Exercise Precipitation titration is a volumetric method based upon the formation of slightly soluble Precipitate: AgNO3(aq) + NaCl(aq) AgCl(s) Construct a titration curve by plotting pAg vs. VAgNO3 added VAgNO3 (ml) [Ag¹] pAg* 0.0 5 10.0 15 Hints: 20 25 30 a 35 40 45 50 oo (a) Before equivalence point: d. b. 1.43 1.30 K sp [CI-] (b) At equivalence point (moles of AgNO3 = moles of NaCl): [Ag*]=√ksp (c) After the equivalence point: Ag (aq) 4.90 [CI]=[NaCl] = 7.00 e. 7.65 [Ag*] = (d) pAg = -log[Ag*] (e) Set the values of Ksp, [NaC1], VNaCI and [AgNO3] as reference cell. What is pAg at the equivalence point of titration? + Cr [AgNO3]*VA AgNO, 50+ V AgNO, (aq) [Ag] = Ksp [NaCl]*V Naci -[AgNO3]*VA AgNO V Naci+V AgNO = 1.6 10-10 titrant: AgNO3, 0.IM analyte: NaC1, 0.05M, 50 ml

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Chapter11: Solving Equilibrium Problems For Complex Systems
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Problem 11.16QAP
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Exercise Precipitation titration is a volumetric method based upon the formation of
slightly soluble Precipitate:
AgNO3(aq) + NaCl(aq)
AgCl(s)
Construct a titration curve by plotting pAg vs. VAgNO, added
VAgNO3 (ml) [Ag¹] pAg*
0.0
5
10.0
15
Hints:
20
25
30
a
35
40
45
50
oo
(a) Before equivalence point:
d.
b. 1.43
1.30
K sp
[CI]
(b) At equivalence point (moles of AgNO3 = moles of NaCl): [Ag*]=√ksp
(c) After the equivalence point:
Ag (aq)
4.90
[CI]=[NaCl] =
7.00
e. 7.65
[Ag*] =
(d) pAg = -log[Ag*]
(e) Set the values of Ksp, [NaC1], VNaCI and [AgNO3] as reference cell.
What is pAg at the equivalence point of
titration?
+ Cr
[AgNO3]*VA AgNO,
50+ V
AgNO,
(aq)
[Ag] =
ksp
[NaCl]*V Naci -[AgNO3]*VA AgNO₂
V Naci+V AgNO,
= 1.6 10-10
titrant: AgNO3, 0.IM
analyte: NaC1, 0.05M, 50 ml
Transcribed Image Text:Exercise Precipitation titration is a volumetric method based upon the formation of slightly soluble Precipitate: AgNO3(aq) + NaCl(aq) AgCl(s) Construct a titration curve by plotting pAg vs. VAgNO, added VAgNO3 (ml) [Ag¹] pAg* 0.0 5 10.0 15 Hints: 20 25 30 a 35 40 45 50 oo (a) Before equivalence point: d. b. 1.43 1.30 K sp [CI] (b) At equivalence point (moles of AgNO3 = moles of NaCl): [Ag*]=√ksp (c) After the equivalence point: Ag (aq) 4.90 [CI]=[NaCl] = 7.00 e. 7.65 [Ag*] = (d) pAg = -log[Ag*] (e) Set the values of Ksp, [NaC1], VNaCI and [AgNO3] as reference cell. What is pAg at the equivalence point of titration? + Cr [AgNO3]*VA AgNO, 50+ V AgNO, (aq) [Ag] = ksp [NaCl]*V Naci -[AgNO3]*VA AgNO₂ V Naci+V AgNO, = 1.6 10-10 titrant: AgNO3, 0.IM analyte: NaC1, 0.05M, 50 ml
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