A solution prepared by mixing 22.4 mL of 0.380 M NaCl and 22.4 mL of 0.380 M KI
ID: 949469 • Letter: A
Question
A solution prepared by mixing 22.4 mL of 0.380 M NaCl and 22.4 mL of 0.380 M KI was titrated with 0.190 M AgNO3 in a cell containing a silver indicator electrode and a saturated calomel reference electrode.
(a) What is [Ag ] when 21.2 mL of 0.190 M AgNO3 have been added? Express your answer as x, where [Ag ] is a quotient having the form Ksp,AgI/x. (b) What is [Ag ] when 67.0 mL of 0.190 M AgNO3 have been added? Express your answer as y, where [Ag ] is a quotient having the form Ksp,AgCl/y. (c) Which of the following expressions shows how cell voltage depends on [Ag ]? (d) The difference between voltages measured when 67.0 mL and 21.2 mL of AgNO3 have been added is 0.3899 V. Using this voltage difference and your answers to questions (a), (b), and (c), calculate the numerical value of the quotient Ksp,AgCl/Ksp,AgI.
Explanation / Answer
Moles of Cl- in the solution=22.4 ml*10^-3 L/ml*0.380mol/L=0.008512 moles
Moles of I- in the solution=22.4 ml*10^-3 L/ml*0.380mol/L=0.008512 moles
a) moles of AgNO3 added=21.2 mL *10^-3L/ml* 0.190 mol/L=0.004028 moles
titration reactions
Cl-+Ag+AgCl(s)………(1)
I-+Ag+AgI(s)…….(2)
Ksp(AgI)=1.5*10^-16
Ksp(AgCl)=1.8*10^-10
Ksp(AgI)<<ksp(AgCl),so AgCl is more soluble,will exist in ion form.Rxn (2) will take place first.AgI will be precipitated
ICE table
[I-]
[Ag+]
[AgI]
initial
0.008512 moles-0.004028=0.004484
0.004028 moles-0.004028=0
0
remaining
0.004484
0
0.004028
equilibrium
0.004484+x
+x
0.004028-x
Ksp(AgI)=[Ag+][I-]=( 0.004484+x)( x)
Ksp(AgI)/ [I-]=[Ag+]=ksp(AgI)/ ( 0.004484+x)
E==Eo-RT/F ln[I-]
rxn between Ag/AgI
Ag++I-=Ag(s)
AgI(s)=Ag+ +I-
E=Eo-RT/F ln Ksp(AgI)/ [Ag+]
E=Eo-RT/F ln Ksp(AgI)/ x…………(3)
b)
[Ag+]=0.01273 moles
[I-]=0.008512 moles
After precipitation of AgI,remaining Ag+=0.01273 moles-0.008512moles=0.004218 moles
Rxn of Ag+ with Cl-,
ICE table
[Cl-]
[Ag+]
[AgCl]
initial
0.008512 moles
0.004218 moles-0.004218=0
0
remaining
0.008512-0.004218=0.004294
-
0.004218
equilibrium
0.004294+y
y
0.004218-y
Ksp(AgI)=[Ag+][I-]=( 0.008512-y)( 0.004028-x)
With excess Ag+ ,AgCl will also begin to precipitate next
[Ag+]=ksp(AgCl)/[Cl-]= ksp(AgCl)/(0.008512)
c)nerst equation for cell voltage of silver indicator electrode,
E=Eo-RT/F ln[Cl-] [for dilute cl- solution acl-=[Cl-]]
Also, [Ag+]=ksp(AgCl)/[Cl-]= ksp(AgCl)/(0.008512)
Gives,[Cl-]= ksp(AgCl)/[Ag+]= ksp(AgCl)/y
E=Eo-RT/F ln ksp(AgCl)/y……………..(4)
d) eqn(3)-eqn(4)/
0.3899V=-RT/F( ln Ksp(AgI)/ x/ ksp(AgCl)/y)
0.3899V=-RT/F ln (ksp(AgI)/ksp(AgCl) *y/x [y,x <<<]
0.3899V=-8.314J/kmol*298K/96485C/mol ln(ksp(AgI)/ksp(AgCl)
0.3899V=0.0256 ln(ksp(AgI)/ksp(AgCl)
ln(ksp(AgI)/ksp(AgCl) =-15.23
(ksp(AgI)/ksp(AgCl) =exp(-15.23)=2.42*10^-7
[I-]
[Ag+]
[AgI]
initial
0.008512 moles-0.004028=0.004484
0.004028 moles-0.004028=0
0
remaining
0.004484
0
0.004028
equilibrium
0.004484+x
+x
0.004028-x
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.