A solution prepared by mixing 21.6 mL of 0.640 M NaCl and 21.6 mL of 0.640 M KI
ID: 1027630 • Letter: A
Question
A solution prepared by mixing 21.6 mL of 0.640 M NaCl and 21.6 mL of 0.640 M KI was titrated with 0.320 M AgNO3 in a cell containing a silver indicator electrode and a saturated calomel reference electrode. (a) What is [Ag'] when 21.1 mL of 0.320 M AgNO3 have been added? Express your answer as x, where [Ag ] is a quotient Number having the form KspAg/x Number (b) What is [Ag ] when 64.4 mL of 0.320 M AgNO3 have been added? Express your answer as y, where [Ag] is a quotient having the form KspAgcly (c) Which of the following expressions shows hoW cell voltage depends on [Ag']? (1) E=10.799-0.059 16 log([Ag +])-0.241 11) E=0.241-10.799-0.059 16 log([AgExplanation / Answer
For the given titration reaction,
Ksp of AgI is lower than Ksp for AgCl. therefore, AgI would precipitate out of solution first.
(a) after 21.1 ml AgNO3 added
initial KI = 0.640 M x 21.6 ml = 13.824 mmol
added AgNO3 = 0.320 M x 21.1 ml = 6.752 mmol
[I-] remained = 7.072 mmol/42.7 ml = 0.166 M
x = 0.166 M
[Ag+] = 8 x 10^-17/0.166 = 4.82 x 10^-16 M
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(b) after 64.4 ml AgNO3 added
Volume AgNO3 needed to completely precipitate ut AgI = 13.824 mmol/0.320 M = 43.2 ml
AgNO3 remaining volume = 64.4 - 43.2 = 21.2 ml
initial NaCl = 0.640 M x 21.6 ml = 13.824 mmol
added AgNO3 = 0.320 M x 21.2 ml = 6.784 mmol
[Cl-] remained = 7.04 mmol/42.8 ml = 0.1645 M
y = 0.1645 M
[Ag+] = 1.8 x 10^-17/0.1645 = 1.09 x 10^-16 M
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(c) Correct expression for cell voltage
E = [0.799 - 0.05916 log(1/[Ag+])] - 0.241
(d) Voltage when 21.1 ml AgNO3 added
E = [0.799 + 0.05916 log(4.82 x 10^-16)] - 0.241
= -0.348 V
Voltage when 64.4 ml AgNO3 added
E = [0.799 + 0.05916 log(1.09 x 10^-16)] - 0.241
= -0.386 V
Difference in voltage = 0.038 V
Ksp(AgCl)/Ksp(AgI) = 4.82 x 10^-16/1.09 x 10-16 = 4.422
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