how to find equivalence point
Using Conductivity to Find Equivalence Point
Valentin Uzunov
(author),
Gavin Koenig
(lab partner),
Luke Weinstein
(lab partner)
CHM 22 Lab 2!"
#$U %&' !*"++
-.$/0-C/ %1/0&UC/%1 /he ob3e4tive o5 this e6peri7ent 8as to 9eter7ine the 4on4entration o5 an a:ueous solution o5 bariu7 h;9ro6i9e ( .a(H)2 ) using titration< Con9u4tivit; o5 the solution 8as use9 as a 7eans o5 i9enti5; the e:uivalen4e point< /he titration in this e6peri7ent involve9 the pre4ipitation rea4tion bet8een +<+2 M suphuri4 a4i9 ( H
2
$
*
) an9 2= 7L solution o5 a:ueous bariu7 h;9ro6i9e ( .a(H)
2
)< /he a4i9 base rea4tion 5or7 a pre4ipitate o5 bariu7 sulphate ( .a$
*
) as sho8n b; the net ioni4 e:uation > rn6 " ? .a
2@
(a:) @ 2 H
A
(a:) @ $*
2A
(a:) @ 2 H
@
(a:) B .a$
*
(s) @ 2 H
2
() > r6n " ? - Vernier Con9u4tivit; #robe atta4he9 to a 4o7puter running Logger
Pro
, 8as i77erse9 in the .a(H)2 solution an9 le5t throughout the 4ourse o5 the titration< /he probe 8as use9 to 7easure keep tra4k o5 4on9u4tivit; o5 solution throughout the 4ourse o5 the titration /h e:uivalen4e point o5 the rea4tion is rea4he9 the lo8est point o5< -:ueous .a(H)2 is an ele4trol;ti4 solution 8ith .a2@ an9 HA ions 5ree in solution< /he 4on9u4tivit; o5 the solution is proportional to the :uantit; o5 ions present in solution< -s H2$* is a99e9 9uring the titration, .a$* an9 8ater are 5or7e9< -s both 4o7poun9s are non ioni4, the total :uantit; o5 5ree ions in solution is re9u4e9 an9 4on9u4tivit; o5 the solution 9rops< /he e:uivalen4e point is rea4he9 8hen the ratio o5 7oles bet8een .a2@ an9 $*2A is e:ual in solution an9 the 4on9u4tivit; is thus the least< -s 7ore H2$* is a99e9 past that point, 4on9u4tivit; beging to in4rease again, as there are no 7ore .a2@ or HA ions le5t to rea4t 8ith the a99e9 H2$* -n alternative 7etho9 5or 9eter7ining the e:uivalen4e point 8as also per5or7e9, involving a t8o linear regressions 5ro7 the 9ata 4olle4te9< /his 8as a44o7plishe9 using the Logger#ro so5t8are inbuilt 5un4tionallit;< Dro7 the volu7e o5 H2$* a99e9 at the e:uivalen4e point an9 the kno8n 7olarit; o5 H2$* use9, 8ith the stoi4hio7etri4 ralationships bet8een .a(H)2 an9 the H2$*, the 7oles o5 .a(H)2 8ere 9eter7ine9 an9 subse:uentl; the unkno8n 7olarit; o5 the .aH2 solution<
&-/- /able "A Con9u4tivit; 9ata 5ro7 titration o5 H2$* an9 .a(H)2 Volu7e o5 H2$* in .urret (7L) Con9u4tivit; ( E$F47 ) ( )
23.70 2580 24.70 2360 25.70 2103 26.80 1832 27.80 1600 28.80 1361 29.80 1148 30.82 925 31.79 727 32.81 520 33.81 325 34.91 113 35.40 41 35.63 56 35.75 67 36.72 339 37.70 591 38.71 820 39.70 1043 40.71 1252 41.7 1446 42.70 1635 43.85 1838 44.78 1988 45.64 2126
/able 2 I 0e5eren4e value an9 7easure7ents 0e5eren4e Measure7ent H2$* at start ( 7L )2<+ .a(H)2 titrate9 ( 7L )2= J:uivalen4e point Volu7e ( 7L ) Dro7 &ata=<*+ Dro7 Line regression=<" H2$* a99e9Dro7 &ata""<+ Dro7 Line regression""<* Cal4ulations H2$* a99e9 ( 7L ) volu7e o5 H2$* at e:uivalen4e point A H2$* 7L at start
,=<"K
−
2,<+
=
""<*K
mL H
2
SO
*
Moles o5 H2$* at e:uivalen4e point Moles o5 .a(H)2 at e:uivalen4e point Molarit; o5 .a(H)2 solution 0J$UL/$ Moles o5 .a(H)2 solution (5ro7 9ata) I <* 6 "+A M .a(H)2 Moles o5 .a(H)2 solution (5ro7 regression anal;sis) I <2 6 "+
A
M .a(H)2 &%$CU$$%1 - titration 8as per5or7e9 in using +<+2 M H2$* to 9eter7ine the 4on4etration o5 .a(H)2 solution< - 7easure o5 the 4on9u4tivit; o5 the rea4tion 7i6ture 8as use9 as an in9i4ator to signal the e:uivalen4e point o5 the titration< +<+""*
mLH
2
SO
*
×
+<+2
molH
2
SO
*
"
Lsolution
=
2<+
×
"+
−
*
molH
2
SO
*
2<,+
×
"+
−
*
mol H
2
SO
*
×
2
mol Ba
(
OH
)
2
2
mol H
2
SO
*
=
2<,+
×
"+
−
*
mol Ba
(
OH
)
2
2<+
×
"+
−
*
molH
2
SO
*
+<+2=
L
=
<2
×
"+
−
MBa
(
OH
)
2
Figure 1:
The graph shows how the conductivity of of the Ba(OH)
solution changes as H
SO
!
is added during the titration" The e#uivalence point on the lower point of conductivity appro$imated at %&"!' mL
20 25 30 35 40 45 50 0 400 800 1200 1600 2000 2400 2800
Conductivity of solution during titration
Volume of 2!"4 # m$ %
C o n d u c t i v i t y # & ! ' c m %
how to find equivalence point
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