CHEM 14CL Study Guide - Summer 2018, Comprehensive Midterm Notes - Acid, Titration, Sodium Hydroxide
12 Oct 2018
School
Department
Course
Professor
CHEM 14CL
MIDTERM EXAM
STUDY GUIDE
Fall 2018
.
(amino acids )
-POIYPWHC acids can donate more than one Proton (have more than one Ka )
-Kai >Kaz >Ka }Usually
Cka = acid dissociation constant .larger Ka= stronger acid )
AMINO AADS
-are all POIYPWTIC acids
-general form of Amino acids
F-
side drain determining identity +functionality of amino acid
amine -@
-CH -
of -Carboxylic acid
-in solution 1depending on PH .may rearrange to azwitterion
(aka iso electric Point :overall has 0onarge but contains
+1
-charges )
R
HZN #Hcoo
-
-different forms of amino acids
!1!HsNtp¥[ OH fully Ptotonated form
' ' Hzlt "
or that
2exists @low PH
0
!2!HzN+ #isoeleotnc form →most common !' 'HL "
or HAH '
^0pH between Pta ,and Pkk and at Iso electric point PH
HR
!3!HZN #o
.
fully depwtonated form
"
L
."
or A-
×exists @high pH
HR
!1!and !3!Can't coexist bc they exist in different PH ranges .usually we will
find either !1!/@or !2!/!3!
-amino acid equilibrium (using Levine )
Halt HL carboxylic acid has a
RRmuch higher Ka
Hsnt 'atw±
A
Hstfdttwo'
+Ht Kai -4.69×10-3 Jthan amine ,so Kai
refers to oootf
HL
[
RR
HzNtdHuV HHZNIHWO
-
+Hit Kaz =1.79×10-10
find more resources at oneclass.com
find more resources at oneclass.com
Titration of 10.00mL of 0.050 MLemoine with 0.05014 NAOH
-there are 6Points of interest in adiprotic acid titration
!1!Initial PH of an amino acid :what's the PH of a0.050 MHztt solution ?
Kai >>Kdz ,so only Kai is important (effect of Kdz can be neglected )
Halt =HL +Ht
I0.050 M- -
c-xtx +×x=how much it dissociates
E(0.05014 -X)×X
Kal =X2
.
,
× =EHT ]=1.3×10-2 (Kai >10-5 ,so you can't use
0.050 -Xapproximation ,Must solve auddmtiz )
PH =1.88
-before we do calculations for addition of NAOH ...
-there are 2rxns blw levine and NAOH
Halt tOH
' →HLTHZO (reaction #I,related to fa ,
)}
write this it asked to
prove it's an intermediate
HL tOH
-
→L
. tHZO (reaction #2,related to Kaz )ion (put it in water )
→we expect 2equivalence Points and 2equivalence Pt .volumes .
Macid ×Vaud =Mbase ×Vbase
moles Ht =moles OH
-
so ,(0.050 M)(10.00mL )=CO .050 M)×Vbase
V. base =Vee =10.00mL ←1st ear .Pt .volume
!2!PH After Addition of 5.00mL of 0.050 MNAOH
Halt it OH -
=HL tHAO
I0.01000L ×0.050 M0.00500L ×0.050N -
-
C-2.5×10-4 -2.5×10-4 +2.5×10-4 -
E2.5×104 02.5×10-4 -
rllimiting reagent
conj .acid /[Halt ]-25×w4mo= =
0.017 M
base pair {001500 LE
total volume
[HLJ =0.017 M
this Is abuffet solution !
half equivalence Point :Point in titration where reaction 1is half completed
-Kai >10-5 ,so We can't use Henderson -Hassel back .Use Kai eanilibnum instead
*when doing hydrolysis equilibrium (Wl strong base Or acid ),look at Kaandkb .
If
both Are smaller than 10-5 ,we can use HH .if Kd or Kb >to -5
,
we do HYDWHSIS
eanilibnwm Cand quadratic )with whichever is bigger .it both are bigger
than to -5
,we can use either .
find more resources at oneclass.com
find more resources at oneclass.com
Document Summary
Ch -of determining identity functionality of amino acid. Ph . may rearrange to a zwitterion ( aka iso electric. Hzn # hcoo different forms of amino acids onarge. " " hzlt or that exists @ low. Points of interest in a diprotic acid titration. Must solve auddmtiz before we do calculations for addition of. # 2 related related to to fa , Kaz ) write prove ion this it"s it an asked to intermediate put it in water we expect. Kai eanilibnum instead when doing hydrolysis equilibrium ( wl strong base. Hydwhsis with whichever is bigger it both are bigger than. Hl is an intermediate ion a chemical species that can behave both as an acid or a base an amino acid. Kea= single ( kw - 1014) intermediate species at is. Tsoelectric point omtsoelectnc which is species also an intermediation intermediate ion ex of intermediate. +4 , where xty is not necessarily bioelectric.
