Jump to.. Map db Sapling Learning A solution prepared by mixing 40.0 mL of 0.380
ID: 1039633 • Letter: J
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Jump to.. Map db Sapling Learning A solution prepared by mixing 40.0 mL of 0.380 M AgNO3 and 40.0 mL of 0.380 M TINOs was titrated with 0.760 M NaBr in a cell containing a silver indicator electrode and a reference electrode of constant potential 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIBr is Kp-3.6 x 106 and the solubility constant of AgBr is K-5.0 x 1?t (a) Which precipitate forms first? O AgBr O TIBr (b) Which of the following expressions shows how the cell potential, E, depends on [Ag'1? (1) E-10.799-0.059 16 log(IAg -0.175 (nj ?-0.175 . [.799.0.059 16 log([Ag')] III) E,10.799-0.0591 6 logl?-- -0.175 Ag Scrol down to view the rest of the question I E 0.175-0.799-0.05916 log (AgExplanation / Answer
(a) Ksp of AgBr is lower than Ksp of TlBr
Therefore, precipitate to form first is,
AgBr
(b) correct nernst equations,
IV
(c) Potential when,
First equivalence point
Volume of NaBr added = 0.380 M x 40 ml/0.760 M = 20 ml
[Ag+] = [Br-]
[Ag+] = sq.rt.(Ksp) = sq.rt.(5 x 10^-13) = 7.1 x 10^-7 M
E = 0.175 - [0.799 - 0.05916 log[1/Ag+]
= 0.175 - [0.799 - 0.05916 log(1/7.1 x 10^-7)]
= -0.260
At second equivalence point
[Br-] = sq.rt.(Ksp(TlBr))
= sq.rt.(3.6 x 10^-6)
= 1.9 x 10^-3 M
[Ag+] = Ksp/[Br-]
= 5 x 10^-13/1.9 x 10^-3
= 2.63 x 10^-10 M
E = 0.175 - [0.799 - 0.05916 log[1/Ag+]
= 0.175 - [0.799 - 0.05916 log(1/2.63 x 10^-10)]
= -0.057
(d) potential when,
1.0 ml NaBr added
[Ag+] remained = (0.380 M x 40 ml - 0.760 M x 1.0 ml)/41 ml = 0.352 M
E = 0.175 - [0.799 - 0.05916 log[1/Ag+]
= 0.175 - [0.799 - 0.05916 log(1/0.352)]
= -0.597
(e) potential when,
13.4 ml NaBr added
[Ag+] remained = (0.380 M x 40 ml - 0.760 M x 13.4 ml)/53.4 ml = 0.094 M
E = 0.175 - [0.799 - 0.05916 log[1/Ag+]
= 0.175 - [0.799 - 0.05916 log(1/0.094)]
= -0.563
(f) potential when,
19.0 ml NaBr added
[Ag+] remained = (0.380 M x 40 ml - 0.760 M x 19.0 ml)/59 ml = 0.013 M
E = 0.175 - [0.799 - 0.05916 log[1/Ag+]
= 0.175 - [0.799 - 0.05916 log(1/0.013)]
= -0.512
(h) potential when,
20.3 ml NaBr added
this is post Ist equivalence point
excess [Tl+] remained = (0.380 M x 40 ml - 0.760 M x 0.228 ml)/60.3 ml = 0.25 M
[Br-] = 3.6 x 10^-6/0.25 = 1.44 x 10^-5 M
[Ag+] = 5 x 10^-13/1.44 x 10^-5 = 3.5 x 10^-8 M
E = 0.175 - [0.799 - 0.05916 log[1/Ag+]
= 0.175 - [0.799 - 0.05916 log(1/3.5 x 10^-8)]
= -0.183
(i) potential when,
33.4 ml NaBr added
this is post Ist equivalence point
excess [Tl+] remained = (0.380 M x 40 ml - 0.760 M x 13.4 ml)/73.4 ml = 0.07 M
[Br-] = 3.6 x 10^-6/0.07 = 5.27 x 10^-5 M
[Ag+] = 5 x 10^-13/5.27 x 10^-5 = 9.5 x 10^-9 M
E = 0.175 - [0.799 - 0.05916 log[1/Ag+]
= 0.175 - [0.799 - 0.05916 log(1/9.5 x 10^-9)]
= -0.150
similarly others can be done
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