A solution prepared by mixing 32.3 mL of 0.660 M NaCI and 32.3 mL of 0.660 M KI

Question

A solution prepared by mixing 32.3 mL of 0.660 M NaCI and 32.3 mL of 0.660 M KI was titrated with 0.330 M AgNO3 in a cell containing a silver indicator electrode and a saturated calomel reference electrode. (a) What is [Ag'] when 33.2 mL of 0.330 M AgNO3 have been added? Express your answer as x, where Agis a quotient having the form KspAglx. Number Number (b) What is [Ag'] when 96.0 mL of 0.330 M AgNO3 have been added? Express your answer as y where [Ag is a quotient having the form KspAgcly 5.56x 10 (c) Which of the following expressions cell voltage de Ag'1? shows how s on 11) E-[0. 799-0.059161g[Ag+]]-0.241 111) E=0.241-[0.799-0.05916|og([Ag')] E0.241-0.799-0.05916log Ag

Explanation / Answer

Titration

(a) Total volume of solution = 32.3 + 32.3 + 33.2 = 97.8 ml

moles of [I-] remained = (0.66 M x 32.3 ml - 0.66 M x 33.2 ml)/97.8 ml = 0.106 M

[Ag+] = x = 8.5 x 10^-17/0.106 M = 8.02 x 10^-16 M

(b) When 96 ml AgNO3 added

moles of AgNO3 = 0.33 m x 96 ml = 31.68 mmol

moles of AgI formed = 21.318 mmol

moles of AgCl formed = 10.362 mmol

[Cl-] remained = 10.956 mmol/160.6 ml = 0.068 M

[Ag+] = x = 1.8 x 10^-10/0.068 = 2.64 x 10^-9 M

c. The equation showing the relationship with [Ag+] concentration and cell voltage,

III.

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