorrectunitsofmeasure.
205
������ �������� ����������
�� 5��4+���1�&6��� 5��4+���1�8��� *1*(97.(&1�(43):(9.;.9>�4+���1�&6��� *1*(97.(&1�(43):(9.;.9>�4+���1�8�
� $-.(-�1&'47&947>�9*89�7*8:19�(&3�'*�:8*)�94)*9*72.3*�.+���1�8��.8�&3�*1*(9741>9*�
�� ��1� �� ����� ��1 �� �!
� $-.(-�8:'89&3(*�.8�&3�*1*(9741>9*�
�� ���!���&3)������!��� ���!���&3)���1�� &!��&3)���1�� &!��&3)������!�
�� $-.(-�9<4�(4254:3)8�&7*�*1*(9741>9*8�
�� .9�-&8�&�-.,-�2*19.3,�54.39�� .9�(439&.38�(4;&1*39�'43)8�� .98�&6:*4:8�841:9.43�(43):(98�&3�*1*(97.(
(:77*39�� .98�&6:*4:8�841:9.43�-&8�&�5��;&1:*�4+��
�� ��8:'89&3(*�.8�(1&88.+.*)�&8�&3�*1*(9741>9*�'*(&:8*
�� ���������!� �� ��������!��� ������ &!� �� ����� &!�
� $-.(-�4+�9-*�+4114<.3,�&6:*4:8�841:9.438�.8�9-*'*89�(43):(947�4+�*1*(97.(.9>�
�� ��89:)*39�<&8�,.;*3�+4:7�:3034<3�841:9.438���&(-841:9.43�<&8�(-*(0*)�+47�(43):(9.;.9>�&3)�9*89*)<.9-�5-*3415-9-&1*.3��#-*�7*8:198�&7*�8-4<3�.3�9-*)&9&�9&'1*�'*14<�
�� � �� � �� � �� �
�&8*)�43�9-*�)&9&�9&'1*��<-.(-�:3034<3�841:9.43(4:1)�'*������ &!��
�� �&! �� ��� �� �! �� ��
�� $-.(-�(4254:3)�.8�&3��77-*3.:8�&(.)�
�� &�8&19 �� &�->)74(&7'43�� &3��77-*3.:8�&(.) �� &3��77-*3.:8�'&8*
�� $-*3�43*�(4254:3)�).8841;*8�.3�<&9*7��9-*�431>548.9.;*�.43�574):(*)�.3�9-*�841:9.43�.8��!���&6���#-.8�(4254:3)�.8�(1&88.+.*)�&8
�� ��7 �� �!�� ��7 �� �!�
�� $-.(-�8:'89&3(*�.8�&1<&>8�&�574):(9�<-*3�&3�77-*3.:8�&(.)�.3�&3�&6:*4:8�841:9.43�7*&(98�<.9-&3��77-*3.:8�'&8*�.3�&3�&6:*4:8�841:9.43�
�� �.;*3�9-*�*6:&9.43�
��������1�,�����!� ��������&6�����1@�&6�
�� ->)74=.)* �� ->)743.:2�� ->54(-147.9* �� 5*7(-147&9*
$-.(-�.43�.8�7*57*8*39*)�'>���
�� &1)*->)* �� &1(4-41�� �77-*3.:8�&(.) �� �77-*3.:8�'&8*
� �3�&6:*4:8�841:9.43�4+�1.9-.:2�->)74=.)*(439&.38�->)74=.)*�.438�&8�9-*�431>�3*,&9.;*�.43.3�9-*�841:9.43���.9-.:2�->)74=.)*�.8�(1&88.+.*)�&8&3
�� (-&3,*8�1.92:8�+742�7*)�94�'1:*�� (-&3,*8�5-*3415-9-&1*.3�+742�(41471*88�94
5.30�� 574):(*8�->)743.:2�.438�&8�9-*�431>
548.9.;*�.438�.3�&3�&6:*4:8�841:9.43�� 574):(*8�->)74=.)*�.438�&8�9-*�431>�3*,&9.;*
.438�.3�&3�&6:*4:8�841:9.43
� �((47).3,�94�9-*��77-*3.:8�9-*47>��&3�&(.)�.8�&8:'89&3(*�9-&9
�� ����!!� �� ���!��� ��1 �� �!�
�� $-.(-�(4254:3)�7*1*&8*8�->)74=.)*�.438�.3�&3&6:*4:8�841:9.43�
�� ��!� �� ����� !�@ �� ��!�@
�� $-.(-�+472:1&�7*57*8*398�&�->)743.:2�.43�
�� �� �� ��� �� ��� �� �
� ��841:9.43�<.9-�&�5��4+����-&8�&�->)743.:2�.43(43(*397&9.43�9*3�9.2*8�,7*&9*7�9-&3�&�841:9.43<.9-�&�5��4+
206
�� 5��94�5�� �� 5���94�5����� 5���94�5�� �� 5����94�5��
�� $-.(-�(-&3,*�.3�5��7*57*8*398�&�-:3)7*)+41).3(7*&8*�.3�9-*�(43(*397&9.43�4+�->)743.:2�.438.3�&�841:9.43�
�� �� �� � �� �
�� $-&9�.8�9-*�5��4+�&�841:9.43�9-&9�-&8�&�->)743.:2.43�(43(*397&9.43����9.2*8�,7*&9*7�9-&3�&841:9.43�<.9-�&�5��4+���
�� �� �� � ��
�� $-.(-�5��.3).(&9*8�&�'&8.(�841:9.43�
�� �� � �� � ��
�� $-.(-�4+�9-*8*�5��3:2'*78�.3).(&9*8�9-*�-.,-*891*;*1�4+�&(.).9>�
�� &�841:9.43�.3�<-.(-�'7429->241�'1:*�.8�'1:*�� &�841:9.43�.3�<-.(-�'742(7*841�,7**3�.8�'1:*�� &�841:9.43�.3�<-.(-�5-*3415-9-&1*.3�.8�5.30�� &�841:9.43�.3�<-.(-�2*9->1�47&3,*�.8�7*)
�� �&8*)�43�9-*�7*8:198�4+�9*89.3,�(41471*88�841:9.438<.9-�.3).(&9478��<-.(-�841:9.43�.8�2489�&(.).(�
�� '7429->7341�'1:* �� '742(7*841�,7**3�� 1.92:8 �� 9->241�'1:*
� $-.(-�.3).(&947�<4:1)�'*89�).89.3,:.8-�'*9<**3�&841:9.43�<.9-�&�5��4+��� �&3)�&�841:9.43�<.9-�&�5�4+� �
�� '742(7*841�,7**3 �� '7429->241�'1:*�� 1.92:8 �� 2*9->1�47&3,*
� $-.(-�.3).(&947��<-*3�&))*)�94�&�841:9.43�(-&3,*8�(4147�+742�>*114<�94�'1:*�&8�9-*�5��4+9-*�841:9.43�.8�(-&3,*)�+742� � �94�����
�� #-*�9&'1*�'*14<�8-4<8�9-*�(4147�4+�9-*�.3).(&94782*9->1�47&3,*�&3)�1.92:8�.3�9<4�8&251*8�4+�9-*8&2*�841:9.43�
�� �� �� � �� �
$-.(-�5��;&1:*�.8�(438.89*39�<.9-�9-*�.3).(&9477*8:198�
�� 2*9->1�47&3,* �� '7429->241�'1:*�� '742(7*841�,7**3 �� 9->241�'1:*
�� $-.(-�.3).(&947�.8�>*114<�.3�&�841:9.43�<.9-�&�5�4+�����
�� ���)4347 �� ���&((*5947�� ��)4347 �� ��&((*5947
� !3*�&(.)�'&8*�9-*47>�)*+.3*8�&�'&8*�&8�&3
�� ���)4347 �� ���&((*5947�� !���)4347 �� !���&((*5947
�� !3*�&19*73&9*�&(.)�'&8*�9-*47>�89&9*8�9-&9�&3�&(.).8�&3
�� �(.)8�&3)�'&8*8�&7*�'49-����&((*59478��� �(.)8�&3)�'&8*8�&7*�'49-����)43478��� �(.)8�&7*����&((*59478��&3)�'&8*8�&7*���
�)43478��� �(.)8�&7*����)43478��&3)�'&8*8�&7*���
�&((*59478�
�� $-.(-�89&9*2*39�)*8(7.'*8�&3�&19*73&9*�9-*47>�4+&(.)8�&3)�'&8*8�
�� ���&3)� ����� ���&3)�"!�@�� �"!�@�&3)�"!�@�� �"!�@�&3)� ���
�� �.;*3�9-*�7*&(9.43�&9�*6:.1.'7.:2�
�"!�@��� �����"!�@��� ���
$-&9�&7*�9-*�9<4�85*(.*8�9-&9�&7*�&(.)8�
�� &((*598�&3���� �� )43&9*8�&3������ &((*598�&3���� �� )43&9*8�&3���
�� �((47).3,�94�43*�&(.)�'&8*�9-*47>��<&9*7�&(98�&8&3�&(.)�<-*3�&3��!�241*(:1*
�� �&�1 �� �&��� ��1! �� ��1!
��� $-.(-�(4254:3)�.8�574):(*)�<-*3���1�&6��.83*:97&1.?*)�'>��&�!���&6��
�� '&8*���&(.)���8&19���<&9*7�� '&8*���8&19���<&9*7���&(.)�� 8&19���&(.)���'&8*���<&9*7�� 8&19���<&9*7���&(.)���'&8*
�� $-.(-�<47)�*6:&9.43�7*57*8*398�&�3*:97&1.?&9.437*&(9.43�
207
�� ��*�8�����!�,�����*!��8��� ��,����!�,�����!� ��� � !��&6�����!��&6����� !��&6�����!�
��� �, !��&6������1�&6����� !��&6���
�,�1�8�
�� $-.(-�*6:&9.43�7*57*8*398�&�3*:97&1.?&9.437*&(9.43�
�� *;&547&9.43 �� ).89.11&9.43�� +.197&9.43 �� 9.97&9.43
��� �:7.3,�<-.(-�574(*88�(&3�����2.11.1.9*78�4+�&��� �����1�&6��841:9.43�'*�:8*)�94�)*9*72.3*�9-*:3034<3�(43(*397&9.43�4+�&�,.;*3�;41:2*�4+ &!��&6��841:9.43�
�� ���� �� ������� ��� �� ����
��� ��89:)*39�(4251*9*8�&�9.97&9.43�'>�&)).3,���2.11.1.9*78�4+� &!��&6��4+�:3034<3(43(*397&9.43�94�����2.11.1.9*78�4+��� �����1�&6���$-&9�.8�9-*�241&7�(43(*397&9.434+�9-*� &!��&6��
�� ����� �� �� ����� ������ �� �������
� � �� ���2.11.1.9*7�8&251*�4+�� !��&6��.83*:97&1.?*)�'>����2.11.1.9*78�4+��� ����!��&6���$-&9�.8�9-*�241&7.9>�4+�9-*�� !��&6��
�� � �2� �� �� �2��� ����2� �� ����2�
��� $-.(-�;41:2*�4+������� &!��&6��*=&(91>3*:97&1.?*8� ���2.11.1.9*78�4+�������� !��&6��
�� �� �2� �� �2��� ����2� �� �����2�
��� $-&9�;41:2*�4+�������� !��&6��.8�3**)*)�94(4251*9*1>�3*:97&1.?*� ����2.11.1.9*78�4+������� &!��&6��
208
�&8*�>4:7�&38<*78�94�6:*89.438����&3)����43�9-*�.3+472&9.43�'*14<�
����������3�43*�97.&1�4+�&3�.3;*89.,&9.43�� ����2.11.1.9*78�4+�����&6��4+�&3�:3034<3�(43(*397&9.43�.89.97&9*)�<.9-������� &!��&6����:7.3,�9-*�9.97&9.43��9-*�949&1�;41:2*�4+� &!��&6��&))*)�&3)9-*�(477*8543).3,�5��;&1:*�4+�9-*�7*&(9.43�2.=9:7*�&7*�2*&8:7*)�&3)�7*(47)*)�.3�9-*�9&'1*'*14<�
��� $7.9*�&�'&1&3(*)�*6:&9.43�9-&9�7*57*8*398�9-.8�3*:97&1.?&9.43�7*&(9.43�
��� �3�&349-*7�97.&1�������2.11.1.9*78�4+�����&6��.8�(4251*9*1>�3*:97&1.?*)�'>�����2.11.1.9*78�4+�9-.8
������ &!��&6����&1(:1&9*�9-*�241&7.9>�4+�9-*�9.97&9*)�&(.)�.3�9-.8�97.&1��%4:7�7*85438*�2:89.3(1:)*�����&�3:2*7.(&1�8*9:5�&3)�9-*�(&1(:1&9*)�7*8:19�
��� �&8*�>4:7�&38<*7�94�9-*�+4114<.3,�6:*89.43�43�9-*�.3+472&9.43�'*14<�
�������3�1.6:.)�<&9*7��&3�*6:.1.'7.:2�*=.898�'*9<**3��!� ��241*(:1*8�����&6��.438��&3)�����!���&6����.438����5*7843�*=5*7.*3(.3,�&(.)�.3).,*89.43�&+9*7�)7.30.3,�942&94�/:.(*�(&3.3,*89�2.10�4+�2&,3*8.&�94�7*):(*�9-*�&(.).9>�4+�9-*�8942&(-�(439*398��#42&94�/:.(*�-&8�&5��;&1:*�4+�����.10�4+�2&,3*8.&��&�2.=9:7*�4+�2&,3*8.:2�->)74=.)*�&3)�<&9*7��-&8�&�5�;&1:*�4+���
�425&7*�9-*�->)74,*3�.43�(43(*397&9.43�.3�942&94�/:.(*�94�9-*�->)74,*3�.43�(43(*397&9.43�.32.10�4+�2&,3*8.&�
209
�&8*�>4:7�&38<*78�94�6:*89.438����9-74:,-����43�9-*�.3+472&9.43�'*14<�
�������89:)*39�:8*)�'1:*�1.92:8�5&5*7�&3)�5-*3415-9-&1*.3�5&5*7�&8�.3).(&9478�94�9*89�9-*�5��4+).89.11*)�<&9*7�&3)�+.;*�&6:*4:8�-4:8*-41)�841:9.438��#-*3�9-*�89:)*39�:8*)�&�5��2*9*7�942*&8:7*�9-*�5��4+�9-*�).89.11*)�<&9*7�&3)�*&(-�841:9.43��#-*�7*8:198�4+�9-*�89:)*39A8�<470�&7*7*(47)*)�.3�9-*�9&'1*�'*14<�
�� �=51&.3��:8.3,�9-*�7*+*7*3(*�9&'1*���.3�9*728�4+�9-*�5��7&3,*�+47�(4147�(-&3,*�<->�1.92:8�.8�����&557457.&9*�94�).++*7*39.&9*�9-*�&(.).9>�1*;*18�4+�942&94�/:.(*�&3)�;.3*,&7�
�� �)*39.+>�9-*�1.6:.)�9*89*)�9-&9�-&8�9-*���� ��->)743.:2�.43�(43(*397&9.43�
��� �&8*)�43�9-*�2*&8:7*)�5��;&1:*8��.)*39.+>�9-*�1.6:.)�9*89*)�9-&9�.8���9.2*8�247*�&(.).(�9-&3;.3*,&7�
210